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I'm looking for general rules for working with inequalities and powers.

Are those general rules correct?

1.You may raise any inequality to an odd power of N - and keep the inequality as it was.

For example: N is an odd number.

$$a>b$$

$$a^n > b^n $$

  1. You may raise to an even power of N - only if:

$$|a| < |b|$$ Then, $$a < b$$

$$a^n < b^n$$

Thanks, and sorry for the bad formatting.

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    $\begingroup$ You are correct. $\endgroup$
    – cr001
    Commented Oct 30, 2015 at 11:39

1 Answer 1

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Suppose $N\in\mathbb{N}$ odd. We can rewrite $N=2n+1$ for some $n\in\mathbb{N}$. Since any polynomial function $\mathbb{R}\to\mathbb{R}:x\mapsto x^{2n+1}$ is strictly increasing (the derivative is $(2n+1)x^{2n}>0\,\,\forall x\neq 0$ and equals $0$ only when $x=0$), if $a<b$, then $a^{2n+1}<b^{2n+1}$ for all $a,b\in\mathbb{R}$.

Suppose $N\in\mathbb{N}$ even. We can rewrite $N=2n$ for some $n\in\mathbb{N}$. If $N=n=0$, then $a^{n}<b^{n}$ is false for any $a,b\neq 0$ (since $x^{0}=1\,\forall x\in\mathbb{R}_{0}$) and not defined at $a=0$ or $b=0$. Suppose then $N\neq 0$. As $\mathbb{R}\to\mathbb{R}:x\mapsto x^{2n}$ is symmetric along the axis $x=0$ and strictly increasing on $\mathbb{R}^{+}$, so that $a^{2n}<b^{2n}$ if $|a|<|b|$.

Your rules are correct.

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  • $\begingroup$ Very interesting, I'll learn this proof as I'm a beginner. Thank you. $\endgroup$
    – Taru
    Commented Oct 30, 2015 at 11:54
  • $\begingroup$ You're welcome. There is a little bit to explain for the last argument (with symmetry,...). $\endgroup$ Commented Oct 30, 2015 at 12:19

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