# Real Number between Zero and One is Greater than Power/Natural Number/Proof 1

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## Theorem

Let $x \in \R$.

Let $0 < x < 1$.

Let $n$ be a natural number.

Then:

- $0 < x^n \le x$

## Proof

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

- $0 < x < 1 \implies 0 < x^n \le x$

### Basis for the Induction

$\map P 1$ is true, since $0 < x < 1 \implies 0 < x^1 \le x$ by definition of exponent of $1$.

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:

- $0 < x < 1 \implies 0 < x^k \le x$

Then we need to show:

- $0 < x < 1 \implies 0 < x^{k + 1} \le x$

### Induction Step

This is our induction step:

\(\ds 0 < x < 1\) | \(\leadsto\) | \(\ds 0 < x^k \le x\) | Induction Hypothesis | |||||||||||

\(\ds \) | \(\leadsto\) | \(\ds 0 < x^{k + 1} \le x \cdot x\) | Real Number Ordering is Compatible with Multiplication | |||||||||||

\(\ds \) | \(\leadsto\) | \(\ds 0 < x^{k + 1} \le x\) | Multiple of Positive Real Number with Number Less Than One is Less Than Real Number |

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

Therefore:

- $\forall n \in \N: 0 < x < 1 \implies 0 < x^n \le x$

Hence the result.

$\blacksquare$