Proving $x^2 < y^2$ by means of the Ordering Axioms

Question

How do I prove $x^2 < y^2$, if $0 \le x < y$ with the ordering axioms? thanks!

Answer 1

Well, you are given $0 \leq x < y $. Hence we have

$$ x^2 < yx $$

Similarly, we obtain $$ xy < y^2 $$

Since $xy = yx \; \; \;(Proof?) $, then by transitivity

$$ x^2 < yx = xy < y^2 \implies x^2 < y^2 $$

In particular, $f(x) = x^2$ is increasing in the first quadrant of the $xy-plane$

Answer 2

First multiply by $x$ and then by $y$:

$$0\leq x<y\implies0\leq x^2<xy$$

$$0\leq x<y\implies 0\leq xy<y^2$$

so

$$0\leq x^2<y^2$$

Answer 3

From what is given, we have $y-x>0$. The sum of positives is also positive, hence and $y+x>0$. The product of positives is positive, hence $y^2-x^2=(y-x)(y+x)>0$.