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Simple Direct Proof

  1. Sep 8, 2010 #1
    Let n be an integer. Prove that if n + 5 is odd, then 3n + 2 is even.

    So the instructions say to use a direct proof. I couldnt figure that method out, so I used a controposition proof and that seemed to work ok. Here are my contraposition steps:

    Assume 3n+2 is odd
    Def of odd: n=2k+1
    n+5=2k+1+5 = 2k+6 = 2(k+3)
    n+5 is even (multiple of 2)
    since negation of conclusion implies hypothesis is false, original statement is true.

    Im pretty sure thats correct, but how could this be done using a direct proof?
  2. jcsd
  3. Sep 8, 2010 #2


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    3n+2 is odd, but you assumed that n itself was odd when you plugged n=2k+1 into n+5
  4. Sep 8, 2010 #3
    Office Shredder is right, your proof if flawed. 3n+2 is odd means for some natural number m, 3n + 2 = 2m + 1

    With that in mind, let’s try a direct proof.
    By definition what does n + 5 being odd mean? It means that: n + 5 = 2m + 1 for some natural m.
    Can we solve for n? If we do, can we plug n into 3n + 2? Can we simplify and pull out a factor of 2? If we can pull out a factor of two, what does that mean?
  5. Sep 8, 2010 #4
    Ahh ok I see! So n=2k-4 and 3n+2=2(k-5) which means 3n+2 is even...thanks alot!
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