Let $x,y$ be positive real numbers. Then $x<y$ iff $x^n<y^n$ for all $n\in \mathbb{N}$

Question

I tried proving it but I'm quite skeptical about it:

$(\Rightarrow )$ Assume $x<y$. Clearly, the base case for the induction is done. Assume $x^k<y^k$ for some $k\in \mathbb{N}$. Now, $x^{k+1}=x^k\cdot x<y^k\cdot y=y^{k+1}$.

$(\Leftarrow )$ Assume for contradiction that $x\not< y$. Then $y\le x$. Now, suppose $y\ne x$. Then $y<x$. It follows that $y^n<x^n$ for every $n\in \mathbb{N}$ from $(\Rightarrow )$. This contradicts our hypothesis. Also, if $y\not< x$ then $x=y$. Then $y^n=x^n$ for every $n\in \mathbb{N}$ which also contradicts our hypothesis.

Is this proof correct?

Answer 1

For an approach without induction note that $$ y^n - x^n = (y-x) \left(y^{n-1}+y^{n-2}x+\dots+yx^{n-2}+x^{n-1} \right).$$

Answer 2

Much more is true.

If $x > 0, y>0$ then $x < y \iff x^n < y^n $ for any $n \in \mathbb{N}$.

The proof both ways depends on $x^n-y^n =(x-y)\sum_{k=0}^{n-1} x^ky^{n-1-k} $ and $x>0, y>0 \implies \sum_{k=0}^{n-1} x^ky^{n-1-k} > 0 $.

Note that this implies that $x-y$ and $x^n-y^n$ have the same sign (+, -, or $0$).