# Thread: Help on a proof - square roots

1. ## Help on a proof - square roots

Hey everyone, I've looked at this one for a while and just can't seem to get started. Anyone care to give me a hint / help out? Thanks!

Suppose that S is an ordered field. Define sqrt(a) to be the number such that (sqrt(a))^2 = a for a >= 0. Prove that if a,b are elements of S and 0 =< a =< b, then sqrt(a) =< sqrt(b)

2. It's not true. If the field is $\mathbb R$ and $\mbox{sqrt }a=-\sqrt a$ then the requirement is clearly satisfied by $\mbox{sqrt }$. However if $0\leq a \leq b$ we have $0\geq\mbox{sqrt }a\geq\mbox{sqrt }b$.

Perhaps you need the extra condition that $0\leq\mbox{sqrt }a$?

3. Originally Posted by mistykz
Hey everyone, I've looked at this one for a while and just can't seem to get started. Anyone care to give me a hint / help out? Thanks!

Suppose that S is an ordered field. Define sqrt(a) to be the number such that (sqrt(a))^2 = a for a >= 0. Prove that if a,b are elements of S and 0 =< a =< b, then sqrt(a) =< sqrt(b)
Assuming that, as Bruno J. pointed out, that you also have $\sqrt{a}> 0$ and $\sqrt{b}> 0$, use a proof by contradiction.

Assume $\sqrt{a}> \sqrt{b}$ and multiply on both sides by $\sqrt{a}$. Then multiply the first inequality by $\sqrt{b}$. Compare the two.