ProofWiki:Sandbox

Theorem
Let $x$ be a non-zero  real number.

Let $k$ be an integer.

Then $x^k$ is well-defined.

That is:
 * $ \forall x \in \R \setminus \left \{ 0 \right \} \forall n, m \in \N : n = m \implies x^n = x^m$

Proof
Fix $x \in \R \setminus \left \{ 0 \right \}$.

Positive Integers
We first prove the theorem for positive integers by induction:

For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle x^n$ is well-defined

Basis for the Induction
$P(1)$ is true, since $x^1 = 1$ by definition.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:


 * $\displaystyle x^k$ is well-defined

Then we need to show:


 * $\displaystyle x^{k + 1}$ is well-defined

Induction Step
This is our induction step:

From definition of integer power:
 * $x^{k + 1} = x \cdot x^{k}$

From Real Multiplication is Well-Defined and Induction Hypothesis:
 * $x \cdot x^{k}$ is well-defined

So $x^{k + 1}$ is well-defined.

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall n \in \N: x^{n}$ is well-defined

Non-Positive Integers
Fix $k \in \Z_{< 0}$

First:
 * $k \in \Z_{< 0} \implies -k \in \N$

From the development above, $x^{-k}$ is well-defined.

From Powers of Group Elements/Negative Index, $\dfrac{1}{x^k}$ is well-defined.

From Multiplicative Inverse of Real Number is Unique, $x^k$ is well-defined.

Finally:
 * $\forall $x \in \R \setminus \left \{ 0 \right \} : x^0 = 1$

So $x^k$ is well-defined for $k = 0$.

Hence the result.