ProofWiki:Sandbox

Theorem
Let $x \in \R$.

Let $0 < x < 1$.

Let $n$ be a natural number.

Then:
 * $0 < x^n \leq x$

Proof
For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle 0 < x < 1 \implies 0 < x^n < x$

Basis for the Induction
$P(1)$ is true, since $0 < x < 1 \implies 0 < x^1 \leq x$ by definition of exponent of $1$.

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:


 * $\displaystyle 0 < x < 1 \implies 0 < x^k \leq x$

Then we need to show:


 * $\displaystyle 0 < x < 1 \implies 0 < x^k \leq x$

Induction Step
This is our induction step:

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

Therefore:
 * $\displaystyle \forall n \in \N: 0 < x < 1 \implies 0 < x^n < x$

Hence the result.