Prove if $x ∈ \mathbb{R}$, such that $0 ≤ x ≤ 1$, and $m,n ∈\mathbb{ N}$, with $m ≥ n$. Then $x^m ≤ x^n$

Question

How to prove the following prop.

Let $x \in \mathbb{R}$, such that $0 \le x \le 1$, and $m,n \in\mathbb{ N}$, with $m \ge n$. Then $x^m \le x^n$.

I don't exactly know where to begin with this proof, any guidance is appreciated.

If $0 \le a \le b$, and $0 \le c$, then $ac \le bc$. – copper.hat Dec 12 '14 at 03:26 — copper.hat, Dec 12 '14 at 03:26

Kim Jong Un · Answer 1 · 2014-12-12T04:08:42.117

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Let $d=m-n$. If $d=0$, there is nothing to prove. If $d>0$, we have $$ x^m-x^n=x^n(x^d-1)=\underbrace{x^n}_{\ge 0}\underbrace{(x-1)}_{\le 0}\underbrace{(x^{d-1}+\cdots+x+1)}_{\ge 1}\leq 0. $$

edited Dec 12 '14 at 04:08

answered Dec 12 '14 at 04:02

Kim Jong Un

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score 1 · Accepted Answer · answered Dec 12 '14 at 03:27

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$0 \leq x^m = x^n\cdot x^{m-n} \leq x^n\cdot 1 = x^n$

answered Dec 12 '14 at 03:27

DeepSea

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I don't get it. – raza arshad Dec 12 '14 at 16:42
$x^{m-n} \leq 1$ since $0 \leq x \leq 1$, and $m-n \geq 0$. – DeepSea Dec 12 '14 at 17:07
how do we get to x^m > x^n though? – raza arshad Dec 12 '14 at 17:11
I just corrected the edit that was done earlier. Its $x^m \leq x^n$ – DeepSea Dec 12 '14 at 17:13
makes much more sense now, thanks. – raza arshad Dec 12 '14 at 17:14

score 0 · Answer 3 · answered Dec 12 '14 at 03:27

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HINT: If $0\leq x\leq 1$ then $xB\leq B$ for any positive $B$.

answered Dec 12 '14 at 03:27

Przemysław Scherwentke

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Prove if $x ∈ \mathbb{R}$, such that $0 ≤ x ≤ 1$, and $m,n ∈\mathbb{ N}$, with $m ≥ n$. Then $x^m ≤ x^n$

3 Answers3