- #1
pzzldstudent
- 44
- 0
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?
(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?