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I did a proof! (I think, please check my reasoning)

  1. Sep 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Suppose that a and b are real numbers. Prove that if [itex]a < b < 0[/itex], then [itex]a^{2} > b^{2}[/itex]



    2. Relevant equations
    Properties of inequalities?


    3. The attempt at a solution
    This is how I did the final proof:

    Given that a and b are real numbers and [itex]a < b < 0[/itex], we notice that both a and b are negative since they are less than zero. It follows that, [itex]ab > b^{2}[/itex] and [itex]a^{2} > ab[/itex]. Because [itex]a^{2} > ab[/itex] and [itex]ab > b^{2}[/itex] it is evident that [itex]a^{2} > b^{2}[/itex]
    Q.E.D.

    Please let me know if my reasoning was, well, reasonable. If not, give me some suggestions for proof. This is a problem out of chapter 3 in Velleman's How To Prove It, Second Edition.
     
  2. jcsd
  3. Sep 13, 2014 #2

    HallsofIvy

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    If you are allowed to use "if a< 0 and b< c then ab> ac" yes, that is a valid proof.
     
  4. Sep 13, 2014 #3

    Fredrik

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    There are only three axioms for < in the definition of the real numbers, and one of them is about the concept of "least upper bound". The other two are:

    1. For all x,y,z, if x<y then x+z<y+z.
    2. For all x,y, if x>0 and y>0 then xy>0.

    You shouldn't use any statements that involve inequalities other than the assumptions that go into the theorem, these two axioms, and theorems that you have already proved in this way. But you are of course allowed to also use the axioms about addition and multiplication, and theorems derived from them. In particular, you can use that ##(-a)^2=a^2##.

    If you would like to prove this theorem in this way, without using any other theorems about inequalities, I suggest the following approach: First use the assumption and axiom 1 to prove that 0<-a, 0<-b and 0<b-a. Then use these results with axiom 2. At some point, you should use that < is a transitive relation. (If x<y and y<z, then x<z).
     
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