Power is Well-Defined/Integer

Theorem

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

Let $k$ be an integer.

Then $x^k$ is well-defined.

Proof

Fix $x \in \R \setminus \set 0$.

Positive Integers

We first prove the theorem for positive integers by induction:

For all $n \in \N$, let $\map P n$ be the proposition:

$x^n$ is well-defined

Basis for the Induction

$\map P 1$ is true, since $x^1 = 1$ by definition.

This is our basis for the induction.

Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

$x^k$ is well-defined

Then we need to show:

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

Induction Step

This is our induction step:

By definition of integer power:

$x^{k + 1} = x \cdot x^k$

From Real Multiplication is Well-Defined and the Induction Hypothesis:

$x \cdot x^k$ is well-defined

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

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

Therefore:

$\ds \forall n \in \N: x^n$ is well-defined

$\Box$

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 Reciprocal of Real Number is Unique, $x^k$ is well-defined.

Finally:

$\forall x \in \R \setminus \set 0: x^0 = 1$

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

Hence the result.

$\blacksquare$