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Proof question from How to Prove It

  1. Jan 3, 2010 #1
    proof question from "How to Prove It"

    3.2.8 Suppose a & b are nonzero real numbers. Prove that if a < 1/a < b < 1/b then a < -1

    I understand intuitively why this is true, but I can't figure out how to prove it. According to the hints at the back of the book it says to prove a < 0, then use to prove a < -1.

    When I go through the inequalities I come up with this:

    a < 1/b
    ab < 1

    1/a < b
    1 < ab

    I know that when you multiply both sides of inequality you have to switch the signs. But if both a and b are negative the signs both switch so I don't really understand how this can be true.
  2. jcsd
  3. Jan 3, 2010 #2


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    Re: proof question from "How to Prove It"

    Just take it one step at a time. If a<1/a then either a is in (0,1) or (-infinity,-1). You can show that, right? Same for b. Now if b is also in (0,1) and a<b can 1/a be less than 1/b?
  4. Jan 4, 2010 #3
    Re: proof question from "How to Prove It"

    Thanks, I understand it now.
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