induction-sample

# induction-sample - n 1(by the induction hypothesis P n = n...

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Induction sample Notice that in the example below, the colored text would apply to any proof by induction. Theorem. For all n N , 0 + 1 + · · · + n = n ( n +1) 2 . Proof. We prove this by induction on n N . For each n N , let P(n) be the statement “0 + 1 + ... + n = n ( n +1) 2 ”. Base case: P (0) . The statement P (0) is simply 0 = 0(0+1) 2 , which is cer- tainly true. Induction step: n ( P ( n ) = P ( n + 1)) . Let n N be arbitrary, and assume that P ( n ) holds, that is, that 0 + 1 + · · · + n = n ( n + 1) 2 . We want to prove P ( n + 1), that is, we want to show that 0 + 1 + · · · + n + ( n + 1) = ( n + 1)(( n + 1) + 1) 2 . Now 0 + 1 + · · · + n + ( n + 1) = ( 0 + 1 + · · · + n ) + ( n + 1) = n ( n + 1)
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Unformatted text preview: n + 1) (by the induction hypothesis P ( n )) = n ( n + 1) + 2( n + 1) 2 = ( n + 2)( n + 1) 2 = ( n + 1)(( n + 1) + 1) 2 . Then P ( n + 1) follows by the transitivity of equality. Since we proved this by assuming P ( n ), we have P ( n ) = ⇒ P ( n + 1). Since n was arbitrary, we have ∀ n ( P ( n ) = ⇒ P ( n + 1)) Since we proved P (0) and ∀ n ( P ( n ) = ⇒ P ( n + 1)), the statement ∀ nP ( n ) follows by the principle of Mathematical Induction, and the theorem follows. 1...
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