If $x>y>0$ why is $\sqrt{x} > \sqrt {y}$?

Question

If $x>y>0$ why is $\sqrt{x} > \sqrt {y}$?

I know this to be true, but I don't understand why because can't the square root of $x$ be negative and the square root of $y$ be positive thus disproving it?

Answer 1

Because $x-y>0$, $\sqrt{x}+\sqrt{y}>0$ and $$x-y=(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}).$$

For $a\geq0$ by definition $\sqrt{a}$ it's non-negative number $b$ for which $b^2=a$.

All this says that $\sqrt{a}\geq0$ by definition.

When we want to define $\sqrt{a}$ we need that $\sqrt{.}$ would be function and that by our definition we don't get a contradiction with laws of math.

If you wish to define $\sqrt{a}$ like non-positive $b$ for which $b^2=a$ then try to check the following law: $$\sqrt{\sqrt{a}}=\sqrt[4]{a}.$$

We see that by new definition laws of math sometimes work and sometimes not work.

This prevents us from exploring our world.

Thus we get the correct definition of $\sqrt{.}$ and from here we can not change anything.