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Math Help - Help on a proof - square roots

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    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)
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    MHF Contributor Bruno J.'s Avatar
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    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?
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    Quote Originally Posted by mistykz View Post
    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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