Proof using axioms of real numbers Using axioms of real numbers prove for
∀x∈R, ∀y∈R,∀n∈N: 



($0 \leq x &lt; y$ ) $\Rightarrow$ $x^n &lt; y^n$

Any help with this? I have no idea how to even begin.
By induction.

Base case $n=1$,

$$0\le x &lt; y\implies x^1&lt;y^1.$$

Induction

$$0\le x &lt; y\implies x^n&lt;y^n,$$

$$0\le x &lt; y\land x^n&lt;y^n\implies 0\le x\,x^{n}&lt;y\,x^{n}&lt;y\,y^{n}.$$
$$ 0 \leq x &lt; y \implies ln(x) &lt; ln(y) \implies k \cdot ln(x) &lt; k \cdot ln(y) \implies x^k &lt; y^k $$

Step 1 : The logarithm is a crecient function in $R^+$.

Step 2 : k is positive.

Step 3 : The exponential is a crecient function in $R$.