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Tricky Mathematical Proof

  1. Feb 8, 2005 #1
    We are given the following statement: The universe is all integers. If a-3b is even, then a+b is even. I began off with saying that 3 cases exist: a is odd and b is even, a is even and b is odd, both a and b are odd, and both a and b are even. After this point I got really confused and lost. Can someone please point me in the right direction?
     
  2. jcsd
  3. Feb 8, 2005 #2
    hint, if a number is even, then it can be written as [tex] 2k [/tex] where k is an integer.

    since they tell you that a - 3b is even, then that means

    [tex] a - 3b = 2k [/tex]

    for some integer k.
     
    Last edited: Feb 8, 2005
  4. Feb 8, 2005 #3

    StatusX

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    think about the difference between those two expressions
     
  5. Feb 8, 2005 #4
    I don't understand what you mean when you say a-3b=2k. How does that help you?
     
  6. Feb 8, 2005 #5
    try to manipulate that equation, so you have a + b on one side, and on the other you have 2* some integer.



    edit: I edited the above post, to try to make it more clear
     
    Last edited: Feb 8, 2005
  7. Feb 8, 2005 #6
    I realize that I just sound really stupid right now; but it's been a while since I've really done any math per say. I don't think any way exists to end up with a+b on one side with 2* something on the other side. The problem exists because of the 3, no matter what happens, there is no way to separate the 3 from the b, and still end up with a+b. Is there?
     
  8. Feb 8, 2005 #7
    start by trying to obtain a + b on the left side. Remeber, anything you do on the left side you also have to do on the right side.

    you can do this by adding 4b to both sides, and simplifying :smile:
     
    Last edited: Feb 8, 2005
  9. Feb 8, 2005 #8
    Why worry about a-3b=2k? Why not just remember that the sum of two even numbers is an even number?

    Doug
     
  10. Feb 8, 2005 #9
    Do you mean a+b = 2(k+2b)?
     
  11. Feb 8, 2005 #10
    where does it say that a and b are both even?
     
  12. Feb 8, 2005 #11
    yep... since k + 2b is an integer, then 2(k + 2b) is even, therefore, a + b is even. The trickiest part of this proof, is just knowing that if a number is even then it can be written as 2 * an integer.
     
  13. Feb 8, 2005 #12
    It doesn't. It just says that if a-3b is even, then a+b is even. Then we are supposed to prove that statement any way possible which makes logical sense.
     
  14. Feb 8, 2005 #13
    It doesn't but (a-3b)+c is even if a-3b and c are both even.

    Doug
     
  15. Feb 8, 2005 #14
    Thank you so much guys, for sticking with me.
     
  16. Feb 8, 2005 #15

    learningphysics

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    Yes. This is what I was thinking too:

    a+b = (a-3b) + (4b)

    Since a-3b is given as even, and since 4b is even... and since the sum of two even numbers is even we know that a+b is even.
     
  17. Feb 8, 2005 #16
    I see what you mean ... doesn't seem like either way is faster, but its good to know both :smile: .


    edit: Actually I take that back, your method is a little faster, in that you don't have any manipulating to do.
     
    Last edited: Feb 8, 2005
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