Are the Sets P and Q Equal? Investigating 3-Tuples in the Natural Numbers

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Discussion Overview

The discussion revolves around the question of whether two sets of 3-tuples in the natural numbers, P and Q, are equal. Set P consists of 3-tuples (a, b, c) satisfying the equation a² + b² = c² with the condition a < b, while set Q consists of 3-tuples (x, y, z) defined by specific formulas involving integers m and n. The participants explore methods to prove or disprove the equality of these sets.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that to prove P = Q, one must show that every element of P is in Q and vice versa.
  • Another participant proposes starting with an arbitrary element from Q and checking if it satisfies the conditions of P, specifically whether x² + y² equals z².
  • It is noted that proving the reverse direction (from P to Q) may be more challenging, requiring the identification of integers m and n that satisfy the conditions for elements in P.
  • A participant mentions that the numbers satisfying a² + b² = c² have been completely classified, suggesting that this classification could aid in the investigation.
  • One participant reports finding a specific example (8, 15, 17) that could serve as a counterexample, indicating that the condition a < b may need to be relaxed.

Areas of Agreement / Disagreement

Participants generally agree on the approach to proving set equality but express differing levels of success in proving the reverse inclusion. The discussion remains unresolved regarding whether P and Q are indeed equal.

Contextual Notes

Participants express uncertainty about the implications of the condition a < b and its effect on the equality of the sets. There is also mention of specific tuples that may challenge the equality, but verification of these claims is pending.

honestrosewater
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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)?
 
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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,
Steven
 
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)?

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
 
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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