# Proof for $\frac{x^n}{x^n}=x^{n-n}$

Does the equality $\forall n,x\not=0: \frac{x^n}{x^n}=x^{n-n}$ come straight from the definition of exponents, or is a more elaborate proof needed?

Please note that I'm not asking about $x^0=1$, but only about the $\frac{x^n}{x^n}=x^{n-n}$ part of the equation.

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Depends on your definition of powers. – Qiaochu Yuan Dec 20 '12 at 20:12
$x^0=1 \ \forall x$ is a convention but it has combinatorical justification i.e. the number of combinations drawing 0 elements with $x$ colors is $1$. – Amihai Zivan Dec 20 '12 at 20:12
You could prove it inductively, otherwise it just follows from exponent rules – Math_Illiterate Dec 20 '12 at 20:20
It also follows from the fact that $\frac{y}{y} = 1$ whenever $y \ne 0$. – Joel Cohen Dec 20 '12 at 20:33
Well, write $n$ of them at both numerator and the denominator and simplify. – ashley Dec 20 '12 at 20:40

$$\large \dfrac{x^n}{x^n}= x^n \cdot \frac{1}{x^n} = x^n \cdot x^{-n} = x^{(n\ + \ - n)} = x^{(n\ - \ n)} = (x^0 = 1).$$

Note: just as $\dfrac{1}{a}$ is the multiplicative inverse of $a$, (and can be represented by $a^{-1}$). The multiplicative inverse of $a$ is the number you need to multiply by to arrive at $1$:
That is, we need $a^{-1}$ to be such that $a\cdot a^{-1} = a^{-1} \cdot a = 1$ (since $1$ is the multiplicative identity for the real numbers: any number multiplied by $1$ remains unchanged). This works for all non-zero $a$ provided $a^{-1} = \dfrac1a.$ Then we have that $a \cdot \dfrac1a = 1$ $$\large \dfrac{1}{x^n} = \underbrace{\frac1x\cdot \frac1x \cdots \frac1x}_{n-times} = \underbrace{x^{-1}\cdot x^{-1} \cdot \cdots x^{-1}}_{n-times} = x^{\overbrace{-1 + -1 + \cdots + -1}^{n - times}} = x^{-n}$$

So I'd say you can prove this simply by using the exponent rules.

You might also want to do so using induction, but might be incorporating the use of rules of exponentiation, implicitly. It's sort of a "which came first, the chicken or the egg" question as to which definition is most primitive.

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I think that for this question it is also good to say why $1/x^n=x^{-n}$ (+1) – Belgi Dec 20 '12 at 20:39
@Belgi: I edited about an hour ago to expand on your suggestion. Is this what you had in mind? – amWhy Dec 20 '12 at 22:57
Thanks for the edit, but now I feel you need to explain what I wrote in my first comment for $n=1$. What I had in mind that you show that one is the inverse of the other by simple showing that when you multiply them you get $1$, which is also shorter to write :) – Belgi Dec 20 '12 at 23:28
@Belgi Hopefully, done! :-) – amWhy Dec 20 '12 at 23:48
Looks good to me :) – Belgi Dec 20 '12 at 23:49

Base Case:

Let n = 0

$\frac{x^{0}}{ x^{0}} = \frac{1}{1} = 1$

$x^{n-n} = x^{0 – 0} = x^{0} = 1$

So the base case checks out.

Inductive Hypothesis:

Want to show that $\frac{x^{n+1}}{x^{n+1}} = x^{(n+1) – (n+1)}$

$\frac{x^{n+1}}{x^{n+1}} = \frac{x^{n} \cdot x}{x^{n} \cdot x} = x^{n-n} \cdot x^{1-1} = x^{(n+1) – (n+1) }$

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Assuming $x \neq 0$,

$$\frac{x^n}{x^n} = 1 = x^0 = x^{n-n}.$$

The only exponent rule used in the above is $x^0 = 1$. Note, this is a special case of the exponent rule

$$\frac{x^a}{x^b} = x^{a-b}.$$

So, if you already knew that rule, then it follows immediately by letting $a = b = n$.

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