How to prove $x^0=1$ without using the fact $x^{m+n}=x^mx^n.$

Question

We all get it $$x^0=x^{1-1}=x^1\cdot x^{-1}=x/x =1$$ as long as $x\neq 0.$ But this supposes the rule $x^{m+n}=x^mx^n$ for $m$, $n\in \mathbf{Z}.$ Unfortunately, each proof of the rule $x^{m+n}=x^mx^n$ that I have seen uses the fact that $x^0=1$ in it so to use the fact that $x^{m+n}=x^mx^n$ in a proof that $x^0=1$ would be a circular argument. So, I suppose that I am asking for one of two things: (1) A proof that $x^0=1$ for non-zero x, using only the field axioms (and maybe definitions like $x^{-1}=1/x$), or (2) A proof that $x^{m+n}=x^mx^n$ for $m$, $n\in \mathbf{Z}$ that does not employ $x^0=1$ in its proof. The latter seems more unlikely. Any insight is appreciated.

Answer 1

Suppose we define $x^n$ in the obvious way for positive integer $n$, but abstain from defining $x^0$. We can prove by induction on $n$ that $x^m\cdot x^n=x^{m+n}$ for all values of $x$, $m$, and $n$. For $n=1$, this is the assertion that $x^{m+1}=x^m\cdot x$, which is true by definition. Assuming the statement is true for a given value of $n$, it follows that $$x^m\cdot x^{n+1}=x^m\cdot (x^n\cdot x)=(x^m\cdot x^n)\cdot x=x^{m+n}\cdot x=x^{m+n+1} \, .$$ Consider it as an exercise to justify why each of these equalities must hold.

Then, if we want to preserve the rule $x^m\cdot x^n=x^{m+n}$ for when $m$ and $n$ could be any integers, then at least in the case $x\neq0$, we must define $x^0$ as $1$, and $x^n$ as $1/x^{-n}$ for negative integers $n$. From these definitions, the proof of $x^{m+n}=x^m\cdot x^n$ is straightforward: just break it into different cases.

This reasoning is valid in any field, provided we understand the term "integer" to mean an element of the following set (which is possibly finite): $$ \{0,1,-1,1+1,-1-1,1+1+1,-1-1-1,\dots\} \, . $$

Answer 2

The main motivation for defining $x^0=1$ is because $0$ is the additive identity, $1$ is the multiplicative identity, and they're intimately tied together by the logarithm. Why? Well, logarithms have the property that $\ln(ab)=\ln(a)+\ln(b)$ which is how the arose initially. This means $\ln (a)=\ln(a1)=\ln(a)+\ln(1)$ and so if we extract $\ln(a)=\ln(a)+\ln(1)$ and subtract $\ln(a)$ from both sides we see that $\ln (1)= 0$. People used tables of logarithms to turn large multiplication problems into easier to computer addition problems. You can see the same idea here because it's essentially $x^ax^b=x^{a+b}$.

However we don't have to define the logarithm to be inverse of the exponential function. With a little calculus we can define the logarithm to be $\ln t = \int_1^t \frac{1}{x} \, dx$ and then prove it has the properties we're looking for like being the inverse of $e^x$ and the aforementioned $\ln(ab)=\ln(a)+\ln(b)$. If you don't know any calculus this equation basically represents the area between the graph and the $x$-axis of the function $f(x)=1/x$ between $1$ and $t$. However if $t=1$ then the starting and end point are both $1$, so the area has to be $0$. Proving things this way is a little challenging but it can be done without appealing the usual definitions using $x^ax^b=b^{a+b}$.

Answer 3

Maybe this is too much machinery for a simple thing but one could do:

1. Define $exp(x) = \lim_{N \rightarrow \infty} 1+ \sum^{N}_{n=1} \frac{x^n}{n!}$ by showing it is well defined for each $x$
2. Show it is monotically increasing and thus there is an inverse, which can be defined as $\log$
3. Define $a^x =e^{\log(a)x}, \forall a>0$
4. Show that one recovers the usual $a^x$ for $a,x \in \mathbb{N}$
5. Show that $a^{x+y}=a^xa^y,\forall x,y \in \mathbb{R}$

Now you get as a theorem, $a^0=1, \forall a>0$

Answer 4

[Most of this is in the comments, mine or those by other users, but I think it's worth an answer.]

You want a proof from the field axioms. There is no mention of exponentiation in the field axioms, just addition and multiplication, so at the very least you are going to have to accept a definition of exponentiation as a supplement to the field axioms. The standard way to define $x^n$ for $x$ in the field and $n$ a positive integer is inductive; define $x^1$ to be $x$, and $x^n$ for $n>1$ to be $x^{n-1}x$. Then, using induction again, $x^{m+n}=x^{m+n-1}x=x^mx^{n-1}x=x^mx^n$. So, from the field axioms, you're only one definition and two inductions away from $x^{m+n}=x^mx^n$. Now $x^m=x^{m+0}=x^mx^0$, so $x^0=1$.

Answer 5

You can not prove this. It is merely a convention, or a definition, not a statement that needs to be proven.

All of the "proofs" in other answers are circular and they use this convention in an implicit way in order to "prove" it.

Answer 6

The proof that $x^0=1$ lies in that $x^0 \times a = a\ \forall a$, and therefore $x^0$ is the identity of multiplication. But since it can be demonstrated that $1$ is the identity, you show that $\forall x\ x^0=1$.

You will hear stories about the limit of $0^x$ as $x -> 0$, but the two ways of moving left on this line, is by division (which can get you to 0, but invilves division by 0), or by taking roots (which does not involve division, but you can't make $x=0$, since it remains positive.

Allowing $0^0$ to be anything else but one simply means that you can write any statement, and it will be true. In essence, for integer exponents, such as $0$, $1$, etc, it is the number of times the base is involved in the product. If you suppose $3 = 3\times 0^0 = 3 \times 8 = 37$, simply because each line has 0 repetitions of 0, and this might be freely set, the lines don't necessarily have to agree arithmetically.

Answer 7

I thought $x^0$ must be a contant which is not zero, for all of $x$. Say it is $a\neq 0$. But, then $a^0=a$ too. Taking $x$-th power of both sides, since $(a^0)^x=a^0$, we get $a^x=a$,$\;\forall x$. But this is possible iff $a=1$.

I pushed down my joke below, because mathematics is not personal. Social media is. My joke was: I hope there is not a gap to be attacked by Gary in this thread. It is just for fans or fun: