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Proving two sets are equal

  1. Apr 12, 2005 #1


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    Sorry if this doesn't come out right- figuring out how to state it is part of my problem.
    Let P be the set of all 3-tuples (a, b, c) in N such that a2 + b2 = c2 and a < b.
    Let Q be the set of all 3-tuples (x, y, z) in N such that x = m(2n + 1), y = m(2n2 + 2n), and z = m(2n2 + 2n + 1) for some m and n in N.
    I want to figure out if P = Q. I know that for all sets S and T, (S = T) is equivalent to (S is a subset of T and T is a subset of S), but I don't know where to begin. Should I try to prove that for all 3-tuples t in N, if t is in P, then t is in Q and if t is in Q, then t is in P? Should I state it a different way (I tried, but it seemed more complicated)?
    Last edited: Apr 12, 2005
  2. jcsd
  3. Apr 12, 2005 #2
    What you want to do is take an arbitrary element of one set and show it belongs in the other set. One way appears easy. Take a 3-tuple in Q (x,y,z). Leaving m and n undefined consider [tex]x^2 + y^2[/tex]. Does it equal [tex]z^2[/tex]. The other way is harder. Take a 3-tuple in P (a,b,c). You know that [tex]a^2+b^2= c^2[/tex]. Can you find m and n integers such that the first two requisites are satisfied? If you can the 3rd requisite is satisfied automatically by the work done earlier.

    Also numbers that satisfy [tex]a^2+b^2 =c^2[/tex] have been completely classified so looking that up may help.

    Good luck,
  4. Apr 12, 2005 #3


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    Yes, what you have stated sounds correct, and is a typical approach. One alternative approach is:

    x is in P ==> x is in Q
    x isn't in P ==> x isn't in Q
  5. Apr 13, 2005 #4


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    Yeah, Q is a subset of P was easy- just manipulation. I couldn't get anywhere trying to prove the other half, so I gave up, went searching, and found (8, 15, 17). Eh. x = d(n2 - m2), y = d2nm, and z = d(n2 + m2), for d in N and relatively prime n > m in N supposedly works (I haven't checked), but by this my counterexample above is actually (15, 8, 17), so I guess I just need to lose a < b - my brain is pudding now, so I'll look at it later. Thanks for the help. :smile:
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