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Prove a Number is Even

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Let a[itex]\in[/itex]Z. Prove that 3a+1 is even if and only if (a+1)/2 [itex]\in[/itex]Z


    2. Relevant equations
    We know that C is an odd number if there exists:

    C=2k+1

    Even:

    C=2k


    3. The attempt at a solution
    I think I figured it out, but I'm terrible at Discrete Math, so I was hoping for some input. We know that:

    3a+1=2k

    If we want 3a+1 to be even. Given the extra constraint we have, I said:

    3[(a+1)/2]+1=2k

    After some simplification:

    (3a+5)/2

    I then factored out a positive 2 and:

    2[(3a+5)/4]

    From here:
    (3a+5)/4=k

    Therefor
    2k=2k

    Am I way off base here or is this actually correct?

    Thanks! I have a bunch of homework problems due at the end of the week, so please be patient with me! This class is ruining my life...
     
  2. jcsd
  3. Feb 15, 2012 #2

    SammyS

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    Yes, you assumed what you were to prove.

    This is an if and only if statement. You have to do both of the following:
    1. Assuming that 3a+1 is even, show that (a+1)/2 is an integer.

    From what you have above, you know that means:
    Assume 3a+1 = 2k, where a and k are integers.

    Show that (a+1)/2 = m for some integer m. (Is this saying that a+1 is even?) I underlined the word "show" because this should be a result. Don't assume this -- at least, not for this part.​
    2. Assuming that (a+1)/2 is an integer, show 3a+1 is even.

    From what you have above, you know that means:
    Assume (a+1)/2 = n, where a and n are integers.

    Show that 3a+1 = 2p, for some integer p.​




     
  4. Feb 15, 2012 #3
    How exactly do I show the (a+1)/2=m? I don't think I can assume that a+1 is even, since I have no basis for that conclusion, because, that would imply that a is odd. Or is that the point? Do I have to make a statement saying that the only way this is true is if a is odd? If so, how exactly do I write that?

    My professor is a super harsh grader, so I want to make sure I do this 100% write. Most of the time, he doesn't even take points off for the proof itself, but the wording and formatting of the proof.
     
    Last edited: Feb 15, 2012
  5. Feb 15, 2012 #4

    Deveno

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    we have two problems in one:

    the first problem is, given that 3a+1 is even, we want to show that (a+1)/2 is an integer.

    the second problem is, given that (a+1)/2 is an integer, we want to show that 3a+1 must be even.

    one approach to problem "one":

    if 3a+1 is even, 3a is odd.

    if a were even, 3a would be even, so a is odd.

    if a is odd, a+1 is even. therefore....

    one approach to problem "two":

    if (a+1)/2 is an integer, a+1 is divisible by 2.

    thus a+1 = 2k, for some integer k (you could use any letter instead of k).

    now write 3a+1 in terms of k:

    3a + 1 = 3(2k - 1) + 1 = ...?

    prove your result is an even integer (hint: "even" = "divisible by 2").
     
  6. Feb 15, 2012 #5
    I understand everything up until you rewrite 3a+1 in terms of of k. Why is there a 2k-1?
     
  7. Feb 15, 2012 #6

    SammyS

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    I would say it a bit differently:

    Once you have
    a+1 = 2k ,​
    I suggest that you add 2a to both sides of the equation.

    Do you see how that will show that 3a+1 is even?
     
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