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Supremum proof help

  1. Oct 3, 2008 #1
    Statement to prove:
    (Note: Q is the set of all rational numbers)
    Let B = {r in Q: r > 0 and r² < 2} and α = sup B. Prove that α² = 2.

    My work on the proof:
    Let B = {r in Q: r > 0 and r² < 2} and α = sup B. Note 1 is in B so B is not empty. By definition of B, 0 is an upper bound of B. Hence the supremum (sup) of B exists.
    Call sup B "α". If α² < 2, then α < √2. So by theorem, there exists an r in B such that
    α ≤ r < √2. Since r < √2, r < 0 which is a contradiction (C!) to r being in B.
    If α² > 2, then α > 2. By density, there exists an r in Q such that
    α > r > √2. Since r > √2, then r² > 2. C! to r B. Therefore, α² = 2. QED.

    Is my logic correct?
     
  2. jcsd
  3. Oct 3, 2008 #2

    statdad

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    Homework Helper

    No = you begin by correctly noting that [tex] 1 \in B [/tex], so how can [tex] 0 [/tex] be an upper bound for the set. This comes back when you state that [tex] r < 0 [/tex]. The notion of a proof by contradiction is the way to go, though.
     
  4. Oct 3, 2008 #3
    So can I say that 1 is an upper bound? I need to state an upper bound in order for the supremum to exist, right? So, how can I name an upper bound?

    So can 2 be an upper bound then since r^2 < 2?
     
  5. Oct 9, 2008 #4
    Redo of proof:

    So I turned this in and got 3/8. I can redo this proof to get 6/8 which would boost my HW grade from a B to an A.

    My original proof was:
    Let B = {r in Q: r > 0 and r² < 2} and (alpha) = sup B. Note 1 is in B so B is not empty. By definition of B, 2 is an upper bound of B. Hence the supremum (sup) of B exists.
    Call sup B "(alpha)". If (alpha)² < 2, then (alpha) < √2. By theorem, there exists an r in B such that (alpha) ≤ r < √2. Since r < √2, r < 0 which is a contradiction (C!) to r being in B.
    If (alpha)² > 2, then (alpha) > 2. By density, there exists an r in Q such that
    (alpha) > r > √2. Since r > √2, then r² > 2. C! to r B. Therefore, (alpha)² = 2. QED.

    Comments my professor wrote on my homework were:
    "Be careful...the theorem works for (beta) < sup or inf < (beta). Here we have (alpha) < √2 or "sup < (beta)" -- the theorem doesn't work. Use density in this case."
    She also wrote, "How did you get r in B?" and that "Yes, density gives r in Q (the rationals)."

    Here is my attempt at a redo:
    Let B = {r in Q: r > 0 and r² < 2} and (alpha) = sup B. Note 1 is in B so B is not empty. By definition of B, 2 is an upper bound of B. Hence the supremum (sup) of B exists.
    Call sup B "(alpha)". If (alpha)² < 2, then (alpha) < √2. By density, there exists an r in B such that (alpha) ≤ r < √2. C! to definition of supremum.
    If (alpha)² > 2, then (alpha) > √2. By density, there exists an r in B such that
    √2 < r < (alpha). In particular, since r > √2, then r > 0 but r² > 2. C! to r being in B. Therefore, (alpha)² = 2. QED.

    Is that better than what I had?
     
  6. Oct 9, 2008 #5
    That is much better.
     
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