**MATH1081 - DISCRETE MATHEMATICS ASSIGNMENT 2020**

T1 1. Consider the following sets: S = {180n + 33 | n ∈ Z} , T = {18n − 3 | n ∈ Z} , and U = {20n + 13 | n ∈ Z} . (a) Show that S is a proper subset of T. Let s ∈ S. By the definition of S, there exists a k ∈ Z such that s = 180k + 33, = 180k + 36 − 3, = 18(10k + 2) − 3, where 10k + 36 is an integer. Hence, s ∈ T and so S ⊆ T. Now, observe that 15 ∈ T since 18(1) − 3 = 15 and 1 ∈ Z. However, if 180n + 33 = 15, 180n = −18, n = − 1 10 . Since n /∈ Z, 15 ∈/ S and hence T * S. Therefore, S is a proper subset of T. (b) Show that S is a proper subset of U. Let s ∈ S. By the definition of S, there exists a k ∈ Z such that s = 180k + 33, = 180k + 20 + 13, = 20(9k + 1) + 13, where 20k + 13 is an integer. Hence, s ∈ U and so S ⊆ U. Now, observe that 13 ∈ U since 20(0) + 13 = 13 and 0 ∈ Z. However, if 180n + 33 = 13, 180n = −20, n = − 1 9 . Since n /∈ Z, 13 ∈/ S and hence U * S. Therefore, S is a proper subset of U. 1 MATH1081 - DISCRETE MATHEMATICS ASSIGNMENT 2020 T1 2 (c) Show that there is no containment relation between T and U. Recall from (a) that 15 ∈ T. However, if 20n + 13 = 15, 20n = 2, n = 1 10 . Since n /∈ Z, 15 ∈/ U and hence U * T. Similarly, recall from (b) that 13 ∈ U. However, if 18n − 3 = 13, 18n = 16, n = 8 9 . Since n /∈ Z, 13 ∈/ T and hence T * U. Therefore, there is no containment relation between T and U. 2. A relation is defined on N by x y if and only if x = y + 4k for some non-negative integer k. Prove that is a partial order. For reflexivity, observe that a = a + 4(0), where 0 is a non-negative integer. Hence, a a for all a ∈ N and thus the relation is reflexive. For antisymmetry, suppose that a b and b a where a, b ∈ N. By the definition of , there exists non-negative integers k and l such that a = b + 4k and b = a + 4l. By substituting b we have a = a + 4k + 4l, 0 = 4k + 4l. Since k and l are both non-negative integers, this can only be true if k = l = 0. Hence, a = b and so the relation is antisymmetric. For transitivity, suppose that a b and b c where a, b, c ∈ N. By the definition of , there exists non-negative integers k and l such that a = b + 4k and b = c + 4l. By substituting b we have a = c + 4l + 4k = c + 4(l + k), where l + k is a non-negative integer. Hence, a c and the relation is transitive. Therefore, is a partial order as required. MATH1081 - DISCRETE MATHEMATICS ASSIGNMENT 2020 T1 3 3. Prove that 17 | 134n+1 + 185n+1 + 3 for all n ∈ N. We will use a proof by induction on n for the statement above. For the base case where n = 0, we have 134(0)+1 + 185(0)+1 + 3 = 131 + 181 + 3 = 34 = 17(2), which is divisible by 2. Hence, the statement is true for n = 0. For the inductive hypothesis, assume the statement is true for n = k, where k ∈ N. Thus, 134k+1 + 185k+1 + 3 = 17A for some positive integer A. Rearranging this equation gives 134k+1 = 17A − 185k+1 − 3. For the inductive step, we need to prove the statement is true for n = k + 1, such that 134(k+1)+1 + 185(k+1)+1 + 3 = 17B, 134k+5 + 185k+6 + 3 = 17B for some positive integer B. Proof. Using index laws, we have 134k+5 + 185k+6 + 3 = 134 · 134k+1 + 185 · 185k+1 + 3. Now, using our assumption in the inductive hypothesis, 134k+5 + 185k+6 + 3 = 134 (17A − 185k+1 − 3) + 185 · 185k+1 + 3. Finally, by expanding and simplifying, 134k+5 + 185k+6 + 3 = 134 · 17A − 134 · 185k+1 − 134 · 3 + 185 · 185k+1 + 3 = 134 · 17A + 185 · 185k+1 − 134 · 185k+1 − 134 · 3 + 3 = 134 · 17A + (185 − 134 ) · 185k+1 − 134 · 3 + 3 = 134 · 17A + 1861007 · 185k+1 − 85680 = 17(134A + 109471 · 185k+1 − 5040) = 17B, where B = 134A + 109471 · 185k+1 − 5040, which is a positive integer. We have shown that the statement is true for the base case n = 0, and that the statement is true for n = k + 1 if it is true for a particular n = k. Therefore, the statement is true for all n ∈ N by the principle of mathematical induction.

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