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

From ProofWiki
Jump to navigation Jump to search

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:

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