Wow, Samuel, I'm very impressed with what you wrote about this.  Yes, do question everything!

What about this simple problem involving fractions.  I can't write it out here as something over something, so I'll use / .  The problem is 4/8.  That's (2*2)/(2*2*2).  I see that I can cancel out two sets of 2s in the numerator and denominator to reduce it to lowest terms - 1/2.   When you think of it your way, though - there are no 2s left on top.   Zero 2s in the numerator should make that number equal to zero.  So you've just shown that 4/8 = 0. 

The fact that n⁰=1 holds is a necessary consequence of the law of exponents:

(n^m)(n^q)=n^m+q

We already know that:

(1/n)xn=(n/n)=1

And

(1/n)=n^-1

And

n¹=n

Then:

(n^-1)(n¹)=(n/n)=1=n⁰

 n⁰ cannot be 0. The law of exponents states that any expression n^m can be expressed in terms of the product of two expressions n^p and n^q. If n⁰=0, that would imply that the product of two non-zero real numbers would be zero. This is contrary to the definition of 0. 

And don't forget that certain derivatives  will be 0

y=5x

for example, dy/dx=5x⁰ =0 if x⁰=0 since dy/dx(xⁿ)=nx^(n-1) 

But yet dy/dx is the rate of change with respect to x. The function clearly changes with x.

If we look at the sequence of expressions, the 'problem' arises because you omitted the implied '1' in the 0 case.  Removing a factor in a chain of multiplications is NOT equivalent to replacing it with zero, it is equivalent to replacing it with unity. Replacement by zero would apply to removing an element in a chain of additions.

3 ^ 3 = 3 X 3 X 3

3 ^ 2= 1 X 3 X 3

3 ^ 1 = 1 X 1 X 3

3 ^ 0 = 1 X 1 X 1

3  ^ (-1) = 1 x 1 X 3^(-1)

3 ^ (-2) = 1 X 3^(-1) X 3^(-1)

Barbara: (2*2)/(2*2*2) = 1/2 because 2/2= 1 just like you said. If 2/2= anything to the zero power, then your argument would work.

As for (n^-1)(n^1)= (n/n)= 1= n^0...........that only makes sense this far: 1/n *n/1= n/n= 1.   after that, the only way that you get that it also equals n^0 is because of the adding of exponents "shortcut" Samuel was talking about: (n^-1)(n^1)= n^0 .

As for: n

cannot be 0. The law of exponents states that any expression

can be expressed in terms of the product of two expressions

and

If

that would imply that the product of two non-zero real numbers would be zero. This is contrary to the definition of 0.  - That is only true if n^0= a non-zero number. It doesn't prove that n^0= a non-zero number.

Bob Spence- I can accept implied 1- but why is it there?

Has to be there in somewhat the same way the 'i' has to there for negative exponents 3^-1 = 1/3^1_.

The fallacy of the OP is that multiplying one number by another changes its value by the value of the multiplier, so if multiplying A by 3² changes A by a factor of 3 twice, so multiplying A by 3¹ changes it by a factor of 3, and multiplying by 3⁰ doesn't change it at all.

When adding a number to the original, for the result to be no change, the number added must be zero.

When multiplying the original by another number, that number has to be 1 for there to be no change, not zero.

The OP made a false analogy from addition in his 'argument'. A null or 'empty' addition is adding zero, a null or empty multiplication is multiplying by 1.

There has to be something on each side of the equation - the blank left when you simply remove the last '3' has no defined value, it is not valid to simply assume it must be zero.

Dear Samuel,

You are not properly understanding the development of mathematics.

Please be patient with me as I try to explain to you some of the history of the development of the mathematical concept of exponentiation by an integer, beginning with the definition of exponentiation by a positive integer (which latter definition you are already familiar with, as evidenced by your explanation above).

To help you see what happened in this development, I will carry on a parallel discussion of the history of the mathematical concept of the complex number system (beginning with the construction of the real lnumber system).

I hope that you will, at the same time, enjoy this history. I think that it is very fascinating.

As you pointed out in your article, notation is designed to communicate things. With regard to positive integers, n, a^n is defined to be (not proven to be, but defined to be) a times itself n times. Or more precisely, a^n is defined (not proven) inductively by the rule: a^1 = a and a^{n + 1} = (a^n) * a.

With this as a rigorously (i.e. unambiguously) defined notion of a^n for positive integers (and, at this point, it must be noted, only for positive integers), mathematical induction is used to prove the exponent laws for positive integer exponents.

So far in the development of the mathematics of exponentiation.

As I go through this explanation, it will be helpful to follow a similar history of the development of a mathematical idea; namely, the mathematical idea of a complex number.

The parallel thread begins with the construction of the real number system (it may come as a surprise to you, but the real number system was constructed by mathematicians; the history of this construction is quite interesting. At the time of the ancient mathematicians for instance there was no real number system. Even the discovery of an irrational number, square root of 2,  was something that the early mathematicians puzzled over. To their mind such numbers did not exist.)

The parallel at this point, with my thread concerning the history of the development of the mathematical concept of exponentiation, is this; we have come to the point where the real numbers have been constructed and we have proven, by sound logical reasoning, that the real number system obeys all the familiar algebraic rules, such as distribution of multiplication over division.

Back to my story about the history of exponentiation. At this point in history, the integers have been constructed including the natural numbers (positive integers), 0 , and the negative numbers (it should be noted that at one time in the history of mathematics no one thought of there being a number 0; it's not an altogether unreasonable position to find oneself in, after all "nothing" does not exist); and exponentiation has only been defined for the positive integers.

Now, wishing to extend the notion of exponentiation to all integers and not just the positive integers in such a way that the familiar, extremely useful, laws of exponentiation still hold (as opposed to having special exceptions as happens with your chosen definition of exponentiation by 0), mathematicians asked themselves the following question. How should we define exponentiation by integers so that we will have two desirable conditions satisfied: (i) the new "extended" definition of exponentiation will agree, in the case of positive integers, with the old definition of exponentiation for positive integers and (ii) the old laws of exponentiation will still hold for the new extended definition of exponentiation.

Back to the story of the development of complex numbers. Mathematicians wanted a number system in which the equation x^2 + 1 = 0 has a solution. They decided to call a solution i. Now they asked themselves, if we add a solution i to the real number system what numbers must we have and how should we extend the old definition of the basic algebraic operations (+, -, *, /) to the new system so that the familiar, extremely useful, algebraic laws still hold for the new system.

Back to the story about exponentiation. In their search for the desired definition, mathematicians reasoned as follows. Suppose that we have a definition that has the two desired properties, (i) and (ii). What must that definition be?

Let's find out what that definition must be (after all, isn't it very enticing to think that we could define exponentiation for integers in such a way that we don't lose any of the familiar, extremely useful, laws of exponentiation? I don't know about you, but that seems to me to be a very enticing prospect. As we shall see, mathematicians found that for which they were looking).

Let m = 0 and suppose that a is a real number. (Remember, we are looking for a definition of exponentiation which will satisfy our desired goals, (i) and (ii).) Again, suppose that we have found such a definition. (What must it be?) Let m = 1.

Now since our hypothetical definition, by assumption, (of course, at this point, for all we know, our goal is not attainable. But, as we shall see, that is not the case. We shall actually achieve our goal.), our definition satisfies the exponential laws, a^{m + n} = (a^m)(a^n). That is to say, a^{0 + 1} = (a^0)(a^1). In other words, a^1 = (a^0)(a^1). By the old defintion of exponentiation for positive integers (and only for positive integers), a^1 = a. Hence, a = (a^0)(a).

Now suppose that a is not equal to 0. Then we can divide both sides of a = (a^0)a by a and obtain 1 = (a^0)1. But by the previously established laws of the algebra of the real number system, (x)1 = x for any real number x. Hence, if we want to define a^0 in such a way that, for any nonzero real number, a, a^0 is a real number, we must have it defined in such a way that (a^0)1 = a^0. Hence, 1 = (a^0)1 implies 1 = a^0.

Back to our story about complex numbers. If, reasoned the mathematician, we are going to obtain a new system of numbers in which x^2 + 1 = 0 has a solution, i, and all the familiar, extremely useful, laws of the algebra of the real number system still hold, then we must be able to add and multiply real numbers together with this new number, i. This gives us numbers of the form a + bi, where a and b are real numbers (we had to be able to multiply the real number b by the new number i, yielding another new number bi; and we had to be able to add the old real number a and the new number bi, yielding a new number a + bi).

Now, assuming that our desired system of complex numbers exists (satisfying all the desired, extremely useful, algebraic laws), what must the definition of addition and multiplication be. Here's the answer:

(1) (a + bi) + (c + di) = (a + c) + (bi + di)  (by the desired properties of commutativity and associativity of addition)

 = (a + c) + (b + d)i (by the desired property of distribution of multiplication over addition).

Note that the resulting number is of the same form as a + bi; namely, a real number (a + c) "plus" the "product" of the real number b + d with the new number i.

(2) (a + bi)(c + di) = (a + bi)c + (a + bi)(di) (by the desired property of dist. of mult. over div.

= ((ac) + (bc)i) + ((ad)i + (bd)(i^2)) (by the desired property of dist. of mult. over div.

Now by assumption i is a solution of x^2 + 1 = 0. So i^2 + 1 = 0. So, i^2 = -1.

Thus, continuing with the string of equalities in (2):

(3) (a + bi)(c + di) =  ((ac) + (bc)i) + ((ad)i + (bd)(i^2)) = ((ac) + (bc)i) + ((ad)(i) + (bd)(-1))

= ((ac) + (bd)(-1)) + ((bc) + (ad))i (by the desired properties of comm. and assoc. of add. and dist. of mult. over add.)

Note, again, the resulting "number" is of the same form as a + bi, a real number (ac) + (bd)(-1) "plus" the "product" of a real number (bc) + (ad) with the new "number" i.

Back to the story of exponentiation. I'll leave it to you as an exercise, but doing the same search for the "Holy Grail" of a definition of exponentiation for integers that satisfies the extremely desirable properties (i) and (ii), we find that the only possible definition for a^{-m}, where m is a positive integer and a is a nonzero real number is the following:

(4) a^{-m} = 1/(a^m).

Note, no other definition fits "our bill" (i.e. satisfies (i) and (ii).

Now, in both cases, in the history of exponentiation and the history of the development of the complex numbers, knowing what it was that mathematicians were searching for, what do you think they did? (This is not a trick question. The answer, at this point, should be pretty clear. Think about it for a second.)

Okay, time's up. This is what they did. They used the only possible definitions that satisfied their desired properties. That is to say, they defined (not proved) exponentiation by integers, in the history of the development of the mathematical concept of exponentiation by integers, and addition and multiplication of complex numbers, in the parallel history of the development of the mathematical concept of the complex number system as follows:

Definition. Exponentiation of a real number a by an integer m is defined as follows:

(i) if m is a positive integer, then a^m is defined inductively by the rule: a^1 = a and a^{k + 1} = (a^k)(a)

(ii) if m = 0 and a is a nonzero real number, then a^0 = 1

(iii) if m is a negative integer, so that m = -n for some positive integer, n, then a^m = 1/(a^n).

Definition: A complex number is an ordered pair, (a,b), of real numbers, written in the form a + bi with addition and multiplication of complex numbers defined as follows:

(i) (a + bi) + (c + di) = (a + c) + (b + d)i

and

(ii) (a + bi)(c + di) = (ac - bd) + (ad + bc)i

Then what do you think that the mathematicians did next. I'll give you a couple of seconds to think about it.

Okay, time's up. Heres the answer. They proved that the definitions which they gave satisfied all the extremely desirable properties that they wante their definition to satisfy.

In other words, they discovered the "Holy Grail" for which they were looking.

The very important point to be noted here is that mathematicians did not prove that a^0 = 1 for a nonzero real number a. Much to the contrary. They defined a^0 to be equal to 1. Why did they do this? I've already given the answer. It's the only definition of a^0 for nonzero real numbers a that still satisfies the extremely desirable exponentiation laws.

So what is the problem with your charge of circular reasoning against mathematicians. The problem is that it assumes that mathematicians pretended to prove that a^0 = 1 by circular reasoning. This is a misunderstanding of the history of mathematics.

It is not uncommon to find people who do not understand the history of the development of mathematics who will tell you that they are proving something like a^0 = 1. This is not what mathematicians did. They defined a^0 = 1 (for reasons which I have already made abundantly clear; at least that's for what I hope).

But there is something that these people do with their "proof". They demonstrate what mathematicians learned in their quests for these "Holy Grails". Namely, they prove that the only definition of a^0 for nonzero real numbers a which satisfies the extremely desirable exponentiation laws is this definition: a^0 = 1.

To put it another way, the problem with your charges against mathematicians is that you don't understand that mathematicians did not purport to prove something about an already existing definition of exponentiation by integers (there was no such definition at the time they searched for their "Holy Grail"; at that time exponentiation was only defined for positive integers). Rather, they "invented" (or, I believe, more precisely), discovered) the only possible definition that served their highly desirable purpose (to still have the desired laws satisfied); they took their discovery, made it their definition, and then proved that this definition "fit the bill" (i.e. in fact had all the properties they desired it to have).

I hope that this gives you a better understanding of the history of scientific research in regards to the development of mathematical ideas, ideas which today are taken for granted, but which, at one time, were not even known to exist.

Sincerely,

Dr. John D. McCarthy, Ph. D.

Professor of Mathematics

My response to your argument against Samuel is similar to my response to "Me", the person who posted the "deludegod" response.

(By the way, the correct notation is d(x^n)/dx = nx^{n -1}. What you wrote means, using established rules of notation, either "the product of the derivative of y with respect to x times x^n is equal to the product of n and x raised to the power n - 1" or "the derivative of y with respect to x at x^n is equal to ditto. But what you want to communicate is "the derivative of x^n with respect to x is equal to ditto. The appropriate notation for "the derivative of x^n with respect to x" is d(x^n)/dx". Excuse me for being somewhat picky, but this is my "business". I'm a professor of mathematics.)

Now, let me return to the main point which I wish to communicate.

Your argument against Samuel involves an appeal to a mathematical theorem:

(1) d(x^n)/dx = n(x^{n - 1}).

Now, using Samuel's definition of exponentiation for positive integers, which agrees with the mathematical definition of exponentiation for positive integers, equation (1) can be established from the product rule for derivatives and mathematical induction (a method of logical deduction which works for an inductive system such as the set of all positive integers). 

Now comes the "rub". In order to establish equation (1) for all integers n, we need to define exponentiation for all integers.  If we don't define exponentiation for all integers first, (before appealing to equation (1)), then we will never be able to establish equation (1) and, thereby, arm ourselves with ammunition to refute Samuel's argument.

So, in order to effectively refute Samuel's argument, along the lines which you have attempted to follow, you need to first define exponentiation for all integers (well, at least for 0, since that is about which Samuel was talking).

I think that you will find this somewhat challenging.

The problem with "Me"'s approach and your approach is that you are both appealing to a law which you are assuming has been established for all integers (well, at least for positive integers and 0). But how could such a law have been established if we have not first defined exponentiation for all integers (well, at least for positive integers and 0)?

Mathematicians have not attempted to follow the route which "Me" and you are following. Instead, as I explained in my response to Samuel and to "Me", they defined exponentiation for all integers in the only way that this concept can be defined so that two extremely desirable properties are obtained: (i) the new "extended" definition for all integers agrees with the old definition for positive integers (Samuel's definition) and (ii) the new extended definition still satisfies the exponentiation laws:

Definition:

(a) for positive integers, m, a^m is defined inductively by the rule: a^1 = a and a^{m + 1} = (a^m)(a)

(b) a^0 = 1 for each nonzero real number a

(c) a^{-m} = 1/(a^m) for each nonzero real number a and each positive integer m

Then they proved that this definition has all the extremely desirable properties, (i) and (ii).

If you want to avoid this, that's your choice. But I don't see how you can get around the fundamental issue of defining exponentiation by all integers (or, at least, by positive integers and 0). Unless you do this you cannot appeal to equation (1). Until you define exponentiation by all integers, you cannot claim that equation (1) has been established. How can one establish equation (1) for all integers if one has not defined exponentiation for all integers.

Sincerely,

Dr. John

My response to your argument against Samuel is similar to my response to "Me", the person who posted the "deludegod" response.

(By the way, the correct notation is d(x^n)/dx = nx^{n -1}. What you wrote means, using established rules of notation, either "the product of the derivative of y with respect to x times x^n is equal to the product of n and x raised to the power n - 1" or "the derivative of y with respect to x at x^n is equal to ditto. But what you want to communicate is "the derivative of x^n with respect to x is equal to ditto. The appropriate notation for "the derivative of x^n with respect to x" is d(x^n)/dx". Excuse me for being somewhat picky, but this is my "business". I'm a professor of mathematics.)

Now, let me return to the main point which I wish to communicate.

Your argument against Samuel involves an appeal to a mathematical theorem:

(1) d(x^n)/dx = n(x^{n - 1}).

Now, using Samuel's definition of exponentiation for positive integers, which agrees with the mathematical definition of exponentiation for positive integers, equation (1) can be established from the product rule for derivatives and mathematical induction (a method of logical deduction which works for an inductive system such as the set of all positive integers). 

Now comes the "rub". In order to establish equation (1) for all integers n, we need to define exponentiation for all integers.  If we don't define exponentiation for all integers first, (before appealing to equation (1)), then we will never be able to establish equation (1) and, thereby, arm ourselves with ammunition to refute Samuel's argument.

So, in order to effectively refute Samuel's argument, along the lines which you have attempted to follow, you need to first define exponentiation for all integers (well, at least for 0, since that is about which Samuel was talking).

I think that you will find this somewhat challenging.

The problem with "Me"'s approach and your approach is that you are both appealing to a law which you are assuming has been established for all integers (well, at least for positive integers and 0). But how could such a law have been established if we have not first defined exponentiation for all integers (well, at least for positive integers and 0)?

Mathematicians have not attempted to follow the route which "Me" and you are following. Instead, as I explained in my response to Samuel and to "Me", they defined exponentiation for all integers in the only way that this concept can be defined so that two extremely desirable properties are obtained: (i) the new "extended" definition for all integers agrees with the old definition for positive integers (Samuel's definition) and (ii) the new extended definition still satisfies the exponentiation laws:

Definition:

(a) for positive integers, m, a^m is defined inductively by the rule: a^1 = a and a^{m + 1} = (a^m)(a)

(b) a^0 = 1 for each nonzero real number a

(c) a^{-m} = 1/(a^m) for each nonzero real number a and each positive integer m

Then they proved that this definition has all the extremely desirable properties, (i) and (ii).

If you want to avoid this, that's your choice. But I don't see how you can get around the fundamental issue of defining exponentiation by all integers (or, at least, by positive integers and 0). Unless you do this you cannot appeal to equation (1). Until you define exponentiation by all integers, you cannot claim that equation (1) has been established. How can one establish equation (1) for all integers if one has not defined exponentiation for all integers.

Sincerely,

Dr. John

Your argument does not escape Samuel's charge of circular reasoning:
 

"3 ^ 3 = 3 X 3 X 3

3 ^ 2= 1 X 3 X 3

3 ^ 1 = 1 X 1 X 3

3 ^ 0 = 1 X 1 X 1

3  ^ (-1) = 1 x 1 X 3^(-1)

3 ^ (-2) = 1 X 3^(-1) X 3^(-1)"

Why did you not follow Samuel's definition:

"3^3 = 3 x 3 x 3

3^2 = 3 x 3

3^1 = 3

3^0 = ?"

You had the opportunity to put a 1 in each time that you removed a 3. So, you assumed that the pattern would continue when you got down to 3^0.

For any finite sequence, x(1),..,x(n), there are infinitely many rules for x(m) which will produce the given finite sequence x(1),...,x(n) but produce different x(n + 1)'s. Why do you choose out of these infinitely many possible sequences,

"3^3 = 3 x 3 x 3; 3^2 = 1 x 3 x 3; 3^1 = 1 x 1 x 3;....." the initial sequence "3^3 = 3 x 3 x 3; 3^2 = 1 x 3 x 3;

3^1 = 1 x 1 x 3;3^0 = 1 x 1 x 1". It evidently leads to the conclusion that you are trying to establish, but what justifies the presumption that the sequence continues in this way? 

This is part of the germ of truth in Samuel's diatribe against foolish mathematicians. He is challenging them on his assumed grounds that they are presuming that the law of exponents which is already established for exponentiation for positive integers also holds for exponentiation by 0. He is saying that they prove that a^0 = 1 by assuming a law of exponentiation for all integers (or, at least, for positive integers and 0) which law they cannot prove without assuming that a^0 =1. If this was what mathematicians were doing, then Samuel would be correct.

Similarly, you are trying to refute Samuel by assuming that your initial sequence for 3^3, 3^2, and 3^1 (which holds for Samuel's definition of exponentiation for positive integers, which definition agrees with the mathematical definition of exponentiation for positive integers) continues on in the same fashion for exponentiation by 0. What justifies this presumption? It has a nice symmetry to it, but not all things obey the symmetry which we might naively expect of them. Why should exponentiation by all integers obey the nice symmetries that exponentiation by positive integers obey?

My response to your attempt to refute Samuel is basically the same as my response to two other attempts to refute Samuel that I have read on this website. Namely, in all three cases, if the refutations are to hold up to scrutiny, the refuters must first define what they mean by exponentiation by all integers (or at least exponentiation by 0). All three refutations appear to assume that this has already been defined and that the definition satisfies the rules which are known to hold for exponentiation by positive integers.

What is your definition of exponentiation by 0?

Samuel charges mathematicians with circular reasoning. He says that they are attempting to prove that a^0 = 1, by assuming that exponentiation by 0 obeys certain laws, which laws they cannot establish without assuming that a^0 = 1.

If this were what mathematicians were doing, Samuel's charge would stick. But this understanding of what mathematicians have done arises from a lack of understanding of the history of the development of the mathematical concept of exponentiation for all integers.

I've explained this history more fully in my response to Samuel. I'll try to be more brief here.

What mathematicians did was that they figured out what was the only possible definition of exponentiation by all integers which would satisfy two extremely desirable properties: (i) the new "extended" definition of exponentiation for all integers agrees with the old definition of exponentiation for positive integers (Samuel's definition) and (ii) the new extended definition still obeys the extremely desirable exponentiation laws.

This definition is "history":

Definition:

(a) for positive integers m a^m is defined inductively by the rule: a^1 = a and a^{m + 1} = (a^m)(a)

(b) a^0 = 1 for all nonzero real numbers a

(c) a^{-m} = 1/(a^m) for all nonzero real numbers a and all positive integers m.

Taking this as the definition of exponentiation for all integers, they then proceeded to prove that this definition "fits the bill" (i.e. it has the two extremely desirable properties (i) and (ii).

If you do not want to take this route to refute Samuel's charge, this is your choice. But, if you wish your refutation of Samuel to stand up, then you first need to define your terms. What is your definition of exponentiation by all integers (or, at least, what is your definition of exponentiation by positive integers and 0)?

Until you give a definition of such exponentiation, you will not be able to establish that exponentiation has the symmetries which your refutation of Samuel's argument depends upon.

Sincerely,

Dr. John D. McCarthy

Professor of Mathematics

Part of the problem between you and the mathematician is grounded in the necessity for defining the terms which we are using in our discussion.

Your post acts as if you have defined exponentiation for all integers (or at least for all positive integers and 0).

Let's look at your attempt to give such a definition:

"One such method of explanation has been the mathematical method of powers. Exponents. You have a base number, like any other number, with another smaller number floating next to it, top right. What does this say to the human race in mathematical language? What did we decide to make that mean? The exponent, the power, the smaller number floating above, is meant to tell you how many times the base number should be multiplied by itself."

Here is where you make your attempt. To summarize this attempt, this is basically what you are saying here: a^n means that you multiply the base a by itself n times.

Let's examine this attempt at a definition.

What is multiplication? By definition, (actually, I don't know what is your definition of multiplication; so I can only speak for mathematicians; I can only tell you what mathematicians mean by multiplication); multiplication is a binary (i.e 2-fold) operation. This means that multiplication is an operation on two numbers which produces a third number. This binary operation can be repeated over and over to yield an operation which takes two or more numbers, say k numbers, and produces a

(k + 1)th number: a(1) * a(2); (a(1) * a(2)) * a(3), etc., etc..

Now comes the first gap in your attempt at a definition. What does a^1 mean, according to your definition? In other words, using your definition of a^n, what does it mean to multiply a number by itself 1 time. Multiplication, by definition, is a binary operation. It takes as input two numbers and produces another number, called the product of the two numbers. Then, if we wish, we can take a third number and mulltiply the first product and this third number, to get the product of three numbers, and so on. But what does it mean to multiply a number by itself 1 time? According to the well-established meaning of the term "multiplication", it doesn't mean anything. If you begin with a number a and you end with the number a, then you haven't done any multiplication at all. Remember, multiplication is a binary operation. You are, of course, free to choose to speak of a as the result of "multiplying a by itself 1 time", so long as you don't pretend that you have proven that a^1 is equal to a  by doing so. Using the usual definition of "multiplication", you have done no such thing. You have simply chosen to use the "multiplication" terminology to describe the result of starting with a and ending with a.

This is in fact what mathematicians have done. They have decided to define (not prove) a^1 to be equal to a. Again, why did they make this decision? The answer is the same answer I gave in my previous response to you. They decided to do this because it is the only definition of a^1 which has an extremely desirable consequence; namely, it is the only definition of a^1 which still satisfies the extremely desirable exponent laws.

You can not charge a definition with being circular reasoning. A definition simply establishes word usage. In a technical subject such as mathematics, we can not afford to be ambiguous about our terms (such as with your attempt at defining a^n as being a multiplied by itself n times. This works for n at least 2, but it does not define a^1. After all, multiplication is a binary operation.)  In order to avoid the peril of ambiguity, mathematicians give a precise, formal, rigorous definition of the technical terms which they use. (These technical definitions may or may not agree with how any particular term is used "out on the street". In particular, they may or may not agree with your definition. Again, in the case of your definition of a^n, you have not yet defined a^1 or a^0; at least not by using the standard meaning of the term "multiplication".)

When a mathematician says a^1 he means a. This is his definition of a^1. He doesn't attempt to prove a^1 = a before defining what a^1 means. He simply defines a^1 to be equal to 1.

What is your definition of a^1? Remember, multiplication is a binary operation. If, according to your definition of multiplication, multiplication is not a binary operation, then please tell me what is your definition of multiplication.

So far, this only addresses your attempt at a definition of a^n with regards to problems with a^1.

Now we get to your discussion about a^0. Again, what do you mean by multiplying a by itself 0 times? Remember, multiplication is a binary operation.

What do you mean by a^0? Remember your definition of a^n has not defined a^0; at least not according to the standard meaning of the term "multiplication".

Mathematicians do not use your definition of exponentiation; (for one thing, its too ambiguous. For another thing it has undesirable consequences; namely, it doesn't satisfy the extremely desirable exponent laws.)

Their definition agrees with your definition for integers n which are at least 2 (integers to which the usual meaning of the term "multiplication" applies). But, understanding that this definition of yours does not define a^1 and a^0 and desiring to satisfy some extremely desirable conditions, they choose to follow a different route to defining a^n for all integers n (including the integers which are at least 2; 1; and 0).

What is their decision regarding a definition of a^n? I have already explained this in my previous response to you. Let me try to briefly summarize it here. They choose to define a^n (not only for integers which are at least 2, but for all integers) by defining it in the only way possible which satisfies two extremely desirable properties (i) the extended definition agrees with the old definition for integers which are at least 2 (your definition) and (ii) the extended definition still satisfies the extremely desirable exponent laws.

If you choose to give a different definition of exponentiation by all integers (or, at least for exponentiation by all integers which are at least 2; 1; and 0), that is your choice. But, I'm just curious. What is your definition? Remember, multiplication is a binary operation.

Sincerely,

Dr. John D. McCarthy

Oops. I should have proofread my second response to you more carefully before posting it.

At one point in this response, I mistakenly wrote, concerning the mathematician's definition of a^1, "He simply defines a^1 to be equal to 1.".

What I intended to write was, "He simply defines a^1 to be equal to a.".

I apologize for any confusion which this misspeak might have caused.

Sincerely,

Dr. John D. McCarthy

 

You are the one who is committing circular reasoning:

"Here is the part that is the most difficult to explain. Once this is a shortcut, it can't really prove one way or the other, 3^0 = 1 or 0. Circular reasoning will enter into this if you try to use it to prove one way or the other. But I'm not trying to prove 3^0 = 0 right now, I'm only trying to disprove 3^0 = 1. Show another possibility. So pay attention as I bring this to even ground. If I am right, then 3^0 = 0. In which case the base numbers are not the same and you cannot use the short cut, but the short cuts own rules! "Oh, but both base numbers are 3!" No, they're not. It just appears that way. If I am right, then 3^0 = 0, which is not at all the same base number as 3. Writing it in 3^0 to begin with was a stupid way. If I am right, then 3^0 = 4^0 and any other number you want to put in front of the zero power! (3^3) * (3^0) is really (3^3) * (X^0), X being any integer. Ergo, they do not, in reality, have the same base. And if that isn't good enough for you, then look at another mathematical process. You can write 3^3 out nicely. By the definition of powers, 3^3 is telling you 3 multiplied by itself 3 times. So 3 * 3 * 3. Looking at the definition of powers, 3^0 means 3 multiplied by itself never. So nothing. So multiplying 3 * 3 * 3 by nothing gives you nothing. Ergo, the short cut still doesn't work, in the case of the raw definitions and meanings of powers. What exponents mean testifies against 3^0 = 1, and testifies for what I say to be true."

Notice, for example, in the above quote, how often your argument depends upon the premise "If I am right". When you are trying to establish that you are right, you don't get to appeal to things which follow from the assumption that you are right. That is, indeed, circular reasoning.

Again, mathematicians are not involved in circular reasoning. They are simply using a technical, precise, unambiguous, formal definition of exponentiation which has extremely favorable consequences, but which does not have the consequence which you want it to have, namely, that a^0 = 0.

Note that you have nowhere given a definition of a^0 from which you logically deduced that a^0 = 0. You apparently believe that you gave such a definition when you said that a^n means to multiply a by itself n times. Remember, however, that multiplication is a binary operation. To multiply a by itself 1 time or by itself 0 times does not mean anything; at least not according to the standard meaning of the term "multiplication" which makes multiplication to be a binary (i.e. 2-fold) operation.

What is your unambiguous definition of a^1? What is your unambiguous definition of a^0? What do you mean by multiplying a by itself 1 time? What do you mean by multiplying a by itself 0 times? (And what would it mean to multiply a by itself -3 times?)

Once you have defined what a^1 and a^0 mean, then I have another question for you. Why did you choose to define a^1 and a^0 this way.

I know what is the mathematicians answer to the last question; the mathematicians definition of a^n, for all integers n, is the only definition of a^n, for all integers which has two extremely desirable properties (i) it agrees with your unambiguous definition of a^n when n is an integer which is at least 2 and (ii) it satisfies the extremely desirable exponent laws.

We are agreed on our choice of definition for exponentiation by integers which are at least 2. But, if you were to ask me for my personal opinion about a definition of exponentiation for all integers, I much prefer the mathematicians notion of exponentiation for all integers to whatever is your notion of exponentiation for all integers. (Note that I have yet to hear from you a definition of exponentiation by 1 or 0. Remember multiplication is a binary operation.) Why do I prefer it to whatever is your definition. Because it has beautiful consequences. It's a much more interesting concept than whatever is your concept.

I do not claim that you can not make sense out of your concept. I imagine that if you try hard enough you will be able to define a concept which you call exponentiation in which concept a^0 will be equal to 0. As I have tried to show you, you have yet to do that. 

But even if you should accomplish this task, I'm sure that mathematicians will stick to studying the concept which they have invented (or, to be more precise, discovered); a very elegant concept which satisfies very elegant laws, a concept which does not have the exceptions which your concept has.

It has been said that one of the chief talents that one needs to be a mathematician is pattern recognition. Mathematicians have discovered a concept which they call exponentiation, a concept which makes sense for all integer powers (in fact, eventually can be made to make sense for all real number powers, a concept which is governed by uniform laws, laws which are very simple and elegant, without exception.

Such a concept, in my humble opinion, is a treasure worth studying.

I am sorry that you seem to be of a different opinion. I think that you are robbing yourself of great intellectual satisfaction.

But, perhaps, this is just a matter of taste.

Sincerely,

Dr. John D. McCarthy

Samuel, you started this article off great:

>>"Another thing one must realize about mathematics is that it's simply a language. The sequence 1 + 1 = 2 appears as words in our minds. They are symbols used to represent values and processes. There are not actually any 1s floating around, nor are there any 2s. There isn't a great plus sign in the sky whose power you summon – to satisfy the deity of equals."

Then you proceeded to make a slew of embarrassing errors and by the end of the article you were saying this:

>>"There may be evidence that a number to the power of zero is truly one, but my main point here today is that I have not seen it."

There is no evidence that n^0=1 for any real number n, Mathematicians define it to be such, for some very good reasons. Your attitude in the second quote completely contradicts what you said in the first quote. This article appears to have been written by two people, one a fairly reasonable individual and the other one extraordinarily ignorant.

________

The rest of this response deals with other issues that are not the main point of my response. My main point is above: mathematicians do not prove exponentiation, they define exponentiation.

Nonetheless, I will continue on with some other points to show you are not only completely misdirected in your entire article, but also just generally clueless.

>>"And also pointing out that, as of now, my side is more justified. "

There is no justification for defining n^0 to be equal to 0, such a definition is completely "red"iculous by your own definition of exponentiation.

>>"The exponent, the power, the smaller number floating above, is meant to tell you how many times the base number should be multiplied by itself. That's what it means, clear and simple. It's a shorter way of writing something out." "3^0 would be 3 zero times. Having nothing to multiply to, we are left with zero, and not with one."

Taking your definition that an exponent tells you "how many times a number is multiplied by itself", let's take an example number:

61,740

Note that we can rewrite this number by factoring out the five:

5*12,348

We can go even further by factoring out the 7's:

5*7*7*7*36

And even further by factoring out the 6's:

5*7*7*7*6*6

Now we can see clearly that, in the number 61,740, we have 5 one time, 7 three times, 6 two times, and 13 zero times, so we can rewrite it using exponents as:

(5^1)*(7^3)*(6^6)*(13^0)

So, in case you haven't kept up, this is the chain of equation we have come to, using only your definition of exponents and elementary factorization:

61,740=5*12,348=5*7*7*7*36=5*7*7*7*6*6=(5^1)*(7^3)*(6^6)*(13^0)

or, taking out the intermediaries:

61,750=(5^1)*(7^3)*(6^6)*(13^0)

Now, as you have said, n^0=0 for any number, so by your logic we get:

61,750=(5^1)*(7^3)*(6^6)*(0)

And of course, as we all agree, if you take any number zero times you get zero:

61,750=0

So a direct consequence of your "more justified side" (btw, what do you mean by "side", are there really other nutjobs who agree with you on this?) is that 61,750=0. Congratulations, your side sucks. I'll stick with the side that actually makes sense, the rational side.

I would have definitely failed your paper.

  A few years ago,I read a book about Zero,it's title " Zero:The biography of a dangerous Idea "by Charles Seife,it was a excellent read.

42. The answer is 42.

Multiplying any number A by any number B zero times leaves A unchanged:

A  *  (B⁰) = A 

Therefore (B⁰) = A/A = 1

Whereas Adding B to A zero times leaves B unchanged:

A + (B * 0) = A

Therefore (B * 0) = A - A  = 0

It's that simple.

Now where the F**k did all that other crap come from???

Some people are really confused.

Reminds me of Theology - a massive amount of discourse to avoid the blindingly obvious that the simple answer is 'God' = 0.

Once again, in order for your argument above, an attempt at refuting Samuel, you have to first defined what you mean by multiplying a number A by a number B "0 times".

By the standard definition of the term "multiplication", multiplication is a binary (i.e. 2-fold) operation. This means that it is an operation that takes 2 things and produces a third. In the case of multiplication the former two things are numbers and the latter third thing is a number, called the product of the former two things.

Now using this binary operation repeatedly, we can multiply two numbers to produce a product and then multiply the product by a third number to produce yet another number called the product of the first two numbers and the third number. 

So, according to the standard definition of the term "multiplication", your phrase "multiply a number A by a number B 0 times" has no meaning.

If you want to have a cogent argument against Samuel, then you have to define what you mean by multiplying a number A by a number B 0 times". Not multiplying at all is, simply put, not multiplying.

The same remarks apply to your phrase "Adding B to A zero times". Once again, by the standard definition of addition, addition is a binary operation. Not adding at all is, simply put, not adding.

So, in both cases, you are apparently using the terms "multiply" and "add" in some other way than these terms are normally used. Hence, in order to have a cogent argument you have to define what you mean by your usage of these terms.

More importantly, as I have pointed out in my response to a number of responses to Samuel, you have to define what you mean by B^0.

Several people, including Samuel, have been carrying on this discussion without defining their terms; without defining what they mean by a^1 and a^0. Samuel at least attempted to define what he meant by a^n. His definition was essentially this: a^n is the result of multiplying the base a "by itself n times". I speak of this as a mere attempt because as with your usage of the terms multiply and add, it fails to define what it means to multiply something by itself 0 times (or even 1 time). Again, multiplication (and addition) are binary operations.

You cannot establish something about anything that is not defined.

Also where did Samuel say that multiplying by B^0 was multiplying by B 0 times? What he said, according to his (undefined) definition of a^n (for n = 1 and n = 0) was that B^0 was multiplying B "by itself" 0 "times". Since he did not define what he meant by multiplying B by itself 0 "times" (or even 1 "time&quot, it is quite possible that according to whatever idea Samuel has in his mind (which he has not communicated to the rest of us by defining his terms), multiplying B by itsel 0 "times" gives an answer of 0.

Perhaps, Samuel means something different by the phrase "multiplying 0 times" than you mean by this phrase. I don't see how I could know whether what he means is different than what you mean, as neither of you said what you mean by this phrase. Once again, multiplication is (according to the standard usage of the term) a binary operation. Not multiplying at all is, simply put, not multiplying.

The post preceding yours is historically correct. Mathematicians do not attempt to work with an undefined concept (such as multiplying 0 times). (You are, of course, free to define what you mean by this terms.) They don't bother to define such a concept. They simply define a^0 = 1. Why do they do this? Because it is the only definition of a^0 that has the extremely desirable property of satisfying the exponent law.

I don't know to what you were referring by the term "crap". If it is to this bit of history, then it is not a bunch of crap. It is the result of cogent development of a mathematical idea, giving an unambiguous definition of an elegant concept, the concept of exponentation by any integer (including not only integers which are at least 2 (which is defined by using the binary operation of multiplication inductively) but also 1, 0, and all negative integers).

(Just as an aside, what would it mean to multiply B by itself a negative number of times.)

Mathematicians do not bother themselves with giving a definition of such concepts as multiplying B by itself 1 time or 0 times or a negative number of times. They simply define their terms. They then go on to prove that the concept which they have defined satisfies all the properties which motivated the definition in the first place.

This is not a bunch of crap. It is a fascinating story of the cogent development of a mathematical concept. I'm sorry that you have a lack of appreciation for an important chapter in man's intellectual development.

Sincerely,

Dr. John D. McCarthy

Your response, above, to Samuel's diatribe against foolish mathematicians is "right on" regarding the history of mathematics. You are correct in pointing out that mathematicians do not attempt to prove that a^0 = 1 (by using some ambiguous notion such as multiplying something by itself 0 times. Note that multiplication is, according to the standard usage of the term, a binary operation. So, according to the standard usage of "multiplication", multiplying something by itself 0 times has no meaning. Also, not multiplying at all is, simply put, not multiplying.)

But, you go on to give an argument which does not escape Samuel's charge of circular reasoning. Speaking about

5 * 7 * 7 * 7 * 6 * 6 * 6 * 6 * 6 * 6, you state:

"we can rewrite it using exponents as:

(5^1)*(7^3)*(6^6)*(13^0)"

Don't you have to assume that 13^0 = 1 to justify this assertion about rewriting?

As far as I can know, Samuel's non-defined definition of multiplying B by itself 0 times, (whatever this definition is, I can't know, because Samuel did not say), implies that B^0 = 0. So according to this definition (I'm giving Samuel the benefit of the doubt that he has a definition in his mind; that he simply did not communicate it so that the rest of us poor souls, who understand multiplication as being a binary operation.), multiplying by B^0 is multiplying by 0; in which case, one can not rewrite as you indicate.

Once again, the argument comes to the point of having defined ones terms. Once you define a^0 to be equal to a, there is no doubt about the assertion that a^0 = a.

I would greatly appreciate it if all the people who are talking about multiplying 0 times would define what they mean. As I understand it, multiplication is a binary operation (not a "0-ary" operation or a "1-ary" operation).

Sincerely,

Dr. John D. McCarthy

You write:

"The fallacy of the OP is that multiplying one number by another changes its value by the value of the multiplier, so if multiplying A by 3² changes A by a factor of 3 twice, so multiplying A by 3¹ changes it by a factor of 3, and multiplying by 3⁰ doesn't change it at all."

Please help me out here. What is your definition of 3^0 that permits you to conclude that multiplying A by 3^0 doesn't change A at all?

You will never be able to prove that multiplying A by 3^0 doesn't change A at all, until you say what is meant by 3^0.

Samuel at least attempted to define 3^0. His definition of a^n is, to paraphrase, a^n is the result of multiplying the base a by itself n times.

Since multiplication is a binary operation, the problem with Samuel's definition is that it does not say what is meant by the phrase (and I quote myself here) "multiplying a by itself 0 times (or even 1 time)". 

I can imagine how you might be defining a^0; but I don't want to make any arguments about your usage of the notation a^0 until you tell me what you mean by a^0.

I only know how mathematicians define a^0; namely, they define a^0 to be 1. Why do they do this? Because it is the only definition which has the extremely desirable property of satisfying the exponent laws.

Let's stick to Samuel's definition of a^n and, by implication, his definition of a^0. According to this definition, a^0 is the result of multiplying a by itself 0 times. (Again, multiplication is a binary operation. So how do you multiply something by itself 0 times (or even 1 time). Not multiplying at all is, simply put, not multiplying.)

Since Samuel never said what he meant by multiplying something by itself 0 times, I can only guess at what he means. Here is my guess (I hope that Samuel will tell me if my guess is wrong and bring me up to speed on what he means.). Perhaps, Samuel means that you write down a next to itself n times: (a)(a)....(a) and if n is equal to 1 so that you've written just a, you define a^1 to be equal to a and, otherwise, you form the product indicated by what you've written and define a^n to be this product. This seems to me to be a reasonable guess as to what Samuel means by the phrase "multiply a by itself n times. Now in the case of n = 0, I assume that you will have not written down anything. If I am correct about this, then what makes Samuel think that he gets anything from a^0?

Your argument above, against Samuel, goes from Samuel's discussion about multiplying B by itself 0 times (whatever that means) to multiplying A by B 0 times (which I assume means to not multiply A by anything and, hence, starting with A ends with A). What justifies asserting that Samuel has to follow along with you from his discussion to yours?

Sincerely,

Dr. John D. McCarthy

You write "(3^0) / (3^0) = 0 / 0, which does not equal 1. Hey, they could assume 3^0 = 1 in their proof, why can't I assume 3^0 = 0 in mine?".

You can assume that 3^0 = 0 in your proof. But just remember anything which you deduce from this assumption is only established to be true if your assumption that 3^0 = 0 is correct. (I cannot know whether it is or is not correct, because you have not said what you mean by multiplying the base a by itself 0 times. According to the standard definition of multiplication, multiplication is a binary operation; so, according to the usual definition  of multiplication multiplying something by itself 0 times has no meaning (it might have meaning according to your definition; but I don't know, because you haven't given your definition), nor does multiplying something by itself 1 time have any meaning (according to the usual definition of multiplication as a binary operation).

You are mistaken that mathematicians assume 3^0 = 1 in their proof. As pointed out by myself and another responder to your argument, mathematicians define a^0 = 1. Why do they do this? Because it is the only definition of a^0 which has the extremely useful property of satisfying the exponent law. It's really satisfying to define (invent? discover?) a concept which satisfies such a simple, elegant law. I'm sure (at least, I hope) that you can appreciate how satisfying that is.

I do not doubt that someone attempted to prove to you that 3^0= 1. I have certainly seen people do that. My reaction to this is that those people don't understand the history of the development of the mathematical concept of exponentiation by all integers.

But these people do serve a useful function with their argument; namely, they prove that a^0 = 1 is the only definition of the concept of exponentiation by all integers which satisfies the exponent laws.

Now it is your privilege to work with a different concept (I'd like to know what it is). I, for one, prefer the mathematician's concept. Even though I don't know what is your concept, I know (or can at least give you the benefit of the doubt) that your concept leads to exceptions to the exponent laws. For this reason, and this reason alone, I much prefer the mathematician's concept of exponentiation by all integers.

I hope that you can at least have some sympathy for why I prefer the mathematician's concept to yours, even if your preference is different than mine.

Sincerely,

Dr. John D. McCarthy

You also are apparently unawares of the mathematical development of the concept of 0 (a mathematical concept which was denied in the early history of mathematics) and the development of the definition of multiplication for this concept.

No one proved that B*0 is equal to 0 by using some undefined notion of multiplying something by itself 0 times.

Rather B* 0 was defined to be 0. Why was this done? It was done because B * 0 = 0 is the only definition of B * 0 which satisfies the distributive law of multiplication over addition, a law which was known to hold for all positive integers by the mathematical method of logical deduction known as the principle of mathematical induction.

The "all that other crap" to which you refer is the result of a studious avoidance of undefined concepts.

What is your definition of multiplication by 0, from which you deduce that B * 0 = 0. I really don't know what it is; I know what the mathematician's definition of this is; I just don't know what is your definition.

Remember, addition is, according to the standard usage of this term, a binary operation (not a "0-ary" or even a "1-ary" operation). Again, not adding at all is, simply put, not adding. Likewise, not multiplying at all is, simply put, not multiplying. If one is not multiplying at all, then what is one doing to get a result for B*0 (or for a^0). If ones says that multiplying B by 0 means not multiplying B at all, then what makes one think that one will get anything at all by doing nothing at all. Likewise, if multiplying a by itself 0 times means not multiplying a at all, then what makes one think that one will get anything at all by doing nothing at all.

Mathematicians avoid having to define such concepts as "multiplying 0 times" or "adding 0 times", by simply defining B * 0 = 0 and a^0 = 1. Why do they do this? Because these are the only definitions that satisfy laws which were already known to hold for positive integers (well before the introduction of 0 and exponentiation).

A good course in the history of mathematics, a field known for not being a bunch of crap, might help you to appreciate the desire to avoid such notions as multiplying 0 times or adding 0 times; or similar notions such as not multiplying at all or not adding at all. As the great philosopher Billy Preston put it "nothing from nothing leaves nothing". Or to put it differently, not doing anything does not do anything. That is a rather self-evident observation. 

(By the way, 0 is not nothing. Nothing does not exist. To say that he did nothing does not mean that he did something called nothing. It means that he did not do anything. 0 is not nothing. Nothing does not exist. 0 exists. 0 is a number. How exactly to define this number (since it appears to be nothing, which its not, because it exists and nothing does not exist) is an interesting challenge in its own right.

Oh, yeah, that's just a bunch of crap.

Sincerely,

Dr. John D. McCarthy
 

Multiplication of A by B is the 'scaling' o f A by the factor B.

Doing anything "zero times" is explicitly equivalent to not doing it.

"Not executing a multiplication" cannot conflict with the definition of multiplication as involving two factors, as there is no multiplication operation taking place.

Exponentiation is NOT defined as a series of multiplications.

Rather:

Exponentiation of A by a positive non-zero integer is defined as corresponding to repeated multiplication by the number A.

Exponentiation of A by a negative non-zero integer is defined as corresponding to repeated division by the number A.

Exponentiation of A by zero is defined as corresponding to no operation at all, making it consistent with both above definitions, ie, neither multiplication or division is actually involved.

IOW, the definitions for the cases where n = 1, 0, and negative, (and non-integer, of course), are chosen so as to be as mathematically consistent as possible with the basic definition for non-zero exponents greater than 1.

Got it yet??

"IOW, the definitions for the cases where n = 1, 0, and negative, (and non-integer, of course), are chosen so as to be as mathematically consistent as possible with the basic definition for non-zero exponents greater than 1."

There is a gentleman that knows what he is talking about. This is exactly how a^1, a^0, and a^{-m}, where m is a positive integer were defined. They were defined in the only way that realizes two extremely desirable goals: (i) the extended definitions agree with the old definitions for positive integers (the definitions which define a^n for an integer n > 1 as the result of the process of exponentiation, the process of "repeated multiplication by a" (rigorously defined by using the principle of mathematical induction) and (ii) the extended definitions still satisfy the exponent laws. It can be logically shown that (a) the mathematician's definition of exponentiation by all integers is the only possible definition to have these two extremely desirable properties and (b) this definition has these two extremely desirable properties. Discovering such a definition laid the groundwork for a cogent concept of exponentiation by all integers, one that "works" not only for integers greater than 1, but for 1, 0, and all negative integers.

Apparently, you think that this groundbreaking intellectual achievement is just a bunch of crap. All I can say is that you are apparently unawares of the gaps in your attempt to define exponentiation. Actually, you have never defined it, as I pointed out in another response to your posts. (Your reference to "corresponds" is at best highly ambiguous; a characteristic of a definition which has no place in rigorous scientific inquiry, the sort of inquiry which lead to the definition (invention?) of the concept of exponentiation by all integers.

Sincerely,

Dr. John D. McCarthy

This is a response to the following post:

Multiplication of A by B is

What the hell are you saying here? 

The first paragraph appears to be agreeing with my analysis, then you say in the second paragraph that I "apparently" think that the process you describe in the previous paragraph is "crap", which is 100% wrong. I basically agree with it, it seems to be a reasonable statement. 

Mathematics is not a scientific enquiry, it is not inductive in the sense used in science. It is based on strict definitions and axioms and theorems developed by strict logical deduction, which means all valid mathematical results are 100% true, meaning consistent with all relevant definitions.

Science, being based on normal induction (not mathematical induction, which is a very specific form of proof)  and observation, can only assign probabilities to its conclusions, based on confidence on the data and the relative accuracy of any theories and 'laws'  applied.

The use of the word 'corresponds' is not part of the definition, it is more a description of why the definition was framed that way.

This is a response to Bob Spence's response to "Quote: "IOW,, submitted by BobSpence1 on November 7, 2009 - 2:56am.

I apologize if I misrepresented you by misunderstanding to what you were referring when you said, in your post "Multiplying any number A by" of November 6, 2009 - 7:42am:

"Now where the F**k did all that other crap come from???

Some people are really confused.

Reminds me of Theology - a massive amount of discourse to avoid the blindingly obvious that the simple answer is 'God' = 0."

I thought that you were referring, among other things, to my lengthy response explaining the history of the development of the mathematical concept of exponentiation by all integers. I thought that you were saying that, among other things posted regarding Samuel, that this lengthy response was "crap" and that, among other people, I was "really confused".

You write, concerning "Quote: IOW,:

"The first paragraph appears to be agreeing with my analysis, then you say in the second paragraph that I "apparently" think that the process you describe in the previous paragraph is "crap", which is 100% wrong. I basically agree with it, it seems to be a reasonable statement."

I don't see how the "first paragraph" in "Quote: IOW" can be construed as agreeing with your analysis. Let me explain why.

First of all, as I pointed out in a previous post, you have not defined what you mean by exponentiation by all integers (or even, what you mean by exponentiation by 1 or 0). So how do you expect someone to understand that you are agreeing with them when they say that the definitions of a^1 and a^0 were "chosen".

Saying that exponentiation corresponds to repeated multiplication does not define exponentiation, not even for integers greater than 1. Even if we were to understand that what you meant by "corresponds" was "is the result of", your definition would only define exponentiation for integers greater than 1. After all, multiplication is a binary operation. So such ideas as "multiplying one time" (when referring to multiplying something by itself, which is what we are talking about in exponentiation by an integer power greater than 1) or "multiplying 0 times" (in the same context) have to be defined; they are not part of the notion of multiplication (of something by itself). If your definition of multiplying 0 times is not multiplying at all, then what makes you think that doing nothing at all gets you anything. (Again, I'm talking about exponentiation, which involves multiplication of something by itself.) As the great philosopher, Billy Preston, said,  "nothing from nothing leaves nothing". These attempts at a definition are not a definition.

The mathematician actually defines his terms. He defines a^1 to be a and a^0 to be 1. As "Quote, IOW." puts it, the mathematician chooses to define a^1 and a^0 the way he defines them.

Your previous posts suggest that you don't agree with this. Why? Because in your previous posts you attempted to prove that a^1 = a and a^0 = 1. Your attempt at proving that a^0 = 1, cannot be regarded as a proof because you never defined what you meant by a^0.

Saying that the effect of multiplication of B by A^0 is not to multiply B by A at all is an assertion about A^0. This assertion cannot be confirmed as being true or false until you say what you mean by A^0. Telling someone the effect of one thing on another thing does not define the former. It only makes an observation about the former. But if you have not defined the former then how can anyone else confirm your observation about the former as being a valid or invalid observation? 

Frankly, I know of no definition of A^0, other than the definition that A^0 is equal to 1, that allows anyone to observe that the effect of multiplying B by A^0 is not to multiply B by A at all. If you have such a definition, then I would be very interested in hearing it. It would definitely be novel; at least to me (and, I believe, the entire mathematical community).

Since you attempted to prove that A^0 is equal to 1, my impression is that you do not agree with the analysis that mathematicians "chose" to "define" A^0 as being 1. Rather, my understanding of your analysis is that you believe that people "concluded" that A^0 is equal to 1. By your attempt at proving that A^0 = 1, I believe that you made your belief apparent to any observer. Why would you attempt to prove that A^0 = 1, if your definition of A^0 is that A^0 is equal to 1? I assume that you understand the fallacy of circular reasoning. Granting this, I, therefore, conclude that you believe that the statement that A^0 is equal to 1 is a statement that one proves, not a definition.

This conclusion of mine is especially confirmed when, after giving your proof, you declare:
 

"It's that simple.

Now where the F**k did all that other crap come from???

Some people are really confused.

Reminds me of Theology - a massive amount of discourse to avoid the blindingly obvious that the simple answer is 'God' = 0."

I think that my conclusions about your position are based on rather sound evidence.

I am willing to grant that you have some definition of A^0 which allows you to observe that multiplication of B by A^0 has the effect of not multiplying B by A at all. I would just like to hear what is your definition.

I agree with your distinction between scientific induction and mathematical deduction. I don't understand why this removes mathematics from the realm of scientific inquiry. Scientists and mathematicians both use mathematical deduction. A research mathematician uses scientific induction in the process of discovering (or conjecturing) the things which he goes on to prove by mathematical deduction. I would say that the mathematician, when he does this, is involved in scientific inquiry.

As for the last sentence in your latest post:
 

"The use of the word 'corresponds' is not part of the definition, it is more a description of why the definition was framed that way."

I can only ask "was framed what way?". Again, I don't know what is your way of framing the definition of A^0. Again, your observing of an affect of A^0 on B does not tell me what is your definition of A^0. Again, since you attempted to prove that A^0 is equal to 1, I conclude, granting you the knowledge of the fallacy of circular reasoning, that you do not define A^0 to be equal to 1. Hence, I'm left without a clue of how it is that you define A^0.

I really would be much more able to follow your analysis if you would simply say what is your way of defining A^0. Again, what is your definition of A^0? What definition of A^0 are you using to support your contention that multiplication of B by A^0 has the effect of not multiplying B by A at all?

I look forward to hearing your response.

Sincerely,

Dr. John D. McCarthy

This is further response to BobSpence1 on November 7, 2009 - 2:56am, a response to "Quote: IOW,

"Mathematics is not a scientific enquiry, it is not inductive in the sense used in science. It is based on strict definitions and axioms and theorems developed by strict logical deduction, which means all valid mathematical results are 100% true, meaning consistent with all relevant definitions.It is based on strict definitions and axioms and theorems developed by strict logical deduction, which means all valid mathematical results are 100% true, meaning consistent with all relevant definitions."

Perhaps, you have a different understanding of scientific induction than is my understanding. Perhaps, your understanding is correct and mine is wrong. But, when you refer to scientific induction, I believe that you are referring to what Wikepedia is referring to in its definition of "induction":

Induction, also known as inductive reasoning or inductive logic, is a type of reasoning that involves moving from a set of specific facts to a general conclusion.^[1] It can also be seen as a form of theory-building, in which specific facts are used to create a theory that explains relationships between the facts and allows prediction of future knowledge.

This seems to me to describe pretty well a branch of mathematics known as applied mathematics. The applied mathematician looks at specific facts (e.g. biological facts) and builds a theory (a mathematical model) which is used to predict the behavior of a biological system. The particular example I am described, an example in biology, is carried on in a field of mathematics known as mathematical biology. Perhaps, you would not regard this as being mathematics. I believe that it is mathematics; its not what is called pure mathematics; but, nevertheless, it is mathematics.

On the other side, I am sure that Einstein would be regarded as a scientist involved in scientific inquiry. But a great part of his scientific inquiry was involved in mathematical reasoning. Some would refer to Einstein as a mathematical physicists. Moreover, there are those mathematicians who refer to themselves as mathematical physicists. Much of their work could be regarded as pure mathematics, with a view towards applied mathematics.

I agree fully with your distinction between scientific induction and mathematical deduction. But I'm not convinced of the line of demarcation which you draw between scientific inquiry and mathematics. I think that mathematicians are involved in scientific inquiry and other scientists are involved in mathematical deduction. We can point to specific activity in both areas which could be put under each column: scientific induction and mathematical deduction.

Actually, I think that it would be more helpful to speak of induction and deduction, rather than to speak of scientific induction and mathematical deduction. Scientific induction is induction applied to the task of scientific inquiry and mathematical deduction is deduction applied to the task of mathematical inquiry, a subset of scientific iniquiry.

I don't think that people would say that Einstein was not involved in scientific inquiry when he carried out his thought experiments which are essentially mathematical arguments, which is why he is sometimes called a mathematical physicist (a special type of scientist).

Sincerely,

Dr. John D. McCarthy

I will apologise for inaccurately referring to the way I used the phrase "corresponds to" - I did use it as part of the definition. Possible "is equivalent to " would be better, meaning "has the same results as". Although "corresponds to" is used in the Wikepedia article, which so far is the best reference I have found so far.

This is from the beginning of the same article, and should further clarify what I see as the definition of exponention:

That aside, between my statements in the earlier post, and the Wikepedia quotes, I see the subject quite adequately and concisely covered.

Do you still seriously have a problem with the statement that "executing an operation 0 times is equivalent to not executing the operation" ?? This seems to one of your core objections to my argument.

In my previous post:

"Multiplying any number A by any number B zero times leaves A unchanged:

A  *  (B⁰) = A 

Therefore (B⁰) = A/A = 1".

This is not a definition, so much as being intended to demonstrate how that interpretation of executing or applying an operation "zero times" leads to that result for B⁰ .

Looking at it again, I can see that there was a bit of a jump there from the definition of integer exponentiation ( Bⁿ ) to the assumption that multiplication of A by Bⁿ is equivalent to multiplying it by B n times.

While I find most of your discussion only confusing and complicating rather than clarifying the issues, I find your responses have usefully driven me to 'clean up' my arguments in an effort to make sure they are as clear and unambiguous as possible.

I may not respond any more, since after examining both my latest arguments and what external references I have had time to dig up, I personally see no difficulties with concepts involved.

Where you somehow see some of my arguments as contradicting others, I see them as looking at the problem from different directions, which is always useful.

The "other crap" I was referring to was all the totally unnecessary and sometimes misconceived elaboration both you and the OP got into, as compared to the basic argument and definitions, as in the Wiki quotes.

In so far as you appear to agree (in your para starting "There is a gentleman that knows what he is talking about." ) with my second expression of the argument, I have no problem with you.

I strongle disagree with the following statement, as obviously does the writer of the Wiki article on exponentiation I have been quoting from:

"Saying that exponentiation corresponds to repeated multiplication does not define exponentiation, not even for integers greater than 1".

Another thing I take issue with is the relevance of the history of the development of a concept, such as zero, to the actual application and current understanding of the concept. It may well be interesting in terms of the history of thought, but progress usually involves significant changes and revisions of such understanding. For example, Chemistry developed from Alchemy, but there is no general value in studying Alchemical principles in understanding modern Chemistry. It may possibly help some individual grasp some particular chemical principle, but is as likely, if not more so, to put them on the wrong track.

Mathematics is a deductive not inductive discipline. 

Math is about going from general principles ( axioms), to all the possible implications of those principles, and does not depend in any way on how accurately those axioms reflect the real world. The closer they reflect reality, the more useful will the conclusions of math be.

There are very esoteric branches of math which have gone off in very counter-intuitive directions, and occasionally we find they match surprisingly well the observations and ideas of some area of science, and then we find them useful.

Applied mathematics is applying mathematics to real world problems - it is a blend of math and empirical problem-solving.

All science employs mathematics and logic as essential tools in all aspects of the study. That neither makes Science a deductive discipline nor Mathematics an inductive one.

Thank you for the clarification of your position.

You write:

"Do you still seriously have a problem with the statement that "executing an operation 0 times is equivalent to not executing the operation" ?? This seems to one of your core objections to my argument."

I don't understand where you are getting that this is one of my core objections to your argument. I don't know of anywhere in my argument where I object to the statement that executing an operation 0 times is equivalent to not executing the operation. I believe that that is evident.

Perhaps, I need to clarify my point. Perhaps, I have been too cryptic. It would not be the first time that I have been too cryptic.

I agree with you that if you don't multiply B by A at all, then you will not affect a change in B.

The point that I was trying to communicate is that the definition of the concept of multiplication does not explain what is meant by the concept of multiplying A by A 0 times. (This is what is relevant to the concept of exponentiation, to the definition of exponentiation; not your observation of the effect of multiplying B by the result A^0 of exponentiating A to the power 0.)

Let's follow your definition of multiplying B by A^0, which is multiplying B by A 0 times. 

According to your argument against Samuel, the result of this multiplying of B by A 0 times is B. Indeed, unless this is the result, your argument against Samuel that A^0 is equal to 1, falls to the ground. Your argument requires that multiplying B by A 0 times is equal to B. That's how you are attempting to argue that A^0 is equal to 1. Your reasoning is that since B times 1 is equal to B and multiplying B by A 0 times is equal to B, therefore A^0 is equal to 1. I believe that this is a fair representation of the "gist" of your argument against Samuel.

So, now let's let B be equal to A. Now, we have agreed that multiplying B by A zero times is equal to B. Therefore, it follows, that multiplying B, A, by A 0 times is equal to B, A.

Now according to the Wikepedia article, A^n is the result of multiplying A repeatedly by A n times. With this definition, it follows that A^0 is the result of multiplying A by A 0 times.

By your observation about the result of multiplying B (in this case, A) by A  0 times, I demonstrated, in the second to last paragraph, that the result of multiplying A by A 0 times is equal to A. 

After all, if it's "good" for any B, then it's "good" for B, when B is equal to A.

Hence, we have just proven, on the basis of your observation about the effect of multiplying B by A 0 times and the Wikepedia's definition of a6n,  that A^0 is equal to A.

I think that I much prefer the mathematician's definition of A^0 to the Wikepedia article's definition of A^0, with which latter definition you apparently agree.

The Wikepedia article's definition of a^n only agrees with the mathematician's definition of a^n when n is an integer greater than 1. If one explains in the Wikepedia article that the expression a^n = a x ... x a (with braces indicating n underneath the expression on the right hand side) when n = 1 is to be interpreted as a^1 = a, then the Wikepedia article has defined a^n for every positive integer n. But only for a positive integer n; not for n = 0. And, if the Wikepedia article did this, then they would be giving the mathematician's definition of a^n for positive integers (accept for the need to define a^n inductively, the only method which I know that achieves the goal of defining something for an arbitrary integer).

I think that I am correct in saying that if you and Wikepedia are going to define a^n as the result of repeatedly multiply a by a n times, then you have to define what it means to do this when n = 0.

Again, frankly, I know of no one who has acheived a definition of exponentiation by 0, in such a way that the exponent laws are still valid, without defining a^0 to be equal to 1.

Again, if you have such a definition, then I would like to see it. The Wikepedia article does not achieve this goal.

Sincerely,

Dr. John D. McCarthy

This is a further response to BobSpence1.

You write:

"All science employs mathematics and logic as essential tools in all aspects of the study. That neither makes Science a deductive discipline nor Mathematics an inductive one."

It was not my intention to argue that Science is a deductive discipline. Nor was it my intent to argue that Mathematics is an inductive discipline.

I agree that mathematics is known for its deductive reasoning. What I fail to see is that this fact excludes mathematical inquiry from the realm of scientific inquiry?

A helpful online article addresses this question as to whether mathematics is a science:

http://andrewlias.blogspot.com/2004/08/is-mathematics-science.html

Speaking of the prevalent style of writing in mathematical journals, the author of this article states:

"The effect of this prevalent style is that everyone but the professional mathematical researcher tends to get a skewed and inaccurate view of how mathematical research is done. This in turn has led some philosophers to conclude that mathematics is not a science, because its methods are (apparently) so different from those of other empirical sciences. I will argue below that, once we go past the facade provided by the writing style and into the way mathematics is done, that this conclusion is in fact quite unwarranted."

"For starters, does mathematics even follow the scientific method? Observation, hypothesis, experimentation, testing, verification?"

"In what may come as a surprise to some, yes, it does. This is where the prevalent style does a disservice to an accurate perception of research mathematics. A mathematician engaging in research does not produce a statement for a theorem and proceed to prove it. She is usually feeling her way in the unknown as much as any scientist. She will consider some specific examples (observations), and try to see if they have a property or not. She will formulate some questions, both general and specific, and try to see how she can answer them for specific cases. She may then attempt a general statement (hypothesis), and proceed to attempt a proof (experimentation); sometimes, if that fails, she will attempt to construct a counterexample (falsification and testing). This process continues until the mathematician finally obtains an argument establishing her hypothesis, or she manages to disprove it (or, finding herself unable to do either, sets it aside and tries something else...)"

The source of mathematical results is not in mathematical deduction. Mathematical deduction is only the standard to which a claim of a mathematical result is held. But, in practice, mathematicians search for knowledge in much the same way as other scientists search for knowledge.

A famous Fields Medalist, William P. Thurston, is known for giving sketches of a mathematical proof which appeal to empirical observations. I saw him once lecture about one of his greatest contributions, which involves the application of hyperbolic geometry, explaining one of the main ideas in terms of the action of a rubber band on a surface of decreasing diameter.

These sketches are not accepted as proofs of his contributions to mathematics, but I think that they say something about the source of his contributions. Theorems do not drop out of abstraction. They are inspired by observations regarding the world in which we live.

Most mathematicians that I have met do not think in terms of mathematical deduction; rather, they prove that their thoughts are correct by using mathematical deduction. But their discovery method is in large part inductive.

Also, the way in which they communicate ideas to one another is often not through mathematical deduction. It has been, I believe rightly, said that you can learn more about a given mathematician's proof by talking to him about the driving ideas behind the proof (which are primarily heuristic not deductive) than by attempting to read the proof without these intuitive ideas (note that intuition is based in experience not deduction).

It has been said that one of the chief talents which a mathematician needs to have is pattern recognition. I have found this to be very true. Pattern recognition is not deductive. Pattern recognition is inductive.

The online article which I mentioned above maintains that mathematics is a science. According to this article, the main difference between mathematics and other sciences is the standard to which knowledge claims are held; namely, proof by mathematical deduction.

Perhaps, you disagree with this. I believe there is at least a germ of truth in it; if not an altogether convincing argument that mathematics is a science.

The famous mathematician, Gauss, called mathematics "the Queen and servant of science". Clearly, he did not mean that mathematics was only a servant of science. But, then, I suppose that Gauss could have been wrong.

Personally, I think that his opinion on the matter was grounded in being a chief contributor to both mathematics and physics, as were many of the mathematicians of his time. I believe that his opinion on the matter was the opinion of a man who was highly educated about both mathematics and science. 

 

I think that the article is probably right. Mathematical practice has probably clouded the scientific nature of mathematical inquiry. Perhaps, as the article seems to hope for, this is changing as mathematicians are beginning to write in such a way as to explain the process of mathematical discoveries, instead of merely reporting the mathematical deductions which verify these discoveries.

Sincerely,

Dr. John D. McCarthy

Re exponentiation:

So you agree that "the statement that executing an operation 0 times is equivalent to not executing the operation." is "evident", yet you still insist that is not a mathematically acceptable starting point to derive a definition for A⁰ ?

I agree that "the concept of multiplication does not explain what is meant by the concept of multiplying A by A 0 times", which why I added the definition derived from the evidently true statement above.

Now since "the result of this multiplying of B by A 0 times is B" is "evidently" equivalent to not applying any operation on B (see above), my argument follows.

You say that "now according to the Wikepedia article, A^n is the result of multiplying A repeatedly by A n times. With this definition, it follows that A^0 is the result of multiplying A by A 0 times." 

That is a misrepresentation of the article. The starting definition Aⁿ explicitly restricts n to positive non-zero integers, and then goes on later to argue for the standard definitions for the cases of n=1, 0, and <0.

It seems like you have not actually read the full article, which seems to me pretty consistent with standard Math.

You later say "the Wikepedia article's definition of a^n only agrees with the mathematician's definition of a^n when n is an integer greater than 1" which further suggests you have not actually read the article. I even quoted the bit where they explicitly declare those things as needing to be defined, and supply the definitions, exactly as required by the mathematicians.

Beyond this point I think you have drifted further off into fairyland, much of it still based on the incorrect reading of what Wiki used as the basic definitions, so it is all pretty pointless to pick apart.

Re mathematics as an inductive discipline:

I will concede that the search for new mathematical truths could be described as inductive, but the evidence sought is of a different nature than that in Science, ie, it is all still internal to the system, not from sources with no necessary connection to the current body of data, as is frequently the case in Science.

I can see how one would explore the properties of various formula and propositions, in a way similar to the way a scientific investigator would see how well various hypotheses fit the observed data. 

But when it comes to verification or testing, that is, at least in principle, should be lead to a firm true/false result, unlike in Science.

I do realize that as Mathematics has expanded, and even before, there are or have been propositions which while seeming to be likely to be true, have eluded proof.

it is also true that many complex proofs and math investigations have become somewhat empirical, requiring computer assistance, and it may well be that it will effectively become (have become?) a largely empirical discipline, in practice, due to the all but impossible task of rigorously verifying many theorems in practice.

And there do seem to be on the fringes of Mathematics, whole areas which totally evade any logical systematization.

So I will concede there are large areas of overlap, but there are still core features that I believe still put Math in a different category.

>

This is a response to BobSpence1:

I apologize for my failure to notice the Wikepedia article and its care in defining x^0. Once again, Wikepedia has shown itself to be a dependable source of information. I have learned a lot from Wikepedia.

n retrospect, I have not the faintest clue why I made a reference to Wikepedia. The definition to which I was referring appears in a different post than the one in which the Wikepedia article to which you refer appeared. The definition to which I was referring is the following definition:

Quote:

 

Exponentiation is a mathematical operation, written aⁿ, involving two numbers, the base a and the exponent n. When n is a positive integer, exponentiation corresponds to repeated multiplication:

just as multiplication by a positive integer corresponds to repeated addition:

I apologize for failure to note that in this definition (whether or not it is from Wikepedia, I do not know) n was stipulated to be a positive integer.

If there remains anything for which I need to apologize, then please let me know. I do not want to misrepresent anyone; I believe that that is an offense.

You write:

"So you agree that "the statement that executing an operation 0 times is equivalent to not executing the operation." is "evident", yet you still insist that is not a mathematically acceptable starting point to derive a definition for A⁰ ?"

Yes, I agree that "the statement that executing an operation 0 times is equivalent to not executing the operation" is "evident". But, yes, I maintain that this is not a mathematically acceptable starting point to derive a definition for A⁰

Please let me try to explain why.

First of all, in discussing A^0 (in any way that agrees with the mathematical definition of A^0) one is not merely "not executing the operation" of multiplication. I agree that when we multiply B by A zero times (i.e. no times), we have not executed the operation of multiplication on B and something else. I even agree that not executing the operation of multiplication on B and something else, leaves B unchanged.

But how does this observation about multiplying B by A zero times define A^0? In forming A^0 we are not taking B = A, or any other "B", and "not executing the operation" of multiplication on it. If we did, as I demonstrated in the post in which you refer to me as going off into fairyland, then we would have, according to this definition of executing an operation 0 times (i.e. "not executing the operation&quot, we would be left with B = A or whatever B we chose to start off with. Unless A or B was 1, we would not be left with 1, which is what we wish to get for A^0. How would a definition of A^0 in terms of "not executing the operation" of multiplication tell us that we were supposed to start off with 1 or we wouldn't get what we wanted to get?

So, yes, I maintain that the idea of "not executing the operation" of multiplication does not provide a starting point for a definition of A^0 (not to mention a mathematically acceptable one). I maintain that it does not even provide an ending point for that definition. 

In forming A^0 by not executing the operation of multiplication, with what are we starting? If we don't start with something before we don't execute the operation of multiplication, then how did we end up with 1? Perhaps, I'm just missing something here. 

In my understanding, the notion of a product applies to two or more things. If the 0 in 3^0 tells one not to execute the operation of multiplication, then what are you starting with and then following the direction not to execute the operation of multiplication so that you end up with 1? How does a definition of 3^0 which involves some notion of "not executing the operation" of multiplication tell you what to start with so that you end up with 1? I'm sorry, but I don't see how the notion of "not executing the operation" of multiplication can tell one that 3^0 = 1. I don't believe that such a definition is a definition, (nor, I believe, does the entire mathematical community). 

The article from Wikepedia that you quote begins with:
"Notice that 3¹ is the product of only one 3, which is evidently 3." What is the definition of the product of only one thing that makes it so evident that the product of only one 3 is 3?

Apparently, the Wikepedia article is using some definition of 3^1 that tells one that 3^1 is the product of only one 3. I don't know what is that definition. Perhaps I should not do this, but I am going to venture a guess. My guess is that the definition that the Wikepedia article is using is that 3^n (yes, I understand, for a positive integer n) means the product of 3^n n times. What does that mean when n is the positive integer 1, given that product refers to the result of executing the operation of multiplication, which operation is an operation which you apply to two or more numbers. I know how mathematicians define a^1. They define a^1 to be equal to 1. They don't say that (by some other definition) 3^1 is the product of only one 3, which is evidently 3. I think that they don't do this for a very good reason; namely, it is not an acceptable starting point for a definition of a^1.

I believe that mathematicians have discovered the acceptable starting point; the one that satisfies all the desirable exponent laws:

Definition:

a^0 = 1 (which definition, I understand, is the definition that the Wikepedia article gives)

a^1 = a

a^{m + 1} = (a^m)(a) for every positive integer m

a^{-m} = 1/(a^m) for every positive integer m

I know of no other definition that defines a^n so that it satisfies all the exponent laws.

If you know of such a definition, then, as I said in an earlier post, I would like to hear it. It would be a novel one for me and, I believe, the entire mathematical community.

I would not be surprised if there was a different definition. Surely there are many logically equivalent ways that a definition can be formulated. Since one can prove that there is only one definition of exponentiation of all integers which satisfies the exponent laws, it will, assuming that it satisfies all the exponent laws, be logically equivalent to the extant definition in mathematics. But, as I said, it would be novel for me. I've never seen a different formulation of the definition. Perhaps, I should not conclude from this that my colleagues have not seen such a definitiion. I would just be very surprised if they had. I expect that if they had, then I would have heard about it; it's not as if this belongs to some particular specialized branch of mathematics.

Sincerely,

Dr. John D. McCarthy

I am NOT saying that "this observation about multiplying B by A zero times define(s) A^0?".

I am saying that when we apply the observation to multiplying a non-zero number by A^0_, we get a simple equation whose solution is that A⁰ = 1.

If you do not feel this amounts to a definition, I am OK with that, I really only claimed most recently that it was a starting point for deciding on a definition, in that it demonstrates a value for B⁰ that is consistent with already defined operations.

In the original context in which I used this expression, it was still not meant to be a definition, so much as a simple illustration that 1 is the most 'logical' value to assign to A^0_.

And I agree that it is ultimately a matter of definition, a definition that should be consistent in a simple and direct way with the basic definition for positive integer exponents greater than 1.

The argument where you tried to make some point about the value of the expression 'A * (B⁰)' when A was either 1 or equal to B still strikes me as wierd. The only condition for my argument to make sense is that A be non-zero.

This is a response to:

Circular Reasoning?

I apologize for the charge of circular reasoning. Your interpretation of Samuel's post is a reasonable interpretation.

I do not, however, think that it is warranted by what Samuel actually said. Perhaps, you might want to ask Samuel if your interpretation of what he said is what he meant.

You write:

"If we follow this pattern that Samuel has established, what do we do when we see, for example, the number:
(3^3)*(4^0)?
Well the base tells us what integer we are mutliplying by, and the exponent tells us the number of times it is used as a multiplier. This is according to Samuel's reasoning.
So translating (3^3)*(4^0) we get that it means we are using 3 as a multiplier three times, and we are using 4 as a multiplier zero times.
Therefore a reasonable way to expand (3^3)*(4^0) based on Samuel's explanation of exponentiation is as follows:
(3^3)*(4^0)=3*3*3"

Based upon what Samuel actually said, I would say that by "(3^3)*(4^0)" Samuel means the following:

(a) The product of (3^3) and (4^0), each of which Samuel defines as follows:

(b) 3^3: the product of 3 by itself 3 times (this language of "by itself" is used repeatedly in Samuel's post)

(c) 4^0: the product of 4 by itself 0 times (the only place that I have been able to see in Samuel's post where Samuel gives us an indication of what he means by multiplying 4 by itself 0 times is when he tells us, concerning 3^0, that "Having nothing to multiply to, we are left with zero, and not with one."

(d) knowing what Samuel means by multiplying 3 by itself 3 times, we can conclude from what Samuel says that 3^3 = 3 * 3 * 3 = 27

(e) I don't know how to interpret Samuel's language, in explaining what is a^n, of multiplying the base a by itself n times, when n = 0.

Hence, I would conclude that we are stuck at this point in determining what Samuel means by (3^3)*(4^0).

Furthermore, if we assume that Samuel has in his mind some definition of multiplying a by itself n times, a definition which includes the case where n = 0, and which has the consequence that a^0 = 0 (which assumption to me seems like a distinct possibility), then by the above points we would conclude that Samuel's definition implies that:

(f) (3^3)*(4^0) = (3 * 3 * 3) * (0) = 0

not that:

(g) (3^3)*(4^0) = 3 * 3 * 3.

Perhaps, I am missing something; but this is all that I have been able to conclude from what Samuel actually says.

Sincerely,

Dr. John D. McCarthy

Perhaps, I need to make myself clearer. Your argument purports to determine that A^0 = 1 by using "this observation about multiplying B by A zero times".

Perhaps I am wrong about this, but I would say that such a claim, the claim that your argument determines that A^0 = 1, an argument that is based solely upon the observation that multiplying B by A zero times, amounts to a claim that "this observation about multiplying B by A zero times". I see no essential difference between a claim that "this observation about multiplying B by A zero times" determines A^0 = 1 and a claim that "this observation about multiplying B by A zero times" defines A^0. In my mind, there is no essential difference between determining the meaning of something (in this case that A^0 = 1) and defining it. If I am wrong about this, then I would replace my question by this one:

How does "this observation about multiplying B by A zero times" determine that A^0 is equal to one? 

You have to first define what you mean by A^0 in order to conclude that "multiplying B by A zero times" is the same thing as multiplying B by A^0. For, after all, how can one confirm that these two things are the same thing if one has not defined an ingredient, A^0, of the second of these two things, so as to be able to know what the second of these two things means. 

This is a response to BobSpence1:

I am NOT saying that "this

It seems that my original point has somehow become obscured in the course of this discussion.

Please let me try once more to communicate my point.

I think that it will be helpful to review the course of events that led to my entry into this discussion.

First, Samuel posts an article in which he falsely claimed that mathematicians were involved in circular reasoning in that they use an assumption about A^0 (namely, that A^0 satisfies the exponent laws, or as Samuel calls it, the "shortcut" ).

Then Bob Spence posted an article: 
 

If we look at the sequence

in which he wrote:

"If we look at the sequence of expressions, the 'problem' arises because you omitted the implied '1' in the 0 case.  Removing a factor in a chain of multiplications is NOT equivalent to replacing it with zero, it is equivalent to replacing it with unity."

Perhaps, I misunderstood the purpose of this response, but understanding it to be an attempt to refute Samuel's charge of circularity, I had the distinct impression that Bob Spence was attempting to prove that A^0 = 1; that is to say, Bob Spence was attempting to demonstrate that A^0 = 1.

In this first post, Bob Spence argues that removing a factor in a chain of multiplication is NOT equivalent to replacing it with zero, it is equivalent to replacing it with unity. This argument is evident in the situation where something is left after removing  a factor in a chain of multiplication. One of Samuel's key challenges is that mathematicians are assuming that a "shortcut" which holds for exponentiation by positive integers need not hold for positive integers and 0. This is actually a cogent challenge. Why should we conclude that since removing a 3 from 3 * 3 * 3 leaves 3 * 3 which is equal to 3 * 3 * 1 that removing 3 from 3 (which leaves us with 0 threes) leaves us with 1? Such a conclusion would be a tacit assumption that a pattern which holds for positive integers holds for positive integers and 0. Making such a tacit assumption will not avoid Samuel's charge of circular reasoning.

More importantly, how is one able to verify this conclusion if 3^0 has not even been defined? 

Moving on in the recount of events, Bob Spence subsequently posted an article:

Multiplying any number A by

In this article, Bob writes:

"Multiplying any number A by any number B zero times leaves A unchanged:

A  *  (B⁰) = A 

Therefore (B⁰) = A/A = 1"


Once again, assuming that this article was offered as a refutation of Samuel's article, I got the distinct impression that Bob Spence was attempting to demonstrate that B^0 = 1 (i.e. to prove that B^0 = 1; to conclude that B^0 = 1).

Again, in this article, there is a tacit assumption; namely, that A * (B^0) is equal to the result of multiplying A by B zero times.

 

Using the laws of arithmetic and Samuel's definition that B^n is the result of multiplying B by itself n times, one can verify this tacit assumption for positive exponents by a generalization (using mathematical induction) of the following chain of equalities:

A * (B^2) = A * (B * B) = (A * B) * B,

verifying the tacit assumption that A * (B^2) is the result of multiplying A by B 2 times.

But how is anyone able to verify this tacit assumption for exponent 0 if B^0 has never been defined.

Again, one of the key challenges of Samuel's argument is that a pattern which holds for positive integer powers need not hold for positive integer powers and the power 0. Samuel's mistake is in assuming that mathematicians assume that the pattern still holds for all positive integers and 0. Mathematicians do no such thing. They define A^0 in such a way that the pattern holds.

This is roughly the point when I came into the discussion. Understanding (perhaps mistakenly) that Bob Spence had made 2 attempts to conclude that A^0 = 1, I asked Bob Spence to answer the following question:

QUESTION: How are you defining A^0?

How are you defining A^0 in such a way that one can verify your tacit assumption that removing 3 from 3 is the same thing as replacing 3 by 1?

How are you defining A^0 in such a way that one can verify your tacit assumption that multiplying B by A zero times is the same thing as multiplying B by A^0?


If Bob Spence does not tell us what he means by A^0, then there is no way for anyone to verify these tacit assumptions.

I look forward to hearing from Bob Spence, how it is that he is defining A^0.

Sincerely,

Dr. John D. McCarthy

This is a second response to BobSpence1:
 

I am NOT saying that "this

Bob Spence wrote:

"The argument where you tried to make some point about the value of the expression 'A * (B⁰)' when A was either 1 or equal to B still strikes me as wierd. The only condition for my argument to make sense is that A be non-zero." 


The "wierd" argument that I was making was the following.

ARGUMENT: Suppose that n is a non-negative integer and two assumptions hold: 

(i) A^n is the result of multiplying A by itself n times.

(ii) B * A^n is the result of multiplying B by A n times

(iii) multiplying B by A zero times is equivalent to not executing the operation of multiplication of B by A.

Then A^0 = A.

PROOF OF ARGUMENT. Let:

(a) B = A.

By assumption (i):

(b) A^0 is the result of multiplying A by itself zero times.

By assumption (ii),

(c) B * A^0 is the result of multiplying B by A zero times.

By (a) and (c):

(d) A * A^0 is the result of multiplying A by A zero times.

Since A is equal to itself, (d) implies that:

(e) A * A^0 is the result of multiplyiing A by itself zero times.

By (b) and (e):

(f) A^0 = A * A^0.

Now by (iii):

(f) the result of multiplying A by itself zero times is A:

By (e) and (f):

(g) A * A^0 = A.

By (f) and (g) and transitivity of equality:

(h) A^0 = A.

END OF ARGUMENT

This was the point of my weird argument.

I understand, from one of your previous posts that the Wikepedia article did not assume (i). For that reason, I suppose that this ARGUMENT is a moot point. But, for what it's worth that was the point of my weird ARGUMENT.

I think that this ARGUMENT yet serves a purpose. It demonstrates that the idea of multiplying something by itself 0 times needs to be defined.

Which is basically the same point that I have been making all along; namely, A^0 has to be defined.

Again, I look forward to hearing from Bob Spence what is his definition of A^0.

Sincerely,


Dr. John D. McCarthy

Once again, what I said was "when we apply the observation to multiplying a non-zero number by A^0_, we get a simple equation whose solution is that A⁰ = 1".

IOW, IF we assume, for the sake of argument, that "multiplying B by (A⁰) is equivalent to not executing any operation on B, and B != 0", then this leads by simple math to the result that (A⁰) = 1. That is all I am saying. That is not a definition, it is a conditional statement. It is not "determining that A⁰ = 1" in definitional or absolute sense at all. How more explicit can I be before you grasp this???

Let me rephrase it yet again. I am saying that IF we define A⁰ as equal to unity, THEN the operation " multiply by A⁰ " is equivalent to  "multiply by 1", which is equivalent to not executing a multiplication at all. 

I am using the phrase "is equivalent to" as meaning that one operation has the same effect under all conditions as some other operation. I specifically use this phrase rather than "is the same as" or "is identical to" to avoid the issue you keep raising.

This then helps to justify, not determine, the decision that A⁰ should be defined as equal to 1, because it is then consistent with such natural, even if not definitive, arguments.

I AM NOT DEFINING, OR ATTEMPTING TO DEFINE, WHAT A⁰ IS!! 

I am just presenting an arguably less than rigorously formal argument that interpreting A0 in this way implies a certain result, which is consistent with the actual official definition.

A major justification for presenting such an 'informal' argument is to help show to people like Samuel why that standard definition is perfectly reasonable and consistent with normal reasoning.

Now please stop misrepresenting or misunderstanding my arguments.

I did not think something with the power of zero could be so powerful.

This is a response to BobSpence1:

Dr. John D. McCarthy

You write:

"Let me rephrase it yet again. I am saying that IF we define A⁰ as equal to unity, THEN the operation " multiply by A⁰ " is equivalent to  "multiply by 1", which is equivalent to not executing a multiplication at all."

What you are saying now is not a rephrasing of what you have been saying up until now. I am willing to grant that it is what you intended to say. But it is not what you said. Why not? Because, now you are doing something you have never done before. What is that? You are finally answering my repeated question to you; namely, how are you defining A^0 to support your assertions about A^0?.

You have finally added something that was never put into any of your posts:

"If we define A^0 as equal to unity".

Finally, you are saying what definition of A^0 you are using to get to your observations.

You're inserting this definition now; you never put it in before. As I said in my earlier emails, I would have found it a lot easier to follow your analysis, if you had simply told me how you were defining A^0. Unfortunately, you never paid me the courtesy of doing this. Things could have been cleared up a lot earlier than this, had you paid me the courtesy of answering a simple question. 

Thank you for finally answering this simple question.

I believe that had you said at the outset, "if we define A^0 as equal to unity, then the operation multiply by A^0 is equivalent to multiply by 1" and then gone into the arguments that you went into, Samuel would have not been challenged by your argument.

If after giving this definition (i.e. A^0 = 1), you had told everybody that your subsequent arguments were motivations for this definition, then I would not have had the impression that you were attempting to conclude that A^0 = 1.

I think that if you go back and read your posts, which I quoted from, then you will see that the natural interpretation of what you were doing was that you were concluding that A^0 = 1.

You repeated one of your arguments in a recent post;

I will apologise for

STATEMENT 1: "Multiplying any number A by any number B zero times leaves A unchanged:

A  *  (B⁰) = A".

I think that it is not unreasonable or something worthy of the charge of misrepresenting you, for someone to see this sentence as an argument about B^0. The last part of the sentence, "A * B^0 = A" is apparently presented here as a result of the first part of the sentence "Multiplying any number A by any number B zero times leaves A unchanged:". 

As I eluded to on more than one occassion, how is anyone able to verify that the last part of this sentence follows from the first part of this sentence if B^0 has not been defined. 

If you had simply answered my simple question, then I could have verified the last part of the sentence.

Then I would have been able to assume that the intent of STATEMENT 1 was the following:

STATEMENT 2: "Multiplying any number A by any number B zero times leaves A unchanged. This is the same result obtained by multiplying B by 1, (i.e. by the above definition of B^0, by B^0):

Since you never defined B^0, why I am being charged with misrepresenting you, when I simply took STATEMENT 1 as being a conclusion about B^0. This seems to me to be a reasonable interpretation of your STATEMENT 1, seeing as you never defined B^0.

STATEMENT 1: "Multiplying any number A by any number B zero times leaves A unchanged:

If you had simply answered my question, which I asked you several times, namely, how are you defining A^0?, then I would not have had the impression which I had.

In my post to which you are responding, I quoted accurately from two of your responses to Samuel. You can go back for yourself and read them. Nowhere did you define what you meant by A^0.

You wrote:

"when we apply the observation to multiplying a non-zero number by A^0_, we get a simple equation whose solution is that A⁰ = 1".
 

Even at the point when you wrote this assertion you had still not yet defined A^0. How was anyone supposed to verify this assertion if you never defined A^0?

 

Finally, you have defined A^0.

If you had presented what appeared to be arguments as mere motivations for defining A^0 to be equal to 1, then I would not have had the impression which I had about your responses to Samuel.


Thank you for finally answering my question.


Sincerely,

Dr. John D. McCarthy

This is a second response to BobSpence1:

Dr. John D. McCarthy

I am willing to grant that I may have misunderstood your argument. But I don't think that I am blameworthy if I did this. 

Here is further support for my understanding of your argument; your post of November 6, 2009 - 7:42am. 

I will let the reader decide for himself if I have misunderstood your argument.

 

Multiplying any number A by

Multiplying any number A by any number B zero times leaves A unchanged:

A  *  (B⁰) = A 

Therefore (B⁰) = A/A = 1

Whereas Adding B to A zero times leaves B unchanged:

A + (B * 0) = A

Therefore (B * 0) = A - A  = 0

It's that simple.

Now where the F**k did all that other crap come from???

Some people are really confused.

Reminds me of Theology - a massive amount of discourse to avoid the blindingly obvious that the simple answer is 'God' = 0.

The statement "It's that simple" has all the appearance of a claim of victory. It has all the appearance of a claim that you have just shown (i.e. proven, deduced, determined, or whatever other word you choose to describe this; I don't know what that word is) that B^0 = 1. The apparent connotation of "It's that simple." is that one does not need any other argument to see that B^0 = 1 than the one you gave. There is no indication in this argument that you are defining B^0 to be equal to 1.

There is, on the other hand, every indication in this argument that you are concluding that B^0 is equal to 1. The "Therefore (B^0) = A/A = 1" has all the appearance of being the end of a logical deduction.

If you had said something like,

PARAGRAPH 1: "Suppose that multiplication by B^0 is the same thing as multiplying A by B zero times. Then A * (B^0) = A and, hence, if A is not equal to 0, B^0 = (A/A) = 1. This gives us motivation to define B^0 to be equal to 1; namely, it is the only definition of B^0 which has the desirable property that multiplication by B^0 is the same thing as multiplying A by B zero times."

I think that if Samuel read what you actually wrote, instead of what you claim is a rephrasing of what you actually wrote, then Samuel would have come to the same conclusion as I came to, and, I am sure, a host of other readers of your article would have come to; namely, that you were providing a deduction that B^0 = 1, not a motivation.

I wish that you had said from the outset that you were defining B^0 is equal to 1 or had said something tantamount to what I said in PARAGRAPH 1 (assuming that this was in fact what you intended to communicate when you responded to Samuel). This would have saved me a lot of time and effort in trying to follow your response to Samuel's post.

Sincerely,

Dr. John D. McCarthy

In your attempted justification of your weird argument, you started with:

(i) A^n is the result of multiplying A by itself n times.

as the first assumption. You are already in error.

That is not the definition of exponentiation, and neither I nor Wikipedia ever used it. The term they employ is repeated multiplication, involving n factors, all equal to A.

By your own words, multiplication is a binary operation. 1 such operation would be A X A, ie n², 2 would be A x A x A = A³, and so on.

IOW, the proper definition, above, implies n-1 actual binary multiplications.

This is why your argument fails. You are not even consistent. A⁰ and A¹ (and A^-1) cannot be defined by (i), since they do not involve any actual multiplication operations.

For someone who comes across as very insistent on precise definitions, to use such a sloppy and downright wrong definition for exponentiation displays very poor thinking skills.

No general definition of exponentiation can refer only to 'multiplication'. Negative exponents use division, and n = -1,0,1 are special cases, since they involve neither multiplication nor division.

Regarding the definition of A^0:

I have previously written:

"Exponentiation of A by a positive non-zero integer is defined as corresponding to repeated multiplication by the number A."

Obviously this incomplete, since I have not spelled out exactly how many multiplications are involved.

In another post:

This is a response to BobSpence1

Regarding the definition of

As a careful reading of your own post, the above mentioned post, will show you, both of these definitions of a^n, which you claim allows you to say that you said what you considered to be the definition of A^0, expressly stipulate that n is a positive integer.

I am willing to grant that you thought that you had said what you considered to be the definition of A^0.

But don't put any blaim on me for missing your definition of A^0. Even if you had said that you considered these two definitions as being the definition of A^0, I still would not have known what was your definition of A^0. Why not? Because, neither of these two definitions define A^0. They only define A^n when n is a positive integer.

Sincerely,

Dr. John D. McCarthy

This is a response to:

 In your attempted

You write:
 

"In your attempted justification of your weird argument, you started with:

(i) A^n is the result of multiplying A by itself n times.

as the first assumption. You are already in error."

You've gotten off to a poor start here. The validity of a "weird" argument does not depend upon the verity of its premises. All that is necessary to make a "weird" argument valid is that the conclusion is a logically necessary consequence of the premise.

Somehow, you managed to get from my argument, which began with a couple of premises, that I was taking the first premise as a definition of A^0. That you took it this way is evident from what you said: 
 

"By your own words, multiplication is a binary operation. 1 such operation would be A X A, ie n², 2 would be A x A x A = A³, and so on.

IOW, the proper definition, above, implies n-1 actual binary multiplications.

This is why your argument fails. You are not even consistent. A⁰ and A¹ (and A^-1) cannot be defined by (i), since they do not involve any actual multiplication operations.

For someone who comes across as very insistent on precise definitions, to use such a sloppy and downright wrong definition for exponentiation displays very poor thinking skills."

Apparently, you think you have me over a barrel here. Apparently, you think that I was taking:

"(i) A^n is the result of multiplying A by itself n times."


It is obvious to anyone who follows the line of my argumentation that I was doing no such thing. That is why I made it very clear that I was taking (i) as one of two premises for a "weird" argument designed to correctly show the logically necessary conclusions of (i) and the other premise (ii). I made this very clear when I said "Suppose ... " as opposed to saying something like, "Now we know (i) ...." or "Since we have previously established (i) ....".

I don't know what kind of thinking that you are using to make the step from my using a premise P in an "IF P, THEN Q." statement to an insinuation that I have taken P as a definition of A^0 (I have done no such thing, as is evident from everything which I have posted in this thread concerning Samuel's post). But I know this, whatever thinking allows someone to make this step does not represent good "thinking skills".

My definition of A^0 was made very clear in what I have written in this thread. Let me repeat it:

DEFINITION: (i) a^0 = 1 for every nonzero real number a

(ii) a^1 = a

(iii) a^{n + 1} = (a^n)(a) for every positive integer n

(iv) a^{-m} = 1/(a^m) for every positive integer m

I have nowhere departed from this definition. How you manage to find the place to accuse me of being inconsistent about this, I haven't a clue. I never said that (i) was a definition of A^n. What I did was, I investigated the consequences of assuming (i). You can't accuse someone of believing that x + 1 = x when he shows correctly shows that x + 1 = x implies that x + 2 = x + 1. You certainly cannot say that he made an error at the beginning of this argument when he says:

PROPOSITION A If x + 1 = x, then x + 2 = x + 1.

There is a very simple completely logically valid proof of this proposition.

I gave a very simple completely logically valid proof of this proposition (the proof which you refer to as a "weird" argument).

Telling someone that he made an error in his proof of PROPOSITION 1 because he began the proof by assuming its hypothesis does not represent good "thinking skills".

If I made a mistake in my posts here, the mistake lies in trying to guess what was your definition of A^0.

Samuel had given the following explanation of the meaning of a^n: 

"The exponent, the power, the smaller number floating above, is meant to tell you how many times the base number should be multiplied by itself." 

You responded to Samuel without challenging this explanation. I wrongly assumed that you were willing to work with this as a definition. I apologize for this.

I guess that I should have simply asked you what was your definition of A^0. Oh yeah, I already did that.

Once again, thank you for finally telling me what was your definition of A^0. It's been very helpful.

Sincerely,

Dr. John D. McCarthy

This is a second response to BobSpence1's article:

Regarding the definition of

You write:

"I have always pointed out that I agree with the 'official' definition."


I don't think that you actually said this until very recently. You did say in an earlier post:

Quote:"IOW,

I basically agree with it, it seems to be a reasonable statement.

I think that a reasonable interpretation of this statement of yours is that you were using a different definition of A^0, one which basically agreed with the 'official' definition. I don't see how this can be taken as giving one your definition. Perhaps, I'm just a bit slow here. It wouldn't be the first time that I had difficulty following someone.

I really don't understand why you never responded to my question "How are you defining A^0?" by writing back "I'm defining A^0 to be equal to 1."

Pointing to a Wikepedia article or one of your earlier posts in which n was a positive integer did not tell me that this was your definition. Call me slow if you wish, but I really don't think that you had given me your definition by doing this.

Frankly, since you never answered by saying something like "I'm defining A^0 to be equal to 1." or "I'm defining A^0 as corresponding (in your words) to the result of repeated multiplication of A."

Presumably this would have meant that you were performing repeated multiplication of A by A (i.e. of A by itself) -1 times (whatever that means).  I say "presumably" (as opposed to "logically&quot, because A^2 is the result of repeated multiplication of A by A one time. (Note that I can only guess what it means to repeat multiplication less than 0 times.) 

But even then I would still be left with not knowing what it means to repeat multiplication by A less than 0 times. If I repeat multiplication of A by A one time, I get A^2. If I repeat multiplication of A by A zero times, I suppose that I get A. But what do I get when I repeat multiplication of A by A less than 0 times; I really don't have a clue.

So, as far as I can see you never gave me your definition of A^0 until you indicated "If A^0 = 1,..".

Even then you had not given me a definition of A^0. Why not? Because, if I took the premise "A^0 = 1" of the "If ..., then...
 statement as your definition of A^0, then I would be making the mistake of concluding that because one used P as the premise of an "If P, then Q." statement, one was indicating that he accepted P as a valid premise.

Once again, so that you don't misconstrue me, I really do appreciate it that you finally came around to telling me what was your definition of A^0.

Thank you.

Sincerely,

Dr. John D. McCarthy

This is a second response to BobSpence1's post:

In your attempted

You write: "For someone who comes across as very insistent on precise definitions, to use such a sloppy and downright wrong definition for exponentiation displays very poor thinking skills." 

I guess that I'm just repeating myself here, but what makes you think that I was using "such a sloppy and downright wrong definition for exponentiation". I was doing no such thing. What I was doing was showing the logical consequences of such a definition.  The definition which I am using for exponentiation is the mathematician's definition, the one that I spelled out in detail in my posts. I won't repeat it here. It is the mathematician's definition.

Using the "sloppy and downright wrong definition for exponentiation" as the premise of an argument regarding the logical consequences of taking this "sloppy and downright wrong definition for exponentiation" as a definition, does not constitute taking this "sloppy and downright wrong definition for exponentiation" as a definition of exponentiation. Apparently, you think that it does; otherwise why would you carry on about how sloppy, downright wrong, and inconsistent I was being? 

Sincerely,


Dr. John D. McCarthy