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Math Help - 'Show that' help

  1. #1
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    'Show that' help

    The question asks:

    Suppose that ab > 0. Show that if a < b, then 1/b < 1/a.

    I know this is correct, but I don't know how to show it. I did this:

    aa^-2 < ba^-2

    1/a < ba^-2

    (b^-2)/a < (a^-2)/b

    Hoping that I could find a way to swap the inequality sign around whilst creating the fractions required. But I can't seem to do it.

    Any assistance would be great!
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Glitch View Post
    The question asks:

    Suppose that ab > 0. Show that if a < b, then 1/b < 1/a.

    I know this is correct, but I don't know how to show it. I did this:

    aa^-2 < ba^-2

    1/a < ba^-2

    (b^-2)/a < (a^-2)/b

    Hoping that I could find a way to swap the inequality sign around whilst creating the fractions required. But I can't seem to do it.

    Any assistance would be great!
    You can consider two cases: (1) a and b are both positive, or (2) a and b are both negative.
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  3. #3
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    Ok. But am I on the right track? Is there a method to proving this?
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by undefined View Post
    You can consider two cases: (1) a and b are both positive, or (2) a and b are both negative.
    We can do it without "consider two cases"...

    1/b<1/a

    1/b-1/a<0

    {a-b}/ab<0

    since ab>0, and a<b, it's obvious!
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  5. #5
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    Thanks! Looks like I was way off! :P

    If I were to write that in an exam, would I have to explain why it's obvious?
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  6. #6
    MHF Contributor Also sprach Zarathustra's Avatar
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    Yes, you explain it in the next form:

    {a-b}<0 since a<b(given)
    ab>0 {given}

    so, {a-b}/ab={-}/{+}={-}<0

    ok?
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  7. #7
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    Yup. Thanks. I really do need more practise with this stuff.
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  8. #8
    MHF Contributor undefined's Avatar
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    Thanks Also sprach Zarathustra, my way was pretty tedious compared with yours.

    Quote Originally Posted by Glitch View Post
    Thanks! Looks like I was way off! :P

    If I were to write that in an exam, would I have to explain why it's obvious?
    You should explain it on an exam, yes, but it's simply that the numerator is negative while the denominator is positive...

    Edit: Too slow.
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  9. #9
    MHF Contributor undefined's Avatar
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    One more thing worth mentioning since "show that" means we're dealing with formal proofs here.

    Quote Originally Posted by Also sprach Zarathustra View Post
    1/b<1/a

    1/b-1/a<0

    {a-b}/ab<0
    These lines are connected implicitly by "if and only if" as in

    1/b<1/a

    \displaystyle \iff 1/b-1/a<0

    \displaystyle \iff {a-b}/ab<0

    For the last line, we don't need to think about whether it's "if and only if" because we only need to go in one direction

    1/b<1/a

    \displaystyle \iff 1/b-1/a<0

    \displaystyle \iff {a-b}/ab<0

    \displaystyle \Longleftarrow ab > 0 \land a < b

    I write this mainly so that nobody gets confused, thinking the proof is not valid because we assumed what we wanted to prove.
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  10. #10
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    I've been meaning to ask, what does that upside-down 'V' symbol mean?
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  11. #11
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Glitch View Post
    I've been meaning to ask, what does that upside-down 'V' symbol mean?
    AND
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  12. #12
    Member Mathelogician's Avatar
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    The simplest way is to multiply 1/ab in to the both sides! and then you wont have to say that this is obvious!!
    Just note that 1/ab>0 since ab>0.
    a<b => a/ab<b/ab => 1/b<1/a .
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  13. #13
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    Given ab > 0 and a < b.


    Case 1: a > 0 and b > 0.

    a < b

    \frac{a}{a} < \frac{b}{a}

    1 < \frac{b}{a}

    \frac{1}{b} < \frac{b}{ab}

    \frac{1}{b} < \frac{1}{a}.


    Case 2: a < 0 and b < 0.

    a < b

    \frac{a}{a} > \frac{b}{a}

    1 > \frac{b}{a}

    \frac{1}{b} < \frac{b}{ab}

    \frac{1}{b} < \frac{1}{a}.


    Q.E.D.
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  14. #14
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    Quote Originally Posted by Glitch View Post
    Thanks! Looks like I was way off! :P

    If I were to write that in an exam, would I have to explain why it's obvious?
    The best thing to do would be to multiply both sids of \frac{a- b}{ab}> 0 by the positive value ab to get a- b> 0, then add b to both sides.
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