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Proof by Contradiction.

  1. May 31, 2015 #1
    1. The problem statement, all variables and given/known data
    Suppose that a and b are nonzero real numbers. Prove that if a< 1/a < b < 1/b then a<-1.

    2. Relevant equations
    Givens: a and b are nonzero real numbers, a< 1/a < b < 1/b, and a≥-1.
    Goal: Arrive at a contradiction.



    3. The attempt at a solution
    Scratch work: First establish whether a<0 or a>0.
    Case i.) a>0 and Case ii.) a<0
    Case i). a>0. a( a<1/a) ⇒ a2<1 and a(a≥-1) ⇒ a2≥-a.
    However a2≥0 and -a<0. Therefore a>0 contradicts the fact that a is at leas greater than or equal to -1.
    Is the argument sound? Thanks for the feedback!
     
  2. jcsd
  3. May 31, 2015 #2

    Mark44

    Staff: Mentor

    Your argument doesn't do anything with b, which could also be either negative or positive.

    Your hypothesis is (in part) that a < 1/a and that b < 1/b. What do these inequalities say about the numbers a and 1/a and b and 1/b?

    Your other hypothesis is that a -1.
     
  4. Jun 1, 2015 #3

    vela

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    No. Take ##a=1/2## for example. Clearly, ##a>0## holds so ##a \ge -1## does as well. Moreover, you have ##a^2 = 1/4 \ge 0## and ##-a = -1/2 < 0##. There's no apparent contradiction.
     
  5. Jun 1, 2015 #4

    HallsofIvy

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    I don't see any where in here where you establish the "fact" that "a is greater than or equal to -1" that you say is contradicted. Of course, in the case that a> 0 it certainly must be greater than -1, but that doesn't seem to be what you are referring to. You do show, correctly, that a2≥-a. But that's trivially true if x> 0. It certainly doesn't contradict "a is greater than or equal to -1"!
     
  6. Jun 14, 2015 #5
    I see I forgot to post the solution here. woops sorry guys.

    If a,b≠0 and a<1/a<b<1/b then a<-1
    We can see that
    a<1/b ⇒ab<1 if b>0 or it implies that ab>1 if b<0
    and
    1/a<b ⇒ab<1 if a<0 or it implies that ab>1 if a>0
    We can conclude that ab<1 when b>0 and a<0 otherwise the inequality a<b does not hold.
    Since a<1/a
    then a2>1 since a<0
    therefore lal>1
    ⇒a>1 or a<-1 bur a<0 thus a<-1.
     
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