# Supremum proof help

1. Oct 3, 2008

### pzzldstudent

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. Oct 3, 2008

No = you begin by correctly noting that $$1 \in B$$, so how can $$0$$ be an upper bound for the set. This comes back when you state that $$r < 0$$. The notion of a proof by contradiction is the way to go, though.

3. Oct 3, 2008

### pzzldstudent

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?

4. Oct 9, 2008

### pzzldstudent

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?

5. Oct 9, 2008

### e(ho0n3

That is much better.