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You are not logged in. #1 20060817 23:52:16
Final question  ProveThis is the last question! Thx! #2 20060818 04:33:31
Re: Final question  ProveEdit: The proof below is invalid, as But I'm leaving it here in case it sparks any other ideas. Without loss of generality, we can let a ≥ b. Note that if a² = .5, then b² = .5. Also note that if b² > .5, it must be that a² < .5. But this means that a < b, contradicting a ≥ b. Thus, we can let a² ≥ .5 ≥ b². Since b^4 must be nonnegative, a^4 + b^4 ≥ .5. QED "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #3 20060818 04:46:26
Re: Final question  ProveSecond try: And Thus, Since ∈² ≥ 0, a^4 + b^4 ≥ .5. QED Edit: Interesingly enough, given a^2 (or just a), you can easily calculuate exactly how much greater than .5 a^4 + b^4 will be, without ever having to know what b is or rasing anything to a 4th power. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #4 20060822 05:40:13
Re: Final question  ProveAnother: . Plugging in: Last edited by krassi_holmz (20060822 05:41:38) IPBLE: Increasing Performance By Lowering Expectations. 