Download Instructor's Solutions Manual: Mathematical Proofs: A by Gary Chartrand, Albert D. Polimeni, Ping Zhang PDF

By Gary Chartrand, Albert D. Polimeni, Ping Zhang

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Since 5a2 + 2a is an integer, 5 | (a2 − 1) and so a2 ≡ 1 (mod 5). (b) We can conclude that b2 ≡ 1 (mod 5). 14 (a) Let n ∈ Z. If n ≡ 0 (mod 3) and n ≡ 1 (mod 3), then n2 ≡ n (mod 3). Proof. Assume that n ≡ 0 (mod 3) and n ≡ 1 (mod 3). Then n ≡ 2 (mod 3). Therefore, n = 3a + 2 for some integer a. Thus n2 − n = = (3a + 2)2 − (3a + 2) = 9a2 + 12a + 4 − 3a − 2 9a2 + 9a + 2 = 3(3a2 + 3a) + 2. Since 3a2 + 3a is an integer, n2 − n ≡ 2 (mod 3) and so n2 ≡ n (mod 3). (b) Let n ∈ Z. Then n2 ≡ n (mod 3) if and only if n ≡ 0 (mod 3) and n ≡ 1 (mod 3).

Then m5 − m = = = k 5 − k and so (k + 1)5 − (k + 1) = k 5 + 5k 4 + 10k 3 + 10k 2 + 5k + 1 − k − 1 (k 5 − k) + 5k 4 + 10k 3 + 10k 2 + 5k = 5x + 5k 4 + 10k 3 + 10k 2 + 5k 5(x + k 4 + 2k 3 + 2k 2 + k). Since x + k 4 + 2k 3 + 2k 2 + k ∈ Z, it follows that 5 | m5 − m , which is a contradiction. Suppose next that n < 0. Then n = −p, where p ∈ N and so 5 | p5 − p . Thus p5 − p = 5y for some integer y. Since n5 − n = (−p)5 − (−p) = −(p5 − p) = −(5y) = 5(−y) and −y ∈ Z, it follows that 5 | n5 − n . 30 Proof.

Since x ∈ A and A ∩ B = ∅, it follows that x ∈ / B. Thus x ∈ (A ∪ B) − B and A ⊆ (A ∪ B) − B. 62 The result is an implication, not a biconditional. The proof is complete after the first paragraph. 63 It is wrong to assume that x − 1 = 3q and y − 1 = 3q for some integer q since x and y need not be equal integers. 64 It is wrong to conclude that x ∈ / B simply because (x, y) ∈ / B × C. It should be since (x, y) ∈ / B ×C and y ∈ C, we have x ∈ / B. 65 Recall that |x − y| = |y − x| for every two real numbers x and y.

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