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Did I skip a major step in this proof? + Theory of this proof

  1. Sep 21, 2011 #1
    1. The problem statement, all variables and given/known data


    [PLAIN]http://img717.imageshack.us/img717/2285/unledxig.png [Broken]

    The problem is to verify (v), that is for some x and y in R, we have (-x)(-y) = xy


    3. The attempt at a solution

    (-x)(-y) = (-1)x * (-1)y by Prop 2.7 iv

    = x*(-1) * (-1)y by M2
    = x*(-1*-1) * y by M1
    = x*(+1)*y
    = xy

    Q.E.D

    Not sure if I did it right.

    Now I am going to throw in another question.

    My prof did a lot of proofs in class in the similiar manner as me (except for being wrong...). The thing is that they all look so silly. Like why do I have to go throw all this trouble just to show that (-1)(-1) = +1? That took like 4 lines. This is like math made unnecessarily hard.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 21, 2011 #2
    Actually wait, I didn't need 4 lines at alll...

    (-x)(-y) = (-1)x(-1)y By Prop.2.7 iv

    =(-1)(-1)xy byM2

    = 1 * xy By M3

    = xy

    Q.E.D.
     
  4. Sep 22, 2011 #3
    Any comment is fine...
     
  5. Sep 22, 2011 #4
    Couldnt agree with you more that this is math made unnecessarily hard, but I think you have to show -1*-1=1
     
  6. Sep 22, 2011 #5
    Yeah I took that for granted...

    I don't know how to do that
     
  7. Sep 22, 2011 #6
    1= 1+0= ?

    does that help
     
  8. Sep 22, 2011 #7
    1 = 1 + 0 = 1 + 1 - 1?

    But this is addition.
     
  9. Sep 22, 2011 #8
    yes....but you can distribute a -1 by prop 4 and m4D
     
  10. Sep 22, 2011 #9
    I have one -1 though
     
  11. Sep 22, 2011 #10
    pull out a negative one
     
  12. Sep 22, 2011 #11
    -1(-1 - 1 + 1)
     
  13. Sep 22, 2011 #12
    now what do you notice...
     
  14. Sep 22, 2011 #13
    1 = 1 + 0 = 1 + 1 - 1 = -1(-1 - 1 + 1)

    1 = -1(-1 + 0)
    1 = -1(-1)

    Is that it? Because this seem really really unnecessary...
     
  15. Sep 22, 2011 #14
    yep...... that is it, and I could not agree more, but pure math people seem to love this kind of stuff because now they now for sure that -1*-1=1 and they are not relying on what others have said before
     
  16. Sep 22, 2011 #15
    How do I include that in my original proof?

    Assume (-x)(-y) = xy

    Conditions [ =(-x)(-y) = (-1)x * (-1)y by Prop 2.7 iv
    = x*(-1) * (-1)y by M2
    = x*(-1*-1) * y by M1
    = stuck here. How do I introduce 1 = 1 + 0 unawkwardly?]

    Result [...]

    Q.E.D
     
  17. Sep 22, 2011 #16
    just call it a lemma and put it before the proof.. of course if you have proven it before you do not need to include but im guessing you havent.

    lemma 1 = -1(-1)

    pf/
    1 = 1 + 0 = 1 + 1 - 1 = -1(-1 - 1 + 1)

    1 = -1(-1 + 0)
    1 = -1(-1)

    pf/
    (-x)(-y) = (-1)x * (-1)y by Prop 2.7 iv

    = x*(-1) * (-1)y by M2
    = x*(-1*-1) * y by M1
    = x*(+1)*y
    = xy

    Q.E.D
     
  18. Sep 22, 2011 #17
    Lemma: (-1)(-1) = 1 or do I have to write out that whole step we did...?
     
  19. Sep 22, 2011 #18
    All the step. Or how is someone who is reading your proof going to know that -1*-1=1 if it hasnt been shown to them.
     
  20. Sep 22, 2011 #19
    But I thought Lemma was the results, not the proof. Isn't it just showing the results that you should believe in?
     
  21. Sep 22, 2011 #20
    you can't believe in anything anymore it must be proven to be true.
     
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