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Simple Proof (Verification)

  1. Sep 3, 2013 #1
    1. The problem statement, all variables and given/known data

    Let n be an integer. Prove that if n^3 is even, then n is even.

    2. Relevant equations



    3. The attempt at a solution

    Suppose n is even. That is n=2m, for some mεZ.
    Then, n^3=(2m)^3=8m^3=2(4m^3).
    Since 4m^3 is an integer, n^3 will be even.

    Now, i proved that if n is even, then n^3 is even. So would this be a valid proof in this context?
     
  2. jcsd
  3. Sep 3, 2013 #2

    Mark44

    Staff: Mentor

    Looks fine.

    Edit: I misread the problem statement as "if n is even then n3 is even." My mistake.
     
    Last edited: Sep 4, 2013
  4. Sep 3, 2013 #3
    Your proof is showing that if n is even then n^3 is even.

    You want to show the converse that if n^3 is even then n is even.

    It is semantics and may seem silly in this sort of proof but in more difficult proofs you have to be careful of what you are saying.

    Edit: Take the factors of n^3 and see what that means to be even.


    Assume n^3 is even and n is an integer. Well, n^3 = n* n* n. Since n^3 is even n*n*n is even which means that either n or n or n is divisible by two. Therefore, n is an integer divisible by two; n is even.
     
    Last edited: Sep 3, 2013
  5. Sep 3, 2013 #4

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    It's good to point out what he did wrong, but I presume you know you are not supposed to supply complete solutions.
     
  6. Sep 3, 2013 #5
    I just wanted to make sure that I wasn't missing something.

    To comply better with PF rules I updated my post
     
  7. Sep 4, 2013 #6

    verty

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    Homework Helper

    So many of these problems are made easier when you look at the contrapositive of the claim. Here it is just that if n is odd, then n^3 is odd.
     
  8. Sep 4, 2013 #7
    Yeah, that's what I was really unsure about, but thanks for clarifying.
     
  9. Sep 4, 2013 #8
    The solution to the actual problem uses the contrapositive method you described, where n is odd, then n^3 is odd. Though, I haven't studied contrapositive proofs yet, so I posted an alternative method above.
     
  10. Sep 4, 2013 #9

    Mark44

    Staff: Mentor

    What you proved was the converse, not the contrapositive.

    If you're uncertain about these terms, maybe this will help.

    Implication P => Q (if P then Q)
    Converse: Q => P
    Contrapositive: ~Q => ~P (if not Q, then not P)

    There's another term for the converse of the contrapositive, but I forget what it is.

    An implication and its contrapositive are always equivalent (i.e., they have exactly the same truth table values), but an implication and its converse are not equivalent.

    Here are some examples.
    Implication: If x = 2, then x2 = 4

    Converse of the implication: If x2 = 4, then x = 2 (not necessarily true, as x could be -2)

    Contrapositive of the implication: If x2 ≠ 4, then x ≠ 2
     
  11. Sep 5, 2013 #10
    Thanks alot Mark44. It makes sense to me now why the solutions manual proved the statement, "if n^3 is even, then n is even" by giving a proof of " if n is odd, then n^3 is odd", which is the contrapositive statement. This would be valid as the implication statement is equivalent to its contrapositive statement, as you pointed out.
     
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