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by armeehn
1157 days ago
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I'm trying to understand 5). If you're claiming that both (A) y > x, and (B) x^y > y^x hold, then x^y > y^x ("will always be larger") holds. You're satisfying your claim by assumption. Nothing new is deduced. However, if only A) needs to be satisfied: y = 3, x = 2 is a counterexample, as x^y = 2^3 = 8 < 9 = 3^2 = y^x. edit: looks like someone had the same thought as me as I was typing my reply! |
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It's hilarious that the FIRST two integers greater than one form a counter example to their "proof".