# Help on a proof - square roots

• Jan 25th 2010, 07:29 PM
mistykz
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)
• Jan 25th 2010, 08:21 PM
Bruno J.
It's not true. If the field is $\displaystyle \mathbb R$ and $\displaystyle \mbox{sqrt }a=-\sqrt a$ then the requirement is clearly satisfied by $\displaystyle \mbox{sqrt }$. However if $\displaystyle 0\leq a \leq b$ we have $\displaystyle 0\geq\mbox{sqrt }a\geq\mbox{sqrt }b$.

Perhaps you need the extra condition that $\displaystyle 0\leq\mbox{sqrt }a$?
• Jan 26th 2010, 05:28 AM
HallsofIvy
Quote:

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 $\displaystyle \sqrt{a}> 0$ and $\displaystyle \sqrt{b}> 0$, use a proof by contradiction.

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