Can a^0 Equal 1? A Simple Proof

[brunopinto90]

I was watching some lectures on complex analysis and in a proof, the lecturer defines 0 = 1-1. Then i was watching a movie and like an epiphanie, i came up with a proof.

a^0 = a^1-1 = a^1 x a^-1 = a/a = 1.

Is it good?

 

[micromass]

Well, then you have to prove:
1) $a^{n+m} = a^n a^m$ for integers $n$ and $m$
2) $a^{-1} = 1/a$
You need to start from somewhere though. So you need to define something.

 

[brunopinto90]

Well the best way to start those proves is with the axioms or definions. a^-1 is just the multiplicative inverse. a^n+m i will have to come up with a proof.

 

[micromass]

So how do you define $a^0$?

 

[brunopinto90]

The definion of a^n is a.a.a.a - n times. So a^1 = a, a^2 = a.a and so on. Is it possible to define a^0 just using the definitions?

 

[micromass]

Most people define it as $a^0 =1$. I'm sure other definitions are possible though.

 

[HallsofIvy]

The definion of a^n is a.a.a.a - n times. So a^1 = a, a^2 = a.a and so on. Is it possible to define a^0 just using the definitions?

I'm not sure what you mean by "possible to define" something "just using the definitions"! What else would you use?

What I would say is this- by using "a^n is defined as "a multiplied by itself n times" we can show that (a^n)(a^m)= a^{n+m}.

Now, if those properties hold for 0 as well, we would have to have a^n= a^{n+ 0}= (a^n)(a^0) so that in order to have that same property, that (a^n)(a^0)= a^{n+ 0} we must define a^0= 1.

That is, in order to keep that property true, we have to define a^0= 1.

 

[mathman]

I'm not sure what you mean by "possible to define" something "just using the definitions"! What else would you use?

What I would say is this- by using "a^n is defined as "a multiplied by itself n times" we can show that (a^n)(a^m)= a^{n+m}.

Now, if those properties hold for 0 as well, we would have to have a^n= a^{n+ 0}= (a^n)(a^0) so that in order to have that same property, that (a^n)(a^0)= a^{n+ 0} we must define a^0= 1.

That is, in order to keep that property true, we have to define a^0= 1.

It does look like a proof that a^0=1. You also must state that this holds only for a\ne 0.

 

[Mark44]

I was watching some lectures on complex analysis and in a proof, the lecturer defines 0 = 1-1. Then i was watching a movie and like an epiphanie, i came up with a proof.

a^0 = a^1-1 = a^1 x a^-1 = a/a = 1.

You need parentheses in the second expression above. a^1-1 means $a^1 - 1$, and not as I'm sure you meant, $a^{1 - 1}$.

If you're not using LaTeX, write this as a^(1 - 1).

Is it good?

 

[HallsofIvy]

It does look like a proof that a^0=1. You also must state that this holds only for a\ne 0.

No, it was an argument why defining a^0= 1 is consistent with other definitions of a to a power. Yes, I should have stated that a could not be 0.

 

[aikismos]

I'd like to be picky and say, I don't think you can say it is defining $a^0$ when it's the result of a proof and an algebraic system. Generally in mathematics, something is a definition by virtue of the fact that it names something that doesn't otherwise have a name. We accept definitions as true assertions because they are essentially substitution or relabeling. The fact that $a^1 \cdot a^{-1} \rightarrow a^0$ according to the theorem $a^m \cdot a^n \rightarrow a^{m+n}$ means that the actual defining comes at the moment one articulates the construction $a^m := ∏a := a \cdot a \cdot ... \cdot a$. Since $a^1 \cdot a^{-1} = a^0$, the only time it would be a definition if it were to be declared equal to something other than $1$. Thoughts from the think tank?

 

[jbriggs444]

I'd like to be picky and say, I don't think you can say it is defining $a^0$ when it's the result of a proof and an algebraic system.

The proof in question establishes that if $a^0$ is meaningful and has a definite value and if the law of exponents is to hold for zero exponents then the value of $a^0$ must be 1 [for all non-zero a]. That stops well short of defining $a^0$. It does not preclude leaving $a^0$ undefined. It also does not preclude defining a^0 as, for instance, $\frac{\pi^2}{a}$. Such a definition would be silly. But being silly does not make a definition invalid.

How were you planning to prove that the law of exponents holds for zero exponents without having a definition for $a^0$?

[...] construction $a^m := ∏a := a \cdot a \cdot ... \cdot a$.

That construction does not establish a value for $a^{-1}$. It does not even establish the value of $a^{m}$ for positive integer m. It suffers from the fact that m is not actually used in the construction.

 

[mathwonk]

here are some old class notes
Positive Integer exponents

We want to define 2^x for all real numbers as a continuous function. We start by saying that 2^1 = 2, and 2^n = a product of n factors of 2, for any positive integer n. I.e. 2^2 = 2(2), 2^3 = 2(2)(2), 2^4 = 2(2)(2)(2), and so on.
Negative integer exponents

But what next? How do we define negative powers of 2? or 2^0? Notice a very important property of the exponential function, it satisfies 2^(n+m) = 2^n2^m for all positive integers n,m. I.e. to get a product of n+m factors of 2, just multiply a product with n factors by a product with m factors. Altogether, there will be n+m factors of 2. This is such a useful property that we would like it to continue to hold for all values of the exponential function. But that limits how we can define the exponential function very much. I.e. if we want to have 2^m = 2^(0+m) = 2^02^m, then we must have 2^0 = 1. And then if we want to have 1 = 2^0 = 2^(n+(-n)) = 2^n2^-n, then we must have 2^-n = 1/2^n. Thus we have no choice about how to define negative and zero powers of 2.Fractional exponents

What about rational powers? If we want to have 2 = 2^1 = 2^(1/2 + 1/2) = (2^1/2)(2^1/2), then (2^1/2) must be a number which gives 2 when multiplied by itself, i.e. we must have (2^1/2) = sqrt(2). Similarly, we must have 2^1/3 = cuberoot(2), and 2^1/n = nth root(2). For 2^m/n we must have 2^m/n =

2^(1/n + 1/n +...+1/n) (m terms) =

(2^1/n)(2^1/n)(...)(2^1/n) (m factors) =

[nth root(2)]^m = nth root(2^m).

As before than we must have 2^-(n/m) = 1/(2^n/m) =

1/[nth root(2^m)]. Thus we are forced to make this definition of a rational power of 2, just by the definition for positive integer powers, plus the basic law 2^(x+y) = 2^x 2^y. This completely determines the exponential function on all rational numbers.

 

[aikismos]

@jbriggs444,

You raise good points. First, I think the question of whether or not $a^0$ is meaningful comes and whether the Law of Exponents holds meaningful is a virtue of it's use WITHIN the algebraic and number systems being applied. So the reasoning goes like this, and without a loss of generality, I'll apply reasoning to integers:

1. The naturals are defined and are closed under addition. (Peano's Axioms or some modern extension)
2. We define the additive identity.
   
3. We define the additive inverse which declares it's existence. (Note at this point we've satisfied the conditions of a group as well as established with a little work the closure of the integers under addition.)
4. We define multiplication as repeated addition.
5. We define the unit fraction as the multiplicative inverse. This way $a^{-1}$ is established before considering the truth or falsity of $a^0 = 1$
   
6. We define exponentiation as repeated multiplication.
7. We define $a^{-m} := (a^m)^{-1}$.
   
8. NOW, given those axioms and definitions, it becomes impossible for $a^0 \neq 1$, and this because the Law of Exponents is really a theorem that can be proved as logically consistent within the formal system given the axioms and definitions above, e.g., Reflexivity of Addition, Closure of the Integers under the operation of addition.
9. The quintessence of the argument is $m + n \rightarrow z \forall m, n, z \in Z \Longrightarrow a^m \cdot a^{-m} \rightarrow a^0$ and $\frac{a^m}{a^m} = 1 \rightarrow a^0 = 1$. In fact, the burden would be on anyone who accepts the assertion $a^0 \neq 1$ to show how they arrived at the conclusion without contradiction the reasoning above.

 

[jbriggs444]

:

1. The naturals are defined and are closed under addition. (Peano's Axioms or some modern extension)
2. We define the additive identity.
   
3. We define the additive inverse which declares it's existence. (Note at this point we've satisfied the conditions of a group as well as established with a little work the closure of the integers under addition.)
4. We define multiplication as repeated addition.
5. We define the unit fraction as the multiplicative inverse. This way $a^{-1}$ is established before considering the truth or falsity of $a^0 = 1$
   
6. We define exponentiation as repeated multiplication.
7. We define $a^{-m} := (a^m)^{-1}$.
   
8. NOW, given those axioms and definitions, it becomes impossible for $a^0 \neq 1$, and this because the Law of Exponents is really a theorem that can be proved as logically consistent within the formal system given the axioms and definitions above, e.g., Reflexivity of Addition, Closure of the Integers under the operation of addition.
9. The quintessence of the argument is $m + n \rightarrow z \forall m, n, z \in Z \Longrightarrow a^m \cdot a^{-m} \rightarrow a^0$ and $\frac{a^m}{a^m} = 1 \rightarrow a^0 = 1$. In fact, the burden would be on anyone who accepts the assertion $a^0 \neq 1$ to show how they arrived at the conclusion without contradiction the reasoning above.

At step 4 you have an interesting choice. How will you define multiplication as repeated addition? Will you use a recursive definition? Will you ground the recursion with $a \cdot 0 = 0$? Or will you ground it with $a \cdot 1 = a$?

At step 5 you will presumably have defined the rationals so that multiplicative inverses are [almost always] guaranteed to exist.

At step 6 you again have an interesting choice. How will you define exponentiation as repeated multiplication? Will you use a recursive definition? Will you ground the recursion with $a^0 = 1$? Or will you ground it with $a^1 = a$? Let us assume that you have chosen the latter course.

The crux is step 8. With the machinery in place up to this point you can prove the law of exponents for strictly positive exponents. And you can prove it for strictly negative exponents. But you have no proof that applies to $a^0$ because you still have no definition and no assigned value for $a^0$.

 

[aikismos]

At step 4 you have an interesting choice. How will you define multiplication as repeated addition? Will you use a recursive definition? Will you ground the recursion with $a \cdot 0 = 0$? Or will you ground it with $a \cdot 1 = a$?

At step 5 you will presumably have defined the rationals so that multiplicative inverses are [almost always] guaranteed to exist.

At step 6 you again have an interesting choice. How will you define exponentiation as repeated multiplication? Will you use a recursive definition? Will you ground the recursion with $a^0 = 1$? Or will you ground it with $a^1 = a$? Let us assume that you have chosen the latter course.

The crux is step 8. With the machinery in place up to this point you can prove the law of exponents for strictly positive exponents. And you can prove it for strictly negative exponents. But you have no proof that applies to $a^0$ because you still have no definition and no assigned value for $a^0$.

Thanks. The intricacies aren't apparent to me at this moment, but I see what you are getting at. I'm going to write some of this out, and then I'll probably follow up. Thanks for the help in thinking!

 

[aikismos]

Okay,

I've revisited the construction of arithmetic under Peano's Axioms, and I now see what you meant by recursive instead of iterative definitions, and the insistence that $a^0 = 1$ is a definition instead of a theorem. If you define exponentiation recursively you are required to have a base case, and that base case will be the definition. So:

$F_{a^S(b)}(a, b) := \{a^0 = 1; a^S(b) = a \cdot (a^b)\} ∀a,b \in ℕ := (0, 1, 2, 3, ...)$

But for any $a^0 \neq 1 \rightarrow a^1 \neq a$ seems to rankle me; maybe it's a linguistic consideration that the definition of definition entails some choice which doesn't lead to an inconsistency.

 

[alva]

\lim_{x\rightarrow 0} {2^x}=1
from the right 2^1=2 2^0.5=1.4 2^0.1=1.07 2^0.01=1.007
from the left 2^-1=0.5 2^-0.01=0.993

BUT the function is not defined in x=0

*********************************************************
${f(x)}={\frac{(x+2)(x-2){x+2}$

$g(x)=(x-2)}$

f(x) and g(x) are not the same function because f(x) is not defined at the point x=-2

If we could say that f(x)=g(x) because the limits are the same, then a^0=1, not by definition but as a proven theorem

I mean f(x)=[(x+2)(x-2)]/(x+2)

 

[jbriggs444]

\lim_{x\rightarrow 0} {2^x}=1
from the right 2^1=2 2^0.5=1.4 2^0.1=1.07 2^0.01=1.007
from the left 2^-1=0.5 2^-0.01=0.993

BUT the function is not defined in x=0

$2^0$ is defined as 1. It is not undefined. There is no need to go fishing for arguments based on limits and continuity in order to decide on a reasonable value when the value is already defined as 1.

The place to fish for arguments is in considering why the sum of an empty list should be the additive identity and why the product of an empty list should be the multiplicative identity. The idea is that 2^1 is the product of an empty list of 2's.

Limits would be useful in coming up with a value for $2^\pi$. Not because $2^\pi$ must be equal to $\lim_{x\rightarrow\pi} {2^x}$ but because that's the choice that makes exponentiation in the reals a continuous extension of exponentiation in the rationals. Defining it otherwise would be silly.

 

[alva]

From https://en.wikipedia.org/wiki/Sinc_function:
"In mathematics, physics and engineering, the cardinal sine function or sinc function, denoted by sinc(x), has two slightly different definitions.[1]
In mathematics, the historical unnormalized sinc function is defined by

......
In either case, the value at x = 0 is defined to be the limiting value sinc(0) = 1.
..........
In both cases, the value of the function at the removable singularity at zero is understood to be the limit value 1."

 

[jbriggs444]

The function f(x) = 2^x does not have a removable singularity at zero. Just like sinc(x) does not have a removable singularity at one.

 

[Redbelly98]

I just skimmed through this thread, and didn't see anybody addressing the case a=0. Is 0⁰ even defined? Using limits, we can come up with conflicting values:

lim_a→0(a⁰)=1
lim_x→0(0^x)=0

 

[HallsofIvy]

$2^0$ is defined as 1. It is not undefined. There is no need to go fishing for arguments based on limits and continuity in order to decide on a reasonable value when the value is already defined as 1.

I agree that this is a definition but one can argue that this is a reasonable definition.

The place to fish for arguments is in considering why the sum of an empty list should be the additive identity and why the product of an empty list should be the multiplicative identity. The idea is that 2^1 is the product of an empty list of 2's.

You mean 2^0.

Limits would be useful in coming up with a value for $2^\pi$. Not because $2^\pi$ must be equal to $\lim_{x\rightarrow\pi} {2^x}$ but because that's the choice that makes exponentiation in the reals a continuous extension of exponentiation in the rationals. Defining it otherwise would be silly.

Yes, and the same thing is true of 2^0= 1.

 

[HallsofIvy]

I just skimmed through this thread, and didn't see anybody addressing the case a=0. Is 0⁰ even defined? Using limits, we can come up with conflicting values:

lim_a→0(a⁰)=1
lim_x→0(0^x)=0

In general, a^x itself is only defined for a> 0.

 

[micromass]

I am perfectly ok with defining $0^0 = 1$. It makes a lot of theorems nicer (such as the binomial theorem and Taylor's theorem). And it also agrees with the set theoretic definition of numbers and operations.

 

[alva]

The function f(x) = 2^x does not have a removable singularity at zero. Just like sinc(x) does not have a removable singularity at one.

?

In general, a^x itself is only defined for a> 0.

?
Are you kidding me?
I am the student and you are the teachers.

I just skimmed through this thread, and didn't see anybody addressing the case a=0. Is 00 even defined? Using limits, we can come up with conflicting values:

lima→0(a0)=1
limx→0(0x)=0

This will be explained in the next course.

 

[Mark44]

The function f(x) = 2^x does not have a removable singularity at zero. Just like sinc(x) does not have a removable singularity at one.

?

In general, a^x itself is only defined for a> 0.

?

Are you kidding me?
I am the student and you are the teachers.

Presumably you are questioning jbriggs's statement that sinc(x) does not have a removable singularity. Since sinc(x) is defined as $\frac{\sin(x)}{x}$, it is clearly undefined at x = 0, but we know that $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$. This represents a "hole" or removable singularity at (0, 1).

I just skimmed through this thread, and didn't see anybody addressing the case a=0. Is 0⁰ even defined? Using limits, we can come up with conflicting values:

lim_a→0(a⁰)=1
lim_x→0(0^x)=0

Fixed the subscripts/superscripts in what you (alva) quoted from Redbelly98.

The above limits are why $0^0$ is considered indeterminate, as well as why HallsOfIvy made his comment about a^x being defined only for a > 0.