T4.1-2 - <?xml version="1.0"...

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(1) = n i 0 i = n(n+1)/2, n 0 for all n N, = n i 0 i = n(n+1)/2 Let P(n): = n i 0 i = n(n+1)/2 We may restate (1) as 2200 nP(n), Universe= N How can we prove such statement? RULE OF INFERENCE . P(0) Premise 2200 n(P(n) -> P(n+1)) Premise ————————— 2200 nP(n) Conclusion If we can prove the two premises (somehow), then we are allowed to say we have proved the conclusion. This rule of inference has the name mathematical induction . 1
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Intuitively, if we prove P for the value 0, then the truth of the second premise says that P is true for the value 1. But if it is true for 1, the second premise says that it is true for 2, and so ad infinitum. In practice P(0) is proved directly. This step is the basis. The second premise is proved by using universal instantiation, and then using the proof method for implications, which is to assume the antecedent and prove the consequent. When we combine these we get a statement like this: Assume P(n) is true for an arbitrary n. Based on this assumption , show that P(n+1) is true. 2
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Example: P(n): = n i 0 i = n(n+1)/2 P(0) is true. (Basis) because = 0 0 i i=0(0+1)/2=0 Assume the truth of P(n). Based on the assumption, prove P(n+1): + = 1 0 n i i = (n+1)(n+2)/2. Proof: + = 1 0 n i i = = n i 0 i + (n+1) = n(n+1)/2 + (n+1) = [n(n+1) + 2(n+1)]/2 = (n+1)(n+2)/2 QED 3
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To prove 2200 nP(n): 1. Identify the propositional function P(n)! 2. Prove P(0) 3. Assume P(n) for arbitrary n 4. Based on that assumption, prove P(n+1) E.g. Prove n < 2 n (Same as 2200 n(n<2 n )) 1. P(n): n < 2 n 2. P(0): 0<2 0 =1 is True 3. Assume n < 2 n 4. Prove: n+1 < 2 n+1 Proof: n < 2 n Induction hypothesis 1 2 n Fact about natural nos. n+1 < 2
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T4.1-2 - &lt;?xml version=&quot;1.0&quot;...

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