hw1[1]

# hw1[1] - 2 5 ² 1 1 = 7 and therefore P 1 does not hold 2.1...

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HOMEWORK 1: SOLUTIONS/HINTS 1.3 We prove this by induction on n . For n = 1 the equality obviously holds. For the induction step n 7! n + 1 de±ne S n := n X k =1 k 3 ; then S n +1 = S n + ( n + 1) 3 = (1 + ::: + n ) 2 + ( n + 1) 3 by induction assumption. As (1 + ::: + n ) = 1 2 n ( n + 1), we ±nally get S n +1 = 1 4 n 2 ( n + 1) 2 + ( n + 1) 3 = 1 4 ( n + 1) 2 ( n + 2) 2 = (1 + ::: + ( n + 1)) 2 : 1.6 Proof by induction; For n = 1 this is clear. By induction assumption 11 n ± 4 n = 7 m for a natural number m: We compute 11 n +1 ± 4 n +1 = 11(11 n ± 4 n ) + 4 n (11 ± 4) = 11 ² 7 m + 7 ² 4 n = 7(11 m + 4 n ) ; which is divisible by 7. 1.8 We only prove part b). For n = 4 we get 4! = 24 > 16 = 4 2 . Let now n ³ 4, then ( n + 1)! = n ! ² ( n + 1) > n 2 ( n + 1) by induction. By part (a) we have that n 2 > ( n + 1), hence ( n + 1)! > ( n + 1) 2 : 1.11 For any n either n or n +5 is even, hence n ( n +5) = n 2 +5 n is always even and therefore n 2 + 5 n + 1 is always odd, so P n is never true. In conclusion, we really have to check the base case. In this example we get 1
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Unformatted text preview: 2 +5 ² 1+1 = 7 and therefore P 1 does not hold. 2.1 Let r = p 3, then r 2 ± 3 = 0. Suppose now that r is rational and therefore is equal to p q for some integers p and q without any common factors. By the Rational Zeros Theorem we can now conclude that p divides 3 and q divides 1. Hence r has to be one of the following numbers ´ 1 ; ´ 3. But these numbers obviously do not satisfy the equation x 2 ± 3 = 0 and therefore cannot be r , which is a contradiction. Consequently, p 3 is not rational. 2.3 Use the Rational Zeros Theorem again or notice that if r = (2 + p 2) 1 2 is rational, then p 2 = r 2 ± 2 would be rational as well, which we know is not the case. 1...
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## This note was uploaded on 02/12/2012 for the course MATH 131a taught by Professor Hitrik during the Spring '08 term at UCLA.

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