What is the decimal (base 10) representation of the binary number ? Show your work and place your final answer in the box.

Transcription

1 Question 1. [10 marks] Part (a) [2 marks] What is the decimal (base 10) representation of the binary number ? Show your work and place your final answer in the box = = Part (b) [2 marks] What is the binary (base 2) representation of the decimal number 41? Show your work and place your final answer in the box. 41 = = 1 + 2(20) = 1 + 2(2 10) = = (1 + 4) = Part (c) [2 marks] Using only base 4, determine the sum of the numbers (1330) 4 and (2211) 4. Show your work and place your final answer in the box Part (d) [4 marks] Using only base 4, determine the product of the numbers (12) 4 and (3202) 4. Show your work and place your final answer in the box Page 1 of 10 cont d...

2 Use the space on this blank page for scratch work, or for any answer that did not fit elsewhere. Clearly label each answer with the appropriate question and part number. Page 2 of 10 cont d...

3 Question 2. Consider the statement [18 marks] r R +, q R +, 3r 2 + q 2 r = q + 1 r q Write a detailed structured proof of the statement. Part marks will be given for having correct elements of the proof structure. ssume r R + ssume q R + ssume r = q Then, 3r 2 + q 2 r = 3r 2 + r 2 r = 4r 2 r = r r + 1 = q + 1 Then, 3r 2 + q 2 r q + 1 Then, 3r 2 + q 2 r = q + 1 r q # [ I (indirect)] Then, q R +, 3r 2 + q 2 r = q + 1 r q # [ I] Then, r R +, q R +, 3r 2 + q 2 r = q + 1 r q # [ I] Page 3 of 10 cont d...

4 Question 3. [18 marks] Recall that a mod b = c means that the remainder of division of a by b is c. When c < b, we can say that a mod b = c can be formulazed as: q N, a = qb + c. Consider the statement The remainder of division of a 2 by 4 is either 1 or 0, for any natural number a. Part (a) [2 marks] Translate the statement into symbolic notation. a N, a 2 mod 4 = 1 a 2 mod 4 = 0 Part (b) [16 marks] Write a detailed structured proof of the statement. Part marks will be given for having correct elements of the proof structure. ssume a N Case 1. a is even Then k N, a = 2k Let k 0 N be such that a = 2k 0 # E Then, a 2 = (2k 0 ) 2 = 4k0 2 Then, q N, a 2 = 4q # k0 2 N Then, a 2 mod 4 = 0 Then, a 2 mod 4 = 0 a 2 mod 4 = 1 # [ I] Case 2. a is odd Then k N, a = 2k + 1 Let k 0 N be such that a = 2k # E Then, a 2 = (2k 0 + 1) 2 = 4k k = 4(k0 2 + k 0) + 1 Then, q N, a 2 = 4q + 1 # k0 2 + k 0 N Then, a 2 mod 4 = 1 Then, a 2 mod 4 = 0 a 2 mod 4 = 1 # [ I] Then, a 2 mod 4 = 0 a 2 mod 4 = 1 Then, a N, a 2 mod 4 = 0 a 2 mod 4 = 1 # Proof by cases, a must be either even or odd # [ I] Page 4 of 10 cont d...

5 Use the space on this blank page for scratch work, or for any answer that did not fit elsewhere. Clearly label each answer with the appropriate question and part number. Page 5 of 10 cont d...

6 Question 4. [18 marks] Recall that for a b can be read as a divides b, and can be formalized as n N, b = na. Consider the following statement: For natural numbers a and b, If 4 divides ab, then 4 divides a or 4 divides b. This statement is equivalent to the symbolic statement: a N, b N, 4 ab 4 a 4 b. Now consider the following argument: ssume a N. ssume b N. ssume 4 ab. Then n N, ab = 4n Let n 0 N be such that ab = 4n 0 Then, a = 4n 0 /b Let k = n 0 /b Then, k N Then, a = 4k Then, k N, a = 4k # [ I] Then, 4 a Then, 4 a 4 b # [ I] Then, 4 ab 4 a 4 b # [ I] Then, b N, 4 ab 4 a 4 b # [ I] Then, a N, b N, 4 ab 4 a 4 b # [ I] # [ E] Part (a) [2 marks] This argument is not a correct proof of the statement. Explain the flaw in the argument. Sample Solution: The incorrect line is the one stating that k N. The proof defined k = n 0 /b, so there is no reason why this must be a natural number (for example, take n 0 = 1 and b = 2) Page 6 of 10 cont d...

7 Part (b) [16 marks] Determine whether the statement a N, b N, 4 ab 4 a 4 b is true or false. If it is true, prove it. If it is false, disprove it. Sample Solution: The statement is false. To show that, we need to prove the netaion. The negation of the statement is: a N, b N, 4 ab (4 a) (4 b) Let a 0 = 2 Then a 0 N Let b 0 = 2 Then b 0 N Then a 0 b 0 = 2 2 = 4, and 4 4 So, 4 a 0 b 0 But (4 2) Then, (4 a 0 ) Then, (4 b 0 ) Then, 4 a 0 b 0 (4 a 0 ) (4 b 0 ) Then, b N, 4 a 0 b (4 a 0 ) (4 b) Then, a N, b N, 4 ab (4 a) (4 b) # [ I] # [ I] # [ I] Page 7 of 10 cont d...

8 Use the space on this blank page for scratch work, or for any answer that did not fit elsewhere. Clearly label each answer with the appropriate question and part number. Page 8 of 10 cont d...

9 [ I] negation introduction ssume. contradiction [ I] conjunction introduction B B [ I] disjunction introduction B B [ E] negation elimination contradiction [ E] conjunction elimination B B [ E] disjunction elimination B B B B [ I] implication introduction (direct) ssume. B B (indirect) ssume B. B [ I] equivalence/bi-implication introduction B B B [ E] implication elimination (Modus Ponens) B B (Modus Tollens) B B [ E] equivalence/bi-implication elimination B B B [ I] universal introduction ssume a D. P (a) x D, P (x) [ I] existential introduction P (a) a D x D, P (x) [ E] universal elimination x D, P (x) a D P (a) [ E] existential elimination x D, P (x) Let a D such that P (a). Standard Equivalences (where P, Q, P (x), Q(x), etc. are arbitrary sentences. ll quantifications are over a domain D.) Commutativity P Q Q P P Q Q P P Q Q P ssociativity P (Q R) (P Q) R P (Q R) (P Q) R Identity P (Q Q) P P (Q Q) P bsorption P (Q Q) Q Q P (Q Q) Q Q Idempotency P P P P P P Double Negation P P DeMorgan s Laws (P Q) P Q (P Q) P Q Distributivity P (Q R) (P Q) (P R) P (Q R) (P Q) (P R) Implication P Q P Q Biconditional P Q (P Q) (Q P ) Renaming (where P (x) does not contain variable y) x, P (x) y, P (y) x, P (x) y, P (y) Quantifier Negation x, P (x) x, P (x) x, P (x) x, P (x) Quantifier Commutativity x, y, S(x, y) y, x, S(x, y) x, y, S(x, y) y, x, S(x, y) Quantifier Distributivity (where S does not contain variable x) S x, Q(x) x, S Q(x) S x, Q(x) x, S Q(x) S x, Q(x) x, S Q(x) S x, Q(x) x, S Q(x) ( x, P (x)) ( x, Q(x)) x, (P (x) Q(x)) ( x, P (x)) ( x, Q(x)) x, (P (x) Q(x)) Page 9 of 10 cont d...

10 Total Marks = 64 Page 10 of 10 End of Solutions