Proving a^0=1: Step-by-Step Guide

[Rijad Hadzic]

I'm trying to prove that a^0 is = 1

So if I define a^1 to be = (a)(1)

and a^n to be = (1)(a)(a)...(a) with the product being taken n times
and a^m to be = (1)(a)(a)...(a) with the product being taken m times

a^n * a^m would then = (1)[(a)(a)...(a) with the product being taken n times * and a^m to be = (1)(a)(a)...(a) with the product being taken m times]

which clearly gives a^n * a^m = a^(n+m)

if m = 0, a^n * a^0 = a^(n+0) = a^(n), so a^0 = 1

for some reason this does make sense to me but I have a feeling the result is not satisfying enough.

 

[jbriggs444]

Iwhich clearly gives a^n * a^m = a^(n+m)

But only for n and m both non-zero positive integers.

 

[PeroK]

$a^0 =1 \ (a \ne 0)$ by definition.

This definition is chosen so that you have:

$a^na^m = a^{n + m}$

 

[Rijad Hadzic]

But only for n and m both non-zero positive integers.

damn it forgot to state this. But my proof still doesn't satisfy me for some reason

 

[Rijad Hadzic]

$a^0 =1 \ (a \ne 0)$ by definition.

This definition is chosen so that you have:

$a^na^m = a^{n + m}$

Are you saying the proof is not valid if I start from a^n*a^m ??

 

[PeroK]

Are you saying the proof is not valid if I start from a^n*a^m ??

I'm saying that, essentially, you cannot prove it. Anymore than you can prove that $0! = 1$.

Sure, if you assume that

$a^na^m = a^{n + m}$

Then $a^0 = 1$ follows from that. But, that's more a motivation for a definition than a proof.

 

[Rijad Hadzic]

I'm saying that, essentially, you cannot prove it. Anymore than you can prove that $0! = 1$.

Sure, if you assume that

$a^na^m = a^{n + m}$

Then $a^0 = 1$ follows from that. But, that's more a motivation for a definition than a proof.

Wouldn't

"So if I define a^1 to be = (a)(1)

and a^n to be = (1)(a)(a)...(a) with the product being taken n times
and a^m to be = (1)(a)(a)...(a) with the product being taken m times
"
allow it to constituted as a proof though, since I'm defining it in that way?

 

[jbriggs444]

Wouldn't

"So if I define a^1 to be = (a)(1)

and a^n to be = (1)(a)(a)...(a) with the product being taken n times
and a^m to be = (1)(a)(a)...(a) with the product being taken m times
"
allow it to constituted as a proof though, since I'm defining it in that way?

A proof of what? That is an adequate definition of a^n for n an integer greater than zero. It says nothing about a^0.

It is also redundant. If you define a^n, you've defined a^m. The "n" and the "m" are dummy variables.

Edit: I do not think I was understanding what you were trying to express.

You want to define a^n as the result of evaluating "1(a)...(a)" where there are n a's. In the case of n=0, this means zero a's and it is just "1".

Under this definition, a^0 = 1 by definition (even when a=0) and there is nothing to prove.

 

[hilbert2]

I think the appropriate way would be to assume that $f(x)=a^x$ is continuous and then show that the sequence

$a^{1/2},a^{1/4},a^{1/8},\dots$

or

$\sqrt{a},\sqrt[4]{a},\sqrt[8]{a},\dots$

has limit 1 when $a\neq 0$.

It's just a matter of choosing some properties you want the exponential function to have, and then showing that the only value of $a^0$ that is logically compatible with those properties is 1.

 

[PeroK]

I think the appropriate way would be to assume that $f(x)=a^x$ is continuous and then show that the sequence

$a^{1/2},a^{1/4},a^{1/8},\dots$

or

$\sqrt{a},\sqrt[4]{a},\sqrt[8]{a},\dots$

has limit 1 when $a\neq 0$.

It's just a matter of choosing some properties you want the exponential function to have, and then showing that the only value of $a^0$ that is logically compatible with those properties is 1.

That's fine, but perhaps a physicist's view. The function $a^x$ for real $x$ is a more advanced construction. You might hope to resolve what $a^0$ should be while you are still dealing with integer powers.

 

[hilbert2]

That's fine, but perhaps a physicist's view. The function $a^x$ for real $x$ is a more advanced construction. You might hope to resolve what $a^0$ should be while you are still dealing with integer powers.

Yes, in my version of the proof the property $a^{1/n} = \sqrt[n]{a}$ is assumed as an "axiom", while a simpler choice could also be possible. I guess we're playing the game of "inventing the exponential for the first time" here, instead of relying on commonly accepted sets of rules.

 

[PeroK]

Yes, in my version of the proof the property $a^{1/n} = \sqrt[n]{a}$ is assumed as an "axiom", while a simpler choice could also be possible. I guess we're playing the game of "inventing the exponential for the first time" here, instead of relying on commonly accepted sets of rules.

That "game" is called pure mathematics!

 

[FactChecker]

Sure, if you assume that

$a^na^m = a^{n + m}$

This just looks to me like the associative law of multiplication: $(aa...a)_{n\text{ times}}(aa...a)_{m\text{ times}} = (aa...a)_{n+m\text{ times}}$

 

[PeroK]

This just looks to me like the associative law of multiplication: $(aa...a)_{n\text{ times}}(aa...a)_{m\text{ times}} = (aa...a)_{n+m\text{ times}}$

That doesn't get you to $a^0=1$.

The issue is, for example, that you could verify this for positive integers:

$a^3 a^2 = a^5$

But, if you try to verify this for any integers you have:

$a^2 a^{-2} = 1$

But, you can't verify that $a^0 =1$ as "a multiplied by itself 0 times" is not immediately defined. You have to define $a^0 =1$ in order for your law of indices to extend to integers.

And that's what is done.

 

[mathman]

$a^{-1}=\frac{1}{a}$. Therefore $a^1\cdot a^{-1}=a^{1-1}=a^0=\frac{a}{a}=1$.

 

[jbriggs444]

$a^{-1}=\frac{1}{a}$

Only if you define it thus.

 

[PeroK]

$a^{-1}=\frac{1}{a}$. Therefore $a^1\cdot a^{-1}=a^{1-1}=a^0=\frac{a}{a}=1$.

If you, a priori, assume that a basic law holds when extend the numbers involved, then:

$a^n b^n = (ab)^n$

Extends to:

$(-1)^{1/2}(-1)^{1/2} =1^{1/2} = 1$

Which is then a "proof" that $-1 = 1$.

 

[nuuskur]

In general, one can think of the expression $A^B$ as the set of all mappings $f:B\to A$. For arbitrary cardinalities it holds that $\left\lvert A^B\right\rvert = \left\lvert A\right\rvert ^\left\lvert B \right\rvert$. Thus $a^0 = \{a\} ^\emptyset = 1_\mathbb N$, as for any set $A$, there is exactly one mapping $f:\emptyset\to A$

One can also use this to semi-prove things like $0! = 1$. Factorial represents the number of permutations, thus there is exactly one "empty permutation". It is safer to define $0! = 1$.

All of this depends on where you are operating. In a group, for instance, we just define $g^0$ to be equal to the identity as the expression $g^0$ doesn't really make sense, otherwise. In a semigroup $s^0$ might be an ill-defined array of symbols.

In the real numbers, one could think $\log _a1 = 0$ iff $a^0 =1$. Which was first, the egg or the chicken? It's a lot of boring debate, to be honest. Let's just say that if the structure permits it, we define $a^0 = 1$ where the meanings of the symbols depend on context.

 

[Infrared]

Maybe this is too basic, but hopefully it helps. Consider the sequence 2,4,8,16,32,\ldots. The n-th term of this sequence is 2^n. In going from the n-th term to the (n+1)-st term, we multiply by 2. If we want this pattern to hold for all integers n, then we are forced to have 2^0=1, 2^{-1}=1/2, etc. so that our sequence is \ldots 1/4,1/2,1,2,4,8,\ldots.

Another way of phrasing this is just that we want 2^{n+1}=2^1\cdot 2^n since 2^{a+b}=2^{a}2^b is a law that we would like to keep.

Probably you can't prove that a^n is the right thing for nonpositive exponents since usually exponentials are defined first for only when the exponent is a positive integer, and then you extend the definition to integer exponents in the above way, and then to rationals, and then to reals.

 

[mathman]

Only if you define it thus.

How else would you define it?

 

[jbriggs444]

How else would you define it?

In a conversation where one is discussing the definition of $a^0$, introducing a definition of $a^{-1}$ seems premature.

 

[Rijad Hadzic]

So at what point can I just define something without proving its true and have a result that's true regardless if the definition is true or not

I guess what a better way to say is when can I make an assumption?

 

[PeroK]

So at what point can I just define something without proving its true and have a result that's true regardless if the definition is true or not

I guess what a better way to say is when can I make an assumption?

The fundamental issue is that when you use some mathematical symbols, you must define what you mean by that arrangement of symbols. Until you know what you mean by those symbols, you cannot start to do mathematics using them. In this case, for example, you might write:

$2^0$

But, what does that mean? There's no immediate way to "multiply 2 by itself 0 times". Unlike $2^1, 2^2, 2^3 \dots$, which have a simple, clear definition.

My recommended approach is to define $2^0 = 1$ before you go any further. Then you know what those symbols mean.

Now, of course, you need to be careful that a definition is consistent with other definitions, and you need to understand the implications of a certain definition.

In this case, the only other candidate might be to define $2^0 = 0$. But, when you look at the way powers work, you see that defining $2^0 =1$ is logical and consistent.

 

[jack action]

My recommended approach is to define $2^0 = 1$ before you go any further (why?). Then you know what those symbols mean.

Now, of course, you need to be careful that a definition is consistent with other definitions, and you need to understand the implications of a certain definition.

In this case, the only other candidate might be to define $2^0 = 0$. But, when you look at the way powers work (you define how powers work afterward?), you see that defining $2^0 =1$ is logical and consistent.

But that means that you basically define what $2^{-1}$ means and then you adjust your "guess" to $2^0 = 1$.

In a conversation where one is discussing the definition of $a^0$, introducing a definition of $a^{-1}$ seems premature.

Not only is defining $a^{-1}$ not premature, it is essential. Otherwise, you are only guessing arbitrarily as @PeroK explained, and you modify your guess as you (finally) define $a^{-1}$. $a^0 = 1$ makes sense only if $a^{-n} =\frac{1}{a^n}$. then a simple limit approach proves the definition of $a^0$. Therefore, I tend to support @mathman 's approach:

$a^{-1}=\frac{1}{a}$. Therefore $a^1\cdot a^{-1}=a^{1-1}=a^0=\frac{a}{a}=1$.

---------------------------------------

If you, a priori, assume that a basic law holds when extend the numbers involved, then:

$a^n b^n = (ab)^n$

Extends to:

$(-1)^{1/2}(-1)^{1/2} =1^{1/2} = 1$

Which is then a "proof" that $-1 = 1$.

Doesn't that only proves that $-1 \times -1 = 1$?

$a^n$ and $a^{-n}$ have both distinct definitions, so stating both «sources» $a$ are the same because they give the same result is as fair as saying that since $\sin\frac{\pi}{2} = 1$ and $\cos 0=1$, then $\frac{\pi}{2} = 0$ must be true.

 

[PeroK]

$a^n$ and $a^{-n}$ have both distinct definitions, so stating both «sources» $a$ are the same because they give the same result is as fair as saying that since $\sin\frac{\pi}{2} = 1$ and $\cos 0=1$, then $\frac{\pi}{2} = 0$ must be true.

That makes no sense to me. Although looking at your avatar might explain things!

 

[jbriggs444]

Not only is defining $a^{-1}$ not premature, it is essential.

Ummm, no. It is not essential. Given a definition for raising a number to an arbitrary positive, non-zero, integer power, one can extend that definition to $a^0$ in one obvious way that preserves the truth of $a^{n+1} = a \times a^n$

There is, of course, an extra bit of freedom in defining 0^0 to be consistent with that equality.

If one is working in the ring of integers, $a^{-1}$ may not be defined.

 

[jack action]

Given a definition for raising a number to an arbitrary positive, non-zero, integer power, one can extend that definition to $a^0$ in one obvious way that preserves the truth of $a^{n+1} = a \times a^n$

Well, that looks more like a proof to me and it is a much more useful answer than your previous ones. And it is certainly better than @PeroK 's answer who basically said «let's create an arbitrary definition and see if it's "logical and consistent"» (which makes no sense to me as a mathematical proof).

I'm glad I made you elaborate your thoughts on the subject, because I like polite exchanges that enrich my life.

 

[mathman]

You could start with $a^{n+1}=a\times a^n$ then $a^n=\frac{a^{n+1}}{a}$ and work backwards to get $a^0=\frac{a^1}{a}=1$. Further continuation to get negative exponents.

 

[jbriggs444]

Well, that looks more like a proof to me

It is not a proof. You don't prove definitions.

 

[jack action]

It is not a proof. You don't prove definitions.

I know Wikipedia is not considered the best reference, but here what they have to say about exponentiation:

Positive exponents
Formally, powers with positive integer exponents may be defined by the initial condition

and the recurrence relation

[...]
Negative exponents
The following identity holds for an arbitrary integer n and nonzero b:

[...]

The identity above may be derived through a definition aimed at extending the range of exponents to negative integers.
For non-zero b and positive n, the recurrence relation above can be rewritten as

By defining this relation as valid for all integer n and nonzero b, it follows that

So the process is to define for positive exponents ($b^1 = b$). Then an identity is derived to extend to negative exponent ($b^n = \frac{b^{n+1}}{b}, n\ge 1$). Note that we use this relation to specifically extend the definition to negative exponents (I like to think this comes before an attempt to understand what $b^0$ could be, but you prove successfully that it is not necessary).

Finally, once this definition is accepted, $b^0=1$ must be true because $\frac{b^{0+1}}{b}=1$. It is not defined as is, it is a consequence of the accepted general definition of what exponentiation is ("It follows that $b^0=1$"). Isn't that a proof? Because I really doubt someone started with «Let's assume $b^0 = 1$ and find an exponentiation definition that includes that definition».