If $0<x<1$ and $n$ is a positive integer, then $x^n < 1$

Question

I'm curious if induction is only/standard way to prove that

If $0<x<1$ and $n$ is a positive integer, then $x^n < 1$

Base Case:

If $n=1$ and $0<x<1$, then it is certainly true that $x<1$.

Induction Hypothesis:

Assume that $x^n < 1$ for some positive integer $n$ where $0<x<1$.

Induction Step:

We need to show that $x^{n+1} < 1$ for some positive integer $n$ and $0<x<1$. This follows immediately from the fact that $x^{n+1}=x^nx^1=x^nx<1$ since $x^n < 1$ and $0<x<1$ so the product must be less than one.

Is this the standard way of proving this result? I have seen this used frequently without proof in many math books and problems.

Answer 1

We have $$ x^n-1=(x-1)(1+x+x^2+\cdots +x^{n-1})<0 $$ since $x-1<0$ and $1+x+\cdots +x^{n-1}>0$ for $0<x<1$.

Answer 2

Your proof is slightly inaccurate.

$$x<1\land x^n<1\implies x^{n+1}<1$$ may be wrong for negative $x$. You should write

$$0\le x<1\land x^n<1\implies x^{n+1}<1.$$

The condition $0\le x$ is crucial to the theorem.

Answer 3

Let $0<x<1$:

Geometric series:

$1+x+x^2+.....x^{n-1}=\dfrac{x^n-1}{x-1}$;

LHS is positive:

Since $x-1<0$ , we conclude $x^n-1<0$, and we are done.