I have already proved $ x < y => x^n < y^n $
I am currently trying to prove the converse statement, but I am getting nowhere. The textbook suggests trying to set up a contradiction, but I can't seem to succeed with this.
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What course are you taking? You could take the $n$th root of both sides, but this would probably not be allowed. – Toby Mak Oct 12 '20 at 09:23
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Proofs. I don't think that would be allowed either. – Drich1002 Oct 12 '20 at 09:24
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4Does this answer your question? If $x^n=y^n$ and $n$ is odd then $x=y$ The correct statement should be for $n$ odd unless $x,y > 0$. – Toby Mak Oct 12 '20 at 09:39
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1You can use Approach0 to check if there are any duplicates before you ask. – Toby Mak Oct 12 '20 at 09:41
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1This is only true, if we restrict to non-negative numbers. – Peter Oct 12 '20 at 09:46
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It is trivial that $x=y\implies x^n=y^n$ and you already know that $x>y\implies x^n>y^n$. In other words, $x\geqslant y\implies x^n\geqslant y^n$. And this is equivalent to $x^n<y^n\implies x<y$.
José Carlos Santos
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If you wish to prove this by way of contradiction, do the following:
Let's start with $x^n < y^n$, and if possible, assume $x \geq y$. Now, raising the latter inequality to the $n^{th}$ power gives $x^n \geq y^n$, which contradicts $x^n < y^n$. Our assumption $x \geq y$ must be incorrect.
So, $x < y$.
One thing that really bothers me is the constraints on $x,y,$ and $n$ - this relation won't hold for all values of $(x,y,n)$, and I encourage you to figure out necessary constraints. A contradictory example is as follows: $$-2 < -1 \nRightarrow (-2)^2 < (-1)^2$$
stoic-santiago
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