Question

Disprove, by finding a suitable counter example, the statement $x^2>y^2$ for all $x>y$.

Answer 1

For negative number The statement $x^2>y^2$ For all x > y is disprove

If $x=-3$ and $y=-4$ Then $-3>-4$

But in case $x^2=(-3{)}^2=9$ and $y^2=(-4{)}^2==16$

So $x^2<y^2$

Then the statement $x^2>y^2$ For all x > y is disprove