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

Theorem

Let $x \in \R$.

Let $0 < x < 1$.

Let $n$ be a natural number.

Then:

$0 < x^n \le x$

Proof 1

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.

$\blacksquare$

Proof 2

Real Number between Zero and One is Greater than Power/Natural Number/Proof 2