COT3100 SI Final Exam Review

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1 1 Symbols COT3100 SI Final Exam Review Jarrett Wendt Spring 2018 You ve learned a plethora of new Mathematical symbols this semester. Let s see if you know them all and what they re used for. How many symbols can you list, give an example for how you would use it (some have multiple uses), and describe how you would read it? Symbol Example Description p q p and q p q p or q p q p xor q p not p alternatively... p p q p implies q p q p if and only if / iff q 0 4 (mod 2) 0 is congruent to 4 modulo 2. This is indeed different than iff. (mod) (mod 3) 1 is congruent to 4 is congruent to 7 modulo 3. With parentheses, the modulo is applied to all of the connected congruences. mod 6 mod 3 = 0 6 modulo 3 is 0. Without the parentheses, this becomes the mod operator such as in programming. You apply it once like a function. 5 Z 5 is an element of the set of all Integers a, b N[a b b = a] for all possible combinations of Natural numbers a and b, a divided by b times b is just a again a Z[3a = 0] there exists an Integer a such that 3 times a is divides 24 also used for = 1.7 The absolute value of 1.7 is 1.7 {1, 9, 5} The magnitude of the set {1, 9, 5} is 3 {a a Z} Set of a s such that a is an element of the Integers N Z The Natural numbers are a subset of the Integers Z + N The positive Integers are a perfect subset of the Natural numbers

2 A B A cross B cross product of A and B cartesian product of A and B 5.5 = 6 The ceiling of 5.5 is = 5 The floor of 5.5 is 5 Don t get tripped up with ceil and floor when dealing with negatives H n = n i=1 1 i The n th -Harmonic number is the summation from i = 0 to n of 1 n H = i Z + 1 i The Infinite Harmonic series is the summation of all Integers i of 1 n n i=1 f(i) The product from i = 1 to n of some function f(i) It s the same as except you re multiplying the sequence instead of adding You may indeed use the same set notation as well Z N = Z The set of negative Integers union the set of Natural numbers is simply the set of all Integers {1, 0, 6.9} N = 0, 1 The intersection of the given set and the Natural numbers is the set containing 1 and 0 Z N = The negative Integers intersect the Natural numbers is the empty set (they have no elements in common) {} Z N = {} Empty curly braces are another way to represent the empty set n i=1... R = n i=1 Ri The transitive closure of relation R is the union of R composed with itself every time from 1 to n Notice how the is bigger when being used like a summation here. n i=1... n i=1... n i=1... Yeah, you can pretty much use any operator like a summation by making it bigger... R arb iff (a, b) R a relates to b if and only if (a, b) is in the relation R. If the relation were named something other than R, say S then we would indeed say asb instead. [ ] 11 [3] R if R = (mod 4) 11 is in the equivalence class of 3 with respect to the equivalence relation created by modulo 4. Sometimes the subscript R may be omitted, in which case the relation should be implicit. Of course, square braces are also used for matrices as well. < > < 3, 9, 5 > Angle brackets are typically used to denote tuples. Parentheses can also be used, but those are typically reserved for ordered pairs i.e. specifically 2-tuples. a b iff a [b] a is equivalent to b if and only if a is in b s equivalence class. Notice here the relation in implicit. Also notice that a [b] b [a].

3 2 Relations Prove if the following relations are equivalence relations, partial ordering relations, or neither. The following will be helpful Relation R on set A is... Reflexive iff a A [(a, a) R] Irreflexive iff a A [(a, a) / R] Symmetric iff a, b A [(a, b) R (b, a) R] Antisymmetric iff a, b A [((a, b) R (b, a) R) a = b] Transitive iff a, b, c A [((a, b) R (b, a) R) (a, c) R] 2.1 R = {(a, b) a, b Z, a = 3b} R is not Reflexive because (1, 1) / R. ordering relation. Thus, R cannot be an equivalence relation or partial 2.2 R = {(a, b) a, b Z, a > 2b b > 2a} R is not Reflexive because (0, 0) / R. ordering relation. Thus, R cannot be an equivalence relation or partial 2.3 R = {(a, b) a, b Z, a 0 (mod b)} R is Reflexive because a Z, 0 a (mod a) therefore a Z, (a, a) R. R is Transitive. Assume for arbitrary a, b, c Z that (a, b) R, (b, c) R. If (a, b) R then a 0 (mod b). Likewise, b 0 (mod c). By the definition of Modulus, this means an = b and bm = c where n, m Z. Substituting for b we get anm = c. Since nm Z we know that a 0 (mod c) by the definition of Modulus. Thus (a, c) R. R is not Symmetric because (10, 5) R but (5, 10) / R. R is Anti-Symmetric. Assume for arbitrary a, b Z that (a, b) R, (b, a) R. If (a, b) R then a 0 (mod b). Likewise, b 0 (mod a). By the definition of Modulus, this means an = b and bm = a where n, m Z. Substituting for b we get anm = a. This can only be true if n = m = 1 which would imply a = b. Thus, (a, b) R, (b, a) R iff a = b. Because R is Reflexive, Transitive, and Anti-Symmetric it is a partial-ordering relation.

4 2.4 R = {(a, b) a, b Z, f(a) = f(b), f : Z Z} R is Reflexive because if (a, a) / R then that would imply f(a) f(a) which cannot be true for any validly defined function. R is Transitive. Assume for arbitrary a, b, c Z that (a, b) R and (b, c) R. If (a, b) R then f(a) = f(b). Likewise, f(b) = f(c). We may substitute for f(b) and get that f(a) = f(c). Thus, (a, c) R. R is Symmetric. If (a, b) R then f(a) = f(b) Clearly, f(b) = f(a). Thus, (b, a) R Since R is Reflexive, Transitive, and Symmetric it is an equivalence relation. 3 Miscellaneous Proofs 3.1 Let a Z, a 1 (mod 3). Prove a 3 1 (mod 9). Direct proof. Since a 1 (mod 3) we may let a = 3n + 1, n Z. a 3 = (3n + 1) 3 = 27n n 2 + 9n + 1 = 9(3n 3 + 3n 2 + n) + 1 = 9c + 1, c = 3n 3 + 3n 2 + n. Clearly, c Z a 3 1 (mod 9) 3.2 Prove (A B A C) (A = (B C )). Proof by contradiction. Assume (A = (B C )). By DeMorgan s Law, this is equivalent to A (B C = ). Since A is nonempty, let x A. x B x C by premise. This implies x B C. Thus, B C which contradicts our assumption that A C =. Therefore the original statement must hold. 3.3 Prove r, s R + (rs > 100) (r > 10 s > 10).

5 Proof by contraposition. The contrapositive of the given statement is r, s R + (r 10 s 10) (rs 100). Let s prove the contrapositive directly. Assume r, s R +, r 10 s 10. r 10 rs 10s multiply both sides by s s 10 10s 100 multiply both sides by 10 rs 10s 100 combine the previous two rs 100 simplify Since the contrapositive is true, the original statement must be true. 3.4 Find the 10 th term in the expansion of (x + 3) 12. Recall the binomial formula: (x + y) n = n ( ) n x n i y i i Notice that the index starts at 0. So the 1 st term in the expansion will be when i = 0. So the 10 th term in the expansion will be when i = 9. n = 12 and y = 3. ( ) 12 x = x Prove the following logical argument: x [p(x) (q(x) r(x))] x [p(x) s(x)] x [r(x) s(x)]

6 x [p(x) (q(x) r(x))] premise (1) x [p(x) s(x)] premise (2) p(c) (q(c) r(c)) (1), Universal Instantiation on c (3) p(c) s(c) (2), Universal Instantiation on c (4) p(c) (4), Simplification (5) s(c) (4), Simplification (6) q(c) r(c) (4), (5), Modus Ponens (7) r(c) (7), Simplification (8) r(c) s(c) (6), (8), Conjunction (9) x [r(x) s(x)] (9), Universal Generalization on c (10) 3.6 Prove n, x N, x > 2 that n x i < x n+1 Induction on n. Base Case: n = 0. 0 x i = x 0 = 1 < 2 < x < x 0+1 Inductive Hypothesis: Assume for an arbitrary value of n = k N that k x i < x k+1 Inductive Step: Prove for n = k + 1 that k+1 x i < x k+2 k+1 x i = k x i + x k+1 < x k+1 + x k+1 by Inductive Hypothesis = 2x k+1 < x x k+1 because x > 2 = x k+2 Thus, we can conclude that n, x N, x > 2, n xi < x n+1.