Help on a proof - square roots

**mistykz** · January 25th 2010, 07:29 PM

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)

**Bruno J.** · January 25th 2010, 08:21 PM

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 $0\leq\mbox{sqrt }a$ ?

**HallsofIvy** · January 26th 2010, 05:28 AM

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.

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