Discrete Mathematics For Computing - Assignment Example

Student Name Disсrеtе Маthеmаtiсs for Соmрuting Institution Name Date Disсrеtе Маthеmаtiсs for Соmрuting Question 1 Answer (a) Truth table for [(pvq) ^ (rvp)] v (rvp) p q P ^ q P v q p q rvp r v q F F F F F T T F T F T T T F T F F T T F F T T T T F T T The expression is neither a tautology nor a contradiction because the values are a combination of True and False for all the premises. When a compound statement is true for every value of all its truth vales are true for every truth-values of its assignments, then the expression is referred to as a tautology, denoted by T0. On the other hand, it is known as a contradiction if the compound statement is false for everyone of such assignment. Answer (b) Use of Logic Expression: [P ^ (p  q)] q p v p p and p^pp p v Fp and p ^ To p p v pTo and p ^ pFo P v To To and p ^ Fo Fo P v (p ^ q) p and p ^ (p v q) p Therefore: P q = () d = We now apply Distributive Law P v (q ^ r) (p v q) ^ (p v r) P^(q v r) (p ^q)v(p ^ r) We now apply Idempotent Law P v pp where pp We now apply Identity Law P v Fop where p ^ Top We now apply Inverse Law P vp To where p^pFo We now apply Domination Law P v ToTo and P ^ FoFo Finally, we apply Absorption Law P v (p ^ q) p, p ^ (p v qp) Question 2 Answer (a). The relation R on the set A is an equivalent relationship since it is symmetric, reflexive and transitive at the same time (Rosen 237). The relation R on a set A is reflexive because (1, 1) R for every member 1A and (2, 2) R for every member 2A. The relation R on a set A is symmetric because (2, 1) R every time (1, 2) R for every member 1, 2A. The relation R on a set A transitive because whenever (1, 2) R and (2,-1) R, then (1,-1c) R for 1, 2, -1A. (b). Matrix representation of the relationship R = Question 3 The geometric progression is 40, 41, 42, 43, 44… 4n. The progression is 1, 4, 16, 64, 256 …. 4n Objective: To prove that 1 + 4 + 16 + 64 +… + 4n = (4 n + 1-1)/3 We split the expression into two parts then we try to confirm if the values of the two sides are equal. x + xy + xy2 +xy3+ …+xbn-1 = x (bn - 1)/b - 1 Considering the sum of the next integer k + 1 [x + xb + xb2 + … +xbk-1+ xbk [x + xb + xb2 + … +xbk-1] + xbk [x + xb + xb2 + … +xbk-1] + xbk = xbn - 1)/b - 1+ xbk xbn - 1) / b – 1 + xbk = (xbk – x + xb k + 1- xbk) / b-1 = (xb k+1-x) / b - 1 = (xb k+1 - 1) / b – 1 It therefore proves that x + xb + xb2 + xbk-1 + xbk = x (bk+1 - 1) / b -1 We take an instance where n = 4 The solution will be: Sum1= 1 + 4 + 16 + 64 + 256 = 341 Now we have an objective to prove that (4 n + 1 - 1) / 3 = 341 Solution N = 4 We substitute n with 4 Sum2 = (4 4+ 1 - 1) / 3 Sum2= (45 - 1) / 3 Sum2 = (1024 - 1) / 3 Sum2 = 1023 / 3 Sum2 = 341 Sum1 = Sum2, hence, the two sides are confirmed to be equal. (b) We have an objective to calculate Sum (4i), where I = 0, 1, 2, 3… 14 and 15 It means we are taking n = 16. Sum (4i) = (4 16+1 -1) / 3 Sum (4i) = (417 - 1) / 3 Sum (4i) = (1073741824 - 1) / 3 = 357913941. We now take Sum (4i) = 357913941 Question 4 (a) The sum of the sequence is S (n) When n = 0, S (0) = 0 When n = 1, S (1) = 1 When n > = 2, S (n) = S (n – 1) + S (n - 2) We find this to be a recursive formula. However, our objective is to produce a non-recursive formula for the same sequence. For any integer n, being equal to or greater than 0, let s(n) be the recursive number of ways in which we are able to write the number n as its total by use of three numbers 1, 2, and 3 in whichever way they are arranged. We will have the following Non-recursive formula: S (n) = S (n − 3) + S (n − 2) – S (n − 5) + 1 Beginning with the numbers S (0) = 1, S (1) = 1, S (2) = 2, S (3) = 3, S (4) = 4, S (5) = 5, S (6) = 7, S (7) = 8, S (8) = 10, S (9) =12, S (10) = 14. (b) S (10) = S (10 - 3) +S (10 - 2) – S (n - 5) + 1 S (10) = S (7) + S (8) – S (5) + 1 S (10) = 8 + 10 - 5 +1 S (10) = 19 - 5 S (10) = 14. Question 5 (a) The original Boolean Expression is xy + xz’ + x’y +x’y’ Digital Circuit that corresponds to this expression is drawn below (b). Writing the expression in a disjunctive normal form, we obtain the expression below x (y + z’) + x’(y + y’) = xx’ (yz + y’z + y’z’ + yz’) c. Karnaugh Map yz y’z y’z’ yz’ x 1 1 0 1 x’ 1 0 0 0 Using the map, we can now simplify the expression by using the points at which the map is indicating a 1. The expression now becomes: x (yz + y’z + yz’) + x’yz = xyz + xy’z +xyz’ + x’yz (d) Digital circuit for the simplified expression Works Cited Rosen, Kenneth. Discrete Mathematics and Its Applications. New York: McGraw-Hill College, 2012. Print. Read More