Proof that a^0 = 1

  • #1
Rijad Hadzic
321
20
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.
 

Answers and Replies

  • #2
jbriggs444
Science Advisor
Homework Helper
11,176
5,760
Iwhich clearly gives a^n * a^m = a^(n+m)
But only for n and m both non-zero positive integers.
 
  • #3
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
22,188
13,599
##a^0 =1 \ (a \ne 0)## by definition.

This definition is chosen so that you have:

##a^na^m = a^{n + m}##
 
  • Like
Likes Demystifier
  • #4
Rijad Hadzic
321
20
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
Rijad Hadzic
321
20
##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
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
22,188
13,599
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.
 
  • Like
Likes jim mcnamara and jbriggs444
  • #7
Rijad Hadzic
321
20
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
Science Advisor
Homework Helper
11,176
5,760
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.
 
Last edited:
  • #9
hilbert2
Science Advisor
Insights Author
Gold Member
1,541
559
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.
 
Last edited:
  • Like
Likes scottdave and Demystifier
  • #10
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
22,188
13,599
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
Science Advisor
Insights Author
Gold Member
1,541
559
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
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
22,188
13,599
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
Science Advisor
Gold Member
7,146
3,037
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
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
22,188
13,599
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
Science Advisor
8,059
540
##a^{-1}=\frac{1}{a}##. Therefore ##a^1\cdot a^{-1}=a^{1-1}=a^0=\frac{a}{a}=1##.
 
  • Like
Likes scottdave, Wes Turner, Delta2 and 1 other person
  • #16
jbriggs444
Science Advisor
Homework Helper
11,176
5,760
##a^{-1}=\frac{1}{a}##
Only if you define it thus.
 
  • #17
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
22,188
13,599
##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
nuuskur
758
588
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.
 
Last edited:
  • Like
Likes ZeGato, dextercioby and Delta2
  • #19
Infrared
Science Advisor
Gold Member
964
538
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
Science Advisor
Homework Helper
11,176
5,760
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
Rijad Hadzic
321
20
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?
 
Last edited:
  • #23
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
22,188
13,599
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.
 
Last edited:
  • Like
Likes Rijad Hadzic, Delta2, Stephen Tashi and 1 other person
  • #24
jack action
Science Advisor
Insights Author
Gold Member
2,464
4,978
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.
 
  • #25
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
22,188
13,599
##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!
 
Last edited:
  • #26
jbriggs444
Science Advisor
Homework Helper
11,176
5,760
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.
 
  • Like
Likes dextercioby and PeroK
  • #27
jack action
Science Advisor
Insights Author
Gold Member
2,464
4,978
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.
 
  • #28
mathman
Science Advisor
8,059
540
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.
 
  • #29
jbriggs444
Science Advisor
Homework Helper
11,176
5,760
Well, that looks more like a proof to me
It is not a proof. You don't prove definitions.
 
  • #30
jack action
Science Advisor
Insights Author
Gold Member
2,464
4,978
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
7d240dbaf6181ae1801474f3d28dcd5504aacae6

and the recurrence relation
22becb6fbb370b056af0dc723f2af7e4db6a034a

[...]
Negative exponents
The following identity holds for an arbitrary integer n and nonzero b:
bc5945fefb607bd5dffd31f93161985362c8e547

[...]

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
50badcf877aaca6046b1f3a8bc15123e6a11943a

By defining this relation as valid for all integer n and nonzero b, it follows that
79324b0f81e42f7b891a815da86599083450324a
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».
 
  • #31
jbriggs444
Science Advisor
Homework Helper
11,176
5,760
Isn't that a proof?
No. It is not a proof. Again, you do not prove definitions. You write them down and prove that they are unambiguous and consistent.
 
  • Like
Likes Stephen Tashi
  • #32
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
22,188
13,599
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).

So, where do you think something like the definition of the derivative, say, came from?

$$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}$$
You can't prove that. All you can do is show that such a definition leads to a logical and consistent concept of a derivative that has the properties you were expecting - with perhaps some surprises along the way.

These things are not arbitrary, they are based on a prior knowledge of mathematics and what you want to achieve. You cannot prove these things. You have to define them and see where those definitions lead.

Note that defining ##2^0## as either:

##2 \cdot 2^0 = 2^1##

or

##2^0 = 1##

Amount to the same thing.

The fundamental issue is that no mathematical symbols have any meaning until you define them. You cannot prove something like ##2^0 = 1## until you have defined what you mean by ##2^0## in the first place. That's part of pure mathematical thinking. If you study pure mathematics, one of the things you learn to do is recognise when something has been fully defined and when it hasn't.

And, as previously mentioned, you have the same issue with ##0! = 1##. You cannot prove that. And, I suggest, most people would expect ##0! = 0##. But, the usefulness of ##0! = 1## becomes apparent when you define the binomial coefficient. So, again, it's far from arbitrary.
 
Last edited:
  • Like
Likes jbriggs444 and Stephen Tashi
  • #33
Stephen Tashi
Science Advisor
7,741
1,527
I know Wikipedia is not considered the best reference, but here what they have to say about exponentiation:
The current Wikipedia article does not develop definitions and theorems dealing with exponents in a mathematically precise way. It is a survey article, similar to the treatment of exponents in high school mathematics texts, which give the "properties" of exponents without specifying which properties are definitions and which are theorems. For example, in the section "Zero Exponent" the statement "##b^0=1##" is made before the supposed deduction in a later section that ##b^0=1##.

On the "talk" page associated with the article you can find debate about whether ##b^0 =1## applies in the case ##b = 0##.

If you are interested in doing proofs in mathematics, you must begin the proof in a very specific context - being clear about what definitions, assumptions, and theorems are established before the proof begins. This is not a simple task because a given topic ( such a exponents) can be treated as a mathematical system in different ways. So you have to specify which particular way you are using.

The mathematically precise approaches to "elementary" topics like exponentiation are complicated and only taught in advanced classes. (For example, what is the proof or definition that tells us it is possible to multiply two irrational numbers? )

The average person (including myself) who contemplates the elementary properties of numbers takes a Platonic view - i.e. we think of the numbers and their "properties" existing independently of any particular mathematical set of definitions and assumptions. This type of intuitive thinking is a necessary tool, but it muddles the process of writing proofs because we are missing the required organization. We have in mind a collection of facts, but we don't know which are definitions, which are assumptions and which are theorems.

In addition to confusion caused by Platonic thinking, there is a confusion that comes from being hypnotized by the magic of symbols. Manipulations with symbols work so well that we begin to think that any nice looking string of symbols automatically has some meaning. If we were asked to prove that ##b^{*</} = 17##, we would rebel by demanding to know the meaning of "##b^{*</}##". However, if we are asked to prove that ##0^0 = 1##, we are tempted to start scribbling out a proof immediately because "##0^0##" is a string of symbols that we pattern-match to other strings like ##2^3##, whose definitions we know.

Some humorous but also serious advice: If you are interested in mathematical proofs, do proofs in intermediate and advanced mathematics where the elementary "properties" of numbers are taken for granted. Only much later in your career should you take on the challenge of developing the elementary properties of numbers as a precise mathematical system. Textbooks on the topic of "Real Analysis" usually have a few chapters devoted to such torture - topics like Dedekind Cuts etc.
 
Last edited:
  • Like
Likes jbriggs444 and PeroK
  • #34
jack action
Science Advisor
Insights Author
Gold Member
2,464
4,978
@jbriggs444 , @PeroK , @Stephen Tashi ,
You write them down and prove that they are unambiguous and consistent.
Considering the OP's question AND the fact that this thread is classified as "B" [Thread Level: Basic (high school)], don't all of you think this is the kind of «proof» the OP is looking for?
 
  • #35
jbriggs444
Science Advisor
Homework Helper
11,176
5,760
@jbriggs444 , @PeroK , @Stephen Tashi ,

Considering the OP's question AND the fact that this thread is classified as "B" [Thread Level: Basic (high school)], don't all of you think this is the kind of «proof» the OP is looking for?
My suspicion is that OP believes that there is an underlying "true" definition for ##a^0## which can be proven correct from first principles. As has been pointed out, such a belief is fundamentally mistaken.
 

Suggested for: Proof that a^0 = 1

  • Last Post
3
Replies
81
Views
9K
  • Last Post
Replies
26
Views
3K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
33
Views
3K
  • Last Post
Replies
3
Views
2K
Replies
28
Views
4K
  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
30
Views
103K
  • Last Post
Replies
16
Views
3K
Top