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Proof by Induction: 2n ≤ n! for All n ≥ 4
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[QUOTE="1MileCrash, post: 4347174, member: 281922"] You didn't use what you were trying to prove, you used what some people call the "induction hypothesis" in your proof itself, which is [i]perfectly[/i] acceptable, and in many cases, required (often it is what drives us to inductively prove in the first place.) HOWEVER, I do understand where your confusion lies. I would like to make the following suggestion to you to avoid this confusion in your proof writing. When making your induction hypothesis, don't say "assume P(n) is true." Say: "Assume that for some k in N, P(k) is true." This distinguishes between the designation "n" to be "some natural number" in the theoem and k to be "[i]THIS natural number, for which I am assuming my theorem holds."[/i] First, notice P(4): 24 = 16 = 4*4 ≤ 4*6 = 4 * 3 * 2 * 1 = 4!. Suppose that for some k in N, P(k) is holds true. Now, consider the case of n=k+1. P(k+1): 2k+1 = 2*2k ≤ 2 * k! ≤ (k+1)*k! = (k+1)! Q.E.D. Basically, I'm saying that you should call the number of your induction hypothesis something different. As mathematicians, we should say what we mean and mean what we say, and you are not really assuming that P(n) is true, because that is your theorem. [/QUOTE]
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Proof by Induction: 2n ≤ n! for All n ≥ 4
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