Proof that a^0 = 1

  • #1
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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.
 

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  • #2
jbriggs444
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Iwhich clearly gives a^n * a^m = a^(n+m)
But only for n and m both non-zero positive integers.
 
  • #3
PeroK
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##a^0 =1 \ (a \ne 0)## by definition.

This definition is chosen so that you have:

##a^na^m = a^{n + m}##
 
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  • #4
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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
 
  • #5
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##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 ??
 
  • #6
PeroK
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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.
 
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  • #7
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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?
 
  • #8
jbriggs444
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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.
 
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  • #9
hilbert2
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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.
 
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  • #10
PeroK
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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.
 
  • #11
hilbert2
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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.
 
  • #12
PeroK
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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!
 
  • #13
FactChecker
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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}}##
 
  • #14
PeroK
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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.
 
  • #15
mathman
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##a^{-1}=\frac{1}{a}##. Therefore ##a^1\cdot a^{-1}=a^{1-1}=a^0=\frac{a}{a}=1##.
 
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  • #16
jbriggs444
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##a^{-1}=\frac{1}{a}##
Only if you define it thus.
 
  • #17
PeroK
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##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##.
 
  • #18
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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.
 
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  • #19
Infrared
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Maybe this is too basic, but hopefully it helps. Consider the sequence [itex]2,4,8,16,32,\ldots[/itex]. The [itex]n[/itex]-th term of this sequence is [itex]2^n[/itex]. In going from the [itex]n[/itex]-th term to the [itex](n+1)[/itex]-st term, we multiply by [itex]2[/itex]. If we want this pattern to hold for all integers [itex]n[/itex], then we are forced to have [itex]2^0=1[/itex], [itex]2^{-1}=1/2[/itex], etc. so that our sequence is [itex]\ldots 1/4,1/2,1,2,4,8,\ldots[/itex].

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

Probably you can't prove that [itex]a^n[/itex] 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.
 
  • #21
jbriggs444
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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.
 
  • #22
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So at what point can I just define something without proving its true and have a result thats true regardless if the definition is true or not

I guess what a better way to say is when can I make an assumption?
 
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  • #23
PeroK
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So at what point can I just define something without proving its true and have a result thats 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.
 
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  • #24
jack action
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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##.
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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.
 
  • #25
PeroK
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##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!
 
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