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Thread: Need help with this (inverse function, proving, real / non-real numbers)

  1. #1
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    Lightbulb Need help with this (inverse function, proving, real / non-real numbers)

    i) Let x be any nonnegative number. By considering an appropriate function, explain briefly that sqrt(x) is a real number.

    ii) Let a and b be nonnegative numbers. Use i) to explain why (sqrt(a)-sqrt(b))^2 is greater than or equal to 0

    iii) Hence deduce that if a and b are nonnegative numbers,then a+b >or= 2(sqrt(ab))

    let a and b be nonnegative numbers. Prove that a+b=2(sqrt(ab)) is and only if a=b
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  2. #2
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    Re: Need help with this (inverse function, proving, real / non-real numbers)

    Hey Math314.

    There are a number of ways you could pursue this. What subjects and theorems are you covering at the moment?
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  3. #3
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    Re: Need help with this (inverse function, proving, real / non-real numbers)

    I'll do (i). The others are similar

    $x > 0$ and since non-negative makes no sense otherwise, $x \in \mathbb{R}$

    let

    $y^2 = x$, i.e. $y$ is a candidate for $y=\sqrt{x}$

    Now for contradiction suppose that

    $y \not \in \mathbb{R} \Rightarrow y=u + \imath v,~~v\neq 0$

    $x = y^2 = (u^2-v^2) + \imath (2uv)$

    we're told that $x \in \mathbb{R} \wedge x > 0$ so

    $(2uv = 0) \wedge (u^2 - v^2 > 0) $

    We supposed for contrdiction that $v \neq 0$ so it must be that $u=0$

    but then $u^2 - v^2 = 0^2 - v^2 = -v^2 \not > 0$

    so we have a contradiction and it must be that

    $y \in \mathbb{R}$
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