How to prove that: let $n$ a natural odd, then prove that $x^n<y^n$ iff $x<y$ I tried it in many ways but I couldn't prove it.
  TASK:
 Let $n$ a natural odd, $\ \ x,y\in\mathbb{R},$ then prove that $x^n&lt;y^n$ iff $x&lt;y$

My Attempt(s): $$\begin{align} &amp; x^n&lt;y^n \\ \Longleftrightarrow{} &amp;  x^n-y^n&lt;0 \\ \Longleftrightarrow{} &amp; (x-y)\sum_{k=0}^{n-1}x^{n-k-1}y^k&lt;0 \end{align}$$ And I'm stuck here.

Thus, I can't solve it with induction either. Any help would be appreciated :)
It is better to prove that $f(x) = x^{n}$ (where $n$ is odd positive integer) is strictly increasing on both intervals $(-\infty, 0]$ and $[0, \infty)$ and this will imply that it is strictly increasing on whole of $\mathbb{R}$. The fact that $f(x)$ is strictly increasing on $[0, \infty)$ is obvious for the very reason that if $x, y \in [0, \infty)$ and $x &lt; y$ then $y$ is strictly positive and $x$ is non-negative so that $$f(x) - f(y) = (x - y)\sum_{k = 0}^{n - 1}x^{n - 1 - k}y^{k} &lt; 0$$ Let's check what happens on interval $(-\infty, 0]$. If $x, y \in (-\infty, 0]$ and $x &lt; y$ then $x$ is strictly negative whereas $y$ is non-positive. Again we have $$f(x) - f(y) = (x - y)\sum_{k = 0}^{n - 1}x^{n - 1 - k}y^{k}$$ Note that the sum $\sum x^{n - 1 - k}y^{k}$ has terms where powers of $x, y$ have same parity (because their powers add up to $(n - 1)$ which is even). So if both the powers are even then the term is obviously non-negative. And if both the powers are odd then the term is product of two non-positive numbers and hence it is again non-negative. Thus each term in the sum $\sum x^{n - 1 - k}y^{k}$ is non-negative. Moreover first term $x^{n - 1}$ is positive because $x$ is negative and $n - 1$ is even. So the whole sum is positive and therefore $$f(x) - f(y) = (x - y)\sum_{k = 0}^{n - 1}x^{n - 1 - k}y^{k} &lt; 0$$
What about considering function f(x) = x^n which is increasing function.

It means that for x &lt; y -> f(x) &lt; f(y).

So x^n &lt; y^n