Given $n \in N$, prove that $f: [0, \infty) \rightarrow [0, \infty) f(x) = x^n$ is increasing without calculus (derivates) I have to prove this using $x_1 &lt; x_2 \rightarrow f(x_1) &lt; f(x_2)$.
I have no idea on how to do this, but what I have tried so far is:
Consider $x_1, x_2  \in [0, \infty)$ such that $x_1 &lt; x_2 \rightarrow f(x_1) &lt; f(x_2)$
$\rightarrow x_1^n &lt; x_2^n$
I do not know what to do next
You have
$$x_2^n - x_1^n = (x_2-x_1) (x_2^{n-1} + x_2^{n-2} x_1 + \cdots + x_2 x_1^{n-2}+ x_1^{n-1}) \gt 0$$ for $x_2 \gt x_1$.
Say $x,y \in [0,\infty) $ and $x\lt y$. Then $$f(x) \lt f(y) \\ \iff x^n \lt y^n \\ $$
Since $\ln x$ is an increasing function, $$\iff n\ln x \lt n \ln y \\ \iff \ln x \lt \ln y \\ \iff x\lt y $$
But that’s of course true.