# Regarding $x < y \Rightarrow x^n < y^n$ proof rigor.

I came across the implication $$x < y \Rightarrow x^n < y^n$$ $$x,y>0, n\in Z^+$$ in a textbook and came up with the following proof.

Proof Since $$x the following chain of inequalities holds. $$x^n

My question now relates to the “solidity” of this proof. I get the feeling that it is a bit vague in its reasoning, yet I cannot see that it isn’t correct... I’m quite new to proving things and can’t always see the difference between intuitive reasoning and rigorous reasoning.

Is my reasoning solid? If not, what is it lacking?

• It looks good to me. Maybe you could mention $x<y\implies1<y/x$ to begin with, and then each step in your induction is a multiplication by $y/x$ on both sides of the old inequality. – String Jan 11 at 11:06
• In a word? Induction. It's perfect for lending some formality to arguments that come with "$\ldots$" somewhere in them. – Theo Bendit Jan 11 at 11:08
• Each of the inequities can be broken down to "If $a>0$ and $b<c$, then $ab<ac$." Now, I'm not sure how much rigor you're looking for. Do you want to prove the fact above via axioms for the real numbers? Or.... – Michael Burr Jan 11 at 11:08
• That seems fine to me. – Matt Samuel Jan 11 at 15:36
• You do want to multiply both sides by the quantity though. – Matt Samuel Jan 11 at 16:36

What's lacking is explicit induction. It's an "et cetera" argument. It's not wrong, but it's not $$100\%$$ rigorous. It's close enough that making it rigorous is easy though, if you know how to do induction.
Ask yourself the following question: Where in the proof do I (tacitly) use the hypothesis that $$x$$ and $$y$$ are positive? Then make that use explicit.
It might also help to rewrite the proof as a proof by induction, inducting on $$n$$. In other words, show that if $$0\lt x\lt y$$ and $$x^n\lt y^n$$, then $$x^{n+1}\lt y^{n+1}$$ (again, making explicit where the positivity hypothesis is used).