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Thread: Have I done this 'show that' problem correctly?

  1. #1
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    Have I done this 'show that' problem correctly?

    The question:

    Show that if $\displaystyle 0 <= a <= b$, then $\displaystyle \sqrt{a} <= \sqrt{b}$

    I did:

    $\displaystyle \sqrt{0} <= \sqrt{a} <= \sqrt{b}$
    $\displaystyle 0 <= \sqrt{a} <= \sqrt{b}$

    Therefore:
    $\displaystyle \sqrt{a} <= \sqrt{b}$

    My textbook doesn't have answers for these types of questions, so I can't check.
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Glitch View Post
    The question:

    Show that if $\displaystyle 0 <= a <= b$, then $\displaystyle \sqrt{a} <= \sqrt{b}$

    I did:

    $\displaystyle \sqrt{0} <= \sqrt{a} <= \sqrt{b}$
    $\displaystyle 0 <= \sqrt{a} <= \sqrt{b}$

    Therefore:
    $\displaystyle \sqrt{a} <= \sqrt{b}$

    My textbook doesn't have answers for these types of questions, so I can't check.
    Well, I think the proof is circular because it relies on being able to take the square root of both sides of an inequality, in other words: for $\displaystyle \displaystyle f(x)=\sqrt{x}$ and for non-negative reals $\displaystyle \displaystyle a$ and $\displaystyle \displaystyle b$ we have $\displaystyle \displaystyle a\le b \implies f(a) \le f(b)$, which is what you're trying to prove (and which is not true for general $\displaystyle \displaystyle f(x)$).

    All we're really saying here is that $\displaystyle \displaystyle f(x)=\sqrt{x}$ is increasing. The "usual" way is to take the derivative and show that it's greater than equal to zero on that interval. This question is posted in the Pre-Calculus subforum; does it makes sense to take the derivative given where you are in the textbook?
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  3. #3
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    Quote Originally Posted by undefined View Post
    This question is posted in the Pre-Calculus subforum; does it makes sense to take the derivative given where you are in the textbook?
    Nope, this chapter is pre-calculus only, and hasn't mentioned derivatives yet.
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Hint:

    $\displaystyle b-a=(\sqrt{b}-\sqrt{a})(\sqrt{b}+\sqrt{a})$
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  5. #5
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    Quote Originally Posted by Glitch View Post
    The question:

    Show that if $\displaystyle 0 <= a <= b$, then $\displaystyle \sqrt{a} <= \sqrt{b}$

    I did:

    $\displaystyle \sqrt{0} <= \sqrt{a} <= \sqrt{b}$
    $\displaystyle 0 <= \sqrt{a} <= \sqrt{b}$

    Therefore:
    $\displaystyle \sqrt{a} <= \sqrt{b}$

    My textbook doesn't have answers for these types of questions, so I can't check.
    I'm not sure if this approach is necessary any more, but nevertheless here it is:

    If $\displaystyle a\leq b~\implies~\sqrt{a} \leq \sqrt{b}$

    Substitute $\displaystyle b = a + k, k \geq 0$ and you'll get:

    $\displaystyle \sqrt{a} \leq \sqrt{a + k} \leq \sqrt{a} + \sqrt{k}$

    Finally you have $\displaystyle 0\leq \sqrt{k}$ which is true because the square-root of a positive number is positive too.
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  6. #6
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    It's not "unecessary", it is wrong. You want to prove that $\displaystyle \sqrt{a}\le \sqrt{b}$, yet the first line of your "proof" begins "$\displaystyle \sqrt{a}\le \sqrt{a+ k}$ which is exactly the same as what you want to prove. And, you are also assuming that $\displaystyle \sqrt{a+ k}\le \sqrt{a}+ \sqrt{k}$ which is a much harder theorem that was was asked here.
    Last edited by HallsofIvy; Jul 12th 2010 at 04:35 AM.
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