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Number Proofs, HELP! **HOMEWORK**

Jimmy KingJimmy King Registered User regular
This is for homework so please, no direct answers. I just can't figure out how to start this in the first place, so clearly I'm not even really trying to prove the right thing. The problem I have is this.

Prove that if x + y is odd then x -y is also odd. The teacher gives a hint of use two cases, one where y is odd and one where y is even. I also have to assume this means integers based on what we are doing in class.

I have no idea what to do after that, though. I make my first case...
Let y be an odd integer.

Then I have tried these 2 things...
x + y = 2k + 1 -> then what? I can prove that x = 2k + 1 - y, but that doesn't get me anywhere
x + (2k +1) = 2m + 1 -> ok, now what? I can solve for x, which proves nothing and just proves that x is even (x=2(m-k)).

Do I take this further and take that even value I got for x = 2(m-k) and plug that in for 2(m-k) - (2k + 1) and show that it creates another odd integer? This is sounding correct now that I type it out.

Then I go back and do the same when y is even, of course.

Posts

  • enlightenedbumenlightenedbum Registered User regular
    edited June 2013
    x + (2k +1) = 2m + 1 -> ok, now what? I can solve for x, which proves nothing and just proves that x is even (x=2(m-k)).

    Do I take this further and take that even value I got for x = 2(m-k) and plug that in for 2(m-k) - (2k + 1) and show that it creates another odd integer? This is sounding correct now that I type it out.

    You've got it. There's no way for 2m - 4k - 1 to be even if m and k are integers. If your teacher is a real stickler, formulate it as 2(m - 2k - 1) + 1 to fit the textbook definition of an odd integer.

    Then do the even case, which is probably going to go in a similar direction.

    enlightenedbum on
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  • NarbusNarbus Registered User regular
    You shouldn't have to do both cases. It's pretty easy to prove that the sum of two even (or two odd) numbers is even, so for the sum of x and y to be odd, either x or (not and) y must be odd. So if you prove the case for y odd, then just change the letters and you have x odd. Otherwise yeah, you're on the right path.

  • Jimmy KingJimmy King Registered User regular
    Awesome, thanks guys.

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