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How do I show that there are no other solution

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://img90.imageshack.us/img90/834/unledml.png [Broken]



    3. The attempt at a solution

    OKay first of all I showed that x is indeed a solution. But I need to state the Axioms of Fields first.

    [PLAIN]http://img194.imageshack.us/img194/2548/unledtao.png [Broken]

    i.e.

    [tex]a + (b - a) = (b - a) + a[/tex] by A2

    [tex]= b + (-a + a)[/tex] by A1

    [tex] = b + 0 [/tex] by A4

    Hence b = b

    Showing that it is the only solution is tough for me because I don't know what that means. I am guessing at the end of the proof I would show that x = b - a?

    Here goes

    [tex]a + x = b[/tex]
    [tex]a + (-a) + x = b + (-a)[/tex]
    [tex]0 + x = b + (-a)[/tex] By A4
    [tex]x = b + (-a)[/tex] By A3

    Now can I just jump to

    [tex]x = b - a[/tex] and say it is the only solution?

    Thanks
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 21, 2011 #2

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    One way to do it is to let x=b-a+h, and show contradiction unless h=0.
     
  4. Sep 21, 2011 #3
    But we haven't learned proof by contradiction yet. So I can't use that.
     
  5. Sep 22, 2011 #4
    Any comments...? I'll take it even if i am wrong.
     
  6. Sep 22, 2011 #5

    Mark44

    Staff: Mentor

    Unfortunately, contradiction is the standard way of proving statements about the uniqueness of something. This technique is not related to the axioms you have to use, so it should be fine to use.
     
  7. Sep 22, 2011 #6
    BUt what was wrong with what I did? I posted this because I wasn't confident
     
  8. Sep 22, 2011 #7
    Without wasting everyone's time...

    Proof

    Let x = b - a + h

    a + x = a + (b - a + h)

    = (a + (-a)) + (b + h) by A2

    = 0 + b + h By A4

    = b + h

    Now we have a + x = b + h

    Stuck...
     
  9. Sep 22, 2011 #8

    Mark44

    Staff: Mentor

    Given a + x = b, you showed that there was a solution, x = b - a.

    Now suppose that y [itex]\neq[/itex] x, is also a solution. Can you take it from there?
     
  10. Sep 22, 2011 #9
    Are you referring to Pallen's method? Where did y come from?
     
  11. Sep 22, 2011 #10

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    What you did was show that b- a [itex]is[/itex] a solution. You did not show that it was the only solution, which was the point of the exercise.
     
  12. Sep 22, 2011 #11
    No I meant this part


    [tex]a + x = b[/tex]
    [tex]a + (-a) + x = b + (-a)[/tex]
    [tex]0 + x = b + (-a)[/tex] By A4
    [tex]x = b + (-a)[/tex] By A3

    Doens't that prove that x can only be x = b + (-a)?
     
  13. Sep 22, 2011 #12
    I dont think so, it shows x is a solution but not the only one. Now assume y is a solution
     
  14. Sep 22, 2011 #13
    Assume y is NOT a solution
    a + y = b

    a + (-a) + y = b + (-a)

    0 + y = b + (-a)

    y = b + (-a)

    y = b - a

    But this contradicts x [tex]\neq [/tex] y

    Woah!
     
  15. Sep 22, 2011 #14
    Assume y is a solution where y does not equal x
    .
    .
    .
    .
     
  16. Sep 22, 2011 #15
    Isn't that what I just showed? If I assume y is a solution I would just go back to my original post and redo that proof again?
     
  17. Sep 22, 2011 #16
    you said assume y is not a solution, yes and no because y is not x so its different but it gets you the same thing so y=x contradiction
     
  18. Sep 22, 2011 #17
    Oh, I should have said y is solution, but my math follows is correct right?
     
  19. Sep 22, 2011 #18
    if y is not a solution does it make any sense to say for example 0=2?
     
  20. Sep 22, 2011 #19
    No I meant like

     
  21. Sep 22, 2011 #20
    yeah that works
     
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