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Proving a equation

  1. Jul 25, 2012 #1
    Can some one show me how this is done?

    Prove that there exists a positive integer x such that x^2=2x.

    Just from the looks, I want to say this this is not possible, hence vacuous proof. I'm sure I'm wrong.
     
  2. jcsd
  3. Jul 25, 2012 #2
    Try setting it up so you can factor it and find the possible values of x.
     
  4. Jul 25, 2012 #3
    Thanks a bunch, I don't know why I did think of 2 :).
     
  5. Jul 25, 2012 #4
    It's easy to see that x=2 is such an integer. Since 2 exists, it's proved.

    Aww, you figured it out before I could type my answer. :(
     
  6. Jul 25, 2012 #5
    Its all good :), I do have another one if you can help.
     
  7. Jul 25, 2012 #6
    Use proof by contradictions to show that for any selection of 3 distinct integers between 0 and 6 that at least one of those numbers will be odd.
     
  8. Jul 25, 2012 #7
    numbers between 0 - 6 : { 0,1,2,3,4,5,6}
    |{ 0,1,2,3,4,5,6}| = 7
    so at least one odd would be 1/7?
    since we are picking 3 times
    does it mean the chance is:
    (1/7)(1/6)(1/5) = 1/(210)
     
  9. Jul 25, 2012 #8
    What?...
     
  10. Jul 25, 2012 #9
    In between 0 and 6 means {1,2,3,4,5} otherwise the statement is false as {2,4,6} is a counter example.
     
  11. Jul 25, 2012 #10

    Mark44

    Staff: Mentor

    The problem is not clearly stated due to the ambiguity of what "between 0 and 6" means. Does "between" here mean "strictly between", with the endpoints not included? Or does "between" here mean that the endpoints are included?

    Robert1986 is assuming that the latter meaning is the one that was intended.
     
  12. Jul 26, 2012 #11
    Easy (I think) ;)
    If you divide the right side by 2 so it becomes x2/2 = x
    Then you can put, for ex. 2
    22/2 = 2
    OR am I wrong? xD
     
  13. Jul 26, 2012 #12
    A little backwards...

    Although the right answer has been mentioned above,

    Assume x != 0

    (I use != to mean does not equal)

    then divide both sides by x!!!

    x2 = 2x
    Becomes
    x2/x = 2x/x

    Cancel the x's

    and you get x = 2. This is a valid solution because x !=0.

    What others have said is to subtract and factor
    x2 = 2x
    Subtract 2x from both sides

    x2 - 2x = 2x - 2x
    Simplify
    x2 -2x = 0

    Factor the x

    x(x-2) = 0

    Zero Product property tells us that either x = 0 or x - 2 = 0

    So we have two solutions x = 0 and x = 2. Since 0 is not POSITIVE, our answer must be x = 2.
     
  14. Jul 27, 2012 #13

    That's certainly a creative approach. Proof by contradiction, you should assume that there are no odd numbers between 0 and 6 and show that it leads to a logical contradiction like 2=6 or something.
     
  15. Jul 27, 2012 #14

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    This is NOT a probability problem. There is a "chance" of picking numbers is relevant only if we were picking numbers at random.

    Even then, your calculations are wrong. The chance of picking any one number is 1/7, not "at least one odd". If you were to choose three numbers from {0, 1, 2, 3, 4, 5, 6}, without replacement, the probability they would all be even would be (4/7)(3/6)(2/5)= 4/35. Therefore the probability of "at least one odd" would be 1- 4/35= 31/35.

    But again, probability has nothing to do with this. The problem was to show that, if pick three numbers from {1, 2, 3, 4, 5}, at least one must be odd. And that is obviously true because there are only two even numbers in that set.
     
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