Quiz2_sol - 12 jk-6 j 30 k-15 1 = 2(6 jk-3 j 15-7 ⇒ 3 mn...

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Solutions to Quiz 2 Question 1 (1) Prove m + 2 is even or n - 5 is odd 3 mn + 1 is odd. (a) If m + 2 is even, we have m + 2 = 2 k m = 2 k - 2 . Thus 3 mn + 1 = 3(2 k - 2) n + 1 = 6 kn - 6 n + 1 = 2(3 kn - 3 n ) + 1 3 mn + 1 is odd. (b) If n - 5 is odd, we have n - 5 = 2 j + 1 n = 2 j + 6 . Thus 3 mn + 1 = 3 m (2 j + 6) + 1 = 6 mj + 18 j + 1 = 2(3 mj + 9 j ) + 1 3 mn + 1 is odd. (2) We could use contrapositive: prove m + 2 is odd and n - 5 is even 3 mn + 1 is even. Since m + 2 is odd and n - 5 is even, let m + 2 = 2 k + 1 m = 2 k - 1 and n - 5 = 2 j n = 2 j + 5 . Thus 3 mn + 1 = 3(2 k - 1)(2 j + 5) + 1 =
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Unformatted text preview: 12 jk-6 j + 30 k-15 + 1 = 2(6 jk-3 j + 15-7) ⇒ 3 mn + 1 is even. Question 2 Assume x / ∈ Q and (-x ) ∈ Q . Then-x = p/q where p,q ∈ Z and q 6 = 0 . Then x =-p/q . Since (-p ) ∈ Z and q 6 = 0 , x is rational which is a contradiction with the premise that x is irrational. Question 3 Disprove: p = 7 and p + 4 = 11 . Both 7 and 11 are prime. 1...
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