Using proof by contradiction to show that $x

Question

Let $x,y$ be positive real numbers. Use a proof by contradiction to prove that $x < y \implies \sqrt x < \sqrt y$. I assumed that x < y and I also assumed that to the contrary that $\sqrt x\ge \sqrt y$. Then if you square both sides you end up getting $x \ge y$ which is a contradiction, but I'm not sure if what I did is valid.

Answer 1

Your proof is correct. If you want to be a bit pedantic you can write that:

since $\sqrt x>0$ and $\sqrt y>0$ than: $$ \sqrt x \ge \sqrt y \Rightarrow \sqrt x \sqrt x\ge \sqrt y \sqrt x \iff x\ge \sqrt y\sqrt x $$ and

$$ \sqrt x \ge \sqrt y \Rightarrow \sqrt x \sqrt y\ge \sqrt y \sqrt y \iff \sqrt x\sqrt y \ge y$$

so (by transitivity) $x\ge y$ that contradicts $x<y$.

Answer 2

Your approach is correct. If you want to use contradiction to prove that $P \implies Q$, you assume $P \land \lnot{Q}$ and look for a contradiction.

If you find a contradiction, then you now know that the negation of your assumption must be true; thus,

$\lnot{(P \land \lnot{Q})}$

Which, using DeMorgan's law, can be rewritten as follows:

$\lnot{P} \lor Q$

Which in turn is equivalent to the following:

$P \implies Q$

Thus, you have proven the implication that you set out prove. In your problem, $P$ is the statement $x < y$, while $Q$ is the statement $\sqrt{x} < \sqrt{y}$, and you followed the above procedure exactly.

Answer 3

If $x < y$, why does it follow that $x^2 < y^2$? If you know that, then your contradiction is fine. But other wise consider why that is true.

It's because $x^2 = x*x < x*y < y*y = y^2$. (And by induction this will hold for all positive integer powers.) Now do your contradiction with confidence. Or do it directly. $x = \sqrt x * \sqrt x ?? \sqrt y * \sqrt y = y$. So x and y and root x and root y must have the same relationship.