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Direct proof confusion?

  1. Sep 6, 2012 #1
    1. The problem statement, all variables and given/known data
    im supposed to use a direct proof to prove that if 1-n^2>0 then 3n-2 is odd for all n∈Z


    2. Relevant equations



    3. The attempt at a solution
    if you let n∈z then suppose that 1-n^2>0 then 1>n^2 but the only inter n such that 1>n^2 is 0. 3x0-2=-2 as -2=2(-1), -2 is even . Hence if 1-n^2>0 then 3n-2 is even when n∈z. The part that im confused on is why in the book does it say for directs proofs to choose an arbitrary value for n and prove that Q(X) is true? In this proof I had to prove it the same way but there was no way I could let N represent an arbitrary value. The only value that worked was when n=0.
     
  2. jcsd
  3. Sep 6, 2012 #2
    The statement you're trying to prove does not make sense. You're saying you have the prove this:
    [itex]1-n^2>0 \Rightarrow 3n-2[/itex] is odd, for ALL [itex]n\in\mathbb{Z}[/itex].

    What about n=0? Then 1-n^2=1>0, so the left side is satisfied. If this were true, then you would have that 3n-2=-2 is odd. But it isn't.

    Moreover, if you're talking about integers, then the only integer which satisfies your left side inequality is n=0, which we've just shown doesn't even work.

    In conclusion... this statement is just wrong. Wrong, wrong, wrong.
     
  4. Sep 6, 2012 #3
    my bad not for all nεZ just that n is an element of Z
     
  5. Sep 6, 2012 #4
    Please copy out the question directly from your textbook, with all the punctuation. I still think you've copied down the question wrong, because n=0 is the only integer for which this holds.
     
  6. Sep 6, 2012 #5
    Its "Let nεZ, Prove that if 1-n^2>0, then 3n-2 is an even integer" That's all it give me and yes 0 is the only integer that satifies 1-n^2>0. But my question is that how can this be considered a direct proof when you have to assume that n represents some arbitrary value in the domain or is this not always the case?
     
  7. Sep 6, 2012 #6
    It's a bit awkward, but you can prove it directly...

    Direct proof:

    Claim 1: n=0 satisfies the inequality.
    Proof: this is trivial.
    Claim 2: n=0 is the only integer which satisfies that inequality.
    Proof: Since [itex]n^2>0[/itex] for all [itex]n\neq 0[/itex], we need only consider positive integers. Then for any n>0, we have [itex]n^2\geq 1[/itex], so [itex]1-n^2\leq 0[/itex]. Hence, n=0 is the only integer which satisfies the inequality.
    Claim 3: n=0 satisfies "3n-2 is even"
    Proof: Trivial.

    This is a direct proof because at no point along the way did I assume a "false" hypothesis in order to come to a contradiction. It may seem as though Claim 2 was done by contradiction, but it wasn't. I am claiming for all [itex]n\neq 0[/itex], we do NOT have [itex]1-n^2>0[/itex], equivalently, we DO have [itex]1-n^2\geq 0[/itex].
     
    Last edited: Sep 6, 2012
  8. Sep 6, 2012 #7
    dont you mean for the last part of claim 2 to be 1-n^2≥0?
     
  9. Sep 7, 2012 #8
    Do you mean the claim, or the proof of the claim? In either case, no, I do not. I'm attempting to prove that for all nonzero integers, the inequality 1-n^2>0 does NOT hold. What does it mean for an inequality to NOT hold? It means that the opposite inequality DOES hold. That is, 1-n^2≤0 holds.
     
  10. Sep 7, 2012 #9

    HallsofIvy

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    The crucial point is that the only integer for which "1- n^2> 0" is n= 0.

    If n= 1 or -1, what is 1- n^2? If n= 2 or -2, what is 1- n^2? If n= 3, or -3, ...

    Personally, I would have started this proof with "If n is an integer such that 1- n^2> 0, then n= 0. Therefore, 3n-2= ....
     
  11. Sep 7, 2012 #10
    Well I reread the section in the book on direct proofs and I happen to gloss otver the part when it said
    '''in order to prove for all x in the domain of S'', and I assumed you needed to let any variable such as x,n etc to represent an arbritrary value in the given domain. Now I know that I need to pay attn. to what values in the domain satisfy the hypothesis.
     
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