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How can I prove rigourously that if for any real numbers $x$ and $y$ (both are positive)

If $x \ge y$ then

This implies $x^n \ge y^n$ for any real $n$ as well. (n is not negative) This is used everywhere for example when n is half or two.

Is there a general result if $n$ is negative?

Matt
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3 Answers3

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First of all, this inequality is only true for $n>0$. If $n$ is negative then exponentiation is order-reversing.

Second, an outline of the proof goes something like this:

  1. First, prove the result for whole number values of $n$. This can be done by induction, making use of the fact that multiplication by a positive number is order-preserving.
  2. Next, use the result above to prove the inequality for $n$ of the form $1/m$, with $m$ a whole number.
  3. Combine the two results to prove it is true for all positive rational exponents.
  4. Use continuity to extend the definition to all positive real exponents.
mweiss
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  • Half greater than quarter and hence quarter greater than 1 by 16. How do you account for this ? – Matt May 07 '17 at 14:52
  • @RaghavSingal Quite true. The result does not require $x,y>1$, I'm not sure what I was thinking. But it does require $n>0$. – mweiss May 07 '17 at 14:58
  • How can we use continuity to prove for irrational as well can you elaborate ? – Matt May 07 '17 at 16:30
  • Depends on how you define $x^r$ for irrational $r$. For example if $r$ is positive than one way to do it is as the supremum of ${x^q | q \in \mathbb{Q}, 0 < q \le r$. If that's your definition, then the inequality you want follows from the inequality for rational exponents. – mweiss May 07 '17 at 21:44
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What if $y<x<-1 \cup n\mod 2 \equiv 0$? In that case, $x^n<y^n$, disproving the function.

Eric Lee
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Taking advantage of the strict monotonicity of $f(x)=\ln{x}$ then for any $n>0$ and for any $x,y>0$ we have WLOG $$ x\ge y \iff \ln{x}\ge\ln{y} \iff n\ln{x}\ge n\ln{y} \iff \ln{x^n}\ge\ln{y^n} \iff x^n \ge y^n. $$

Laars Helenius
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