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Math 1271 3071 Assignment 3 Solutions

Math 1271 3071 Assignment 4 Solutions
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Discrete Mathematics  (Mathematics 1271)

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Academic year: 2023/2024
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Math 1271 / 3071: Assignment 3 Solutions

Show all your work with all necessary steps.

  1. (4 points) Express the number (12) 5 in base 10 as mn where m, n are positive integers such that m, n do not have common divisor larger than 1, that is, gcd(m, n) = 1. Solution: Let x = (12) 5. Then (100) 5 x = (1234) 5. (100) 5 x−x = (1234) 5 −(12) 5 = (1222) 5. 25 x − x = 187, x = 187/24. You can also use geometric series to find the number. (12) 5 = 5 + 2 + 3/5 + 4/ 52 + 3/ 53 + 4 / 44 + ... = 7 + 3/ 5 · (1 + 1/25 + 1/ 252 + ...) + 4/ 25 · (1 + 1/25 + 1/ 252 + ...) = 7 + 3/ 4 · 1 /(1 − 1 /25) + 4/ 25 · 1 /(1 − 1 /25) = 7 + 5/8 + 1/6 = 187/ 24.

  2. (4 points) Let xn be a sequence of real numbers defined as x 1 = 1 and xn+1 = 1 + x 1 n for n ≥ 1. Use the Principle of Mathematical Induction to show that 1 < xn < 2 for all n ≥ 1. Solution: Let P (n) : 1 < xn < 2. Basis step: x 1 = 1, 1 < x 1 < 2. So P (1) is true. Inductive step: For any n ≥ 1, assume P (n) is true (or you can use strong induction, P (1), ..., P (n) are all true), that is, 1 < xn < 2. Consider xn+1 = 1 + x 1 n. We know 1 < 1 + 1/ 2 < 1 + 1/xn < 1 + 1/1 = 2. That is, 1 < xn+1 < 2. So P (n + 1) is true. By mathematical induction, 1 < xn < 2 for all n ≥ 1.

  3. (4 points) Use mathematical induction to show that 1 + 3 + 5 + · · · + (2n − 1) = n 2 for all positive integers n. Solution: We want to show P (n) : 1 + 3 + 5 + · · · + (2n − 1) = n 2 for all n ≥ 1. Basis step: P (1): 1 = 1 2 is true. Inductive step: Assume P (n) is true, that is, 1 + 3 + ... + (2n − 1) = n 2 for every n ≥ 1. Now 1+3+...+(2n−1)+(2n+1) = n 2 +(2n+1) = (n+1) 2. Thus P (n+1) is true. By mathematical induction, 1 + 3 + 5 + · · · + (2n − 1) = n 2 for all n ≥ 1.

  4. (4 points) Show that in any Boolean algebra B, x + x′y = x + y for any x, y ∈ B. Indicate which law is used for each step. Solution: x + x′y = (x + x′)(x + y) (Distributive Law or B3(a)) = 1(x + y) (Complement Law or B5(a)) = x + y (Identity Law or B4(b)).

  5. (6 points) The Boolean expression α(x, y, z) = xy + xy′z + (y + z)′. Evaluate (a) α(0, 0 , 0). (b) α(1, 1 , 1). Solution: α(0, 0 , 0) = 00 + 00′0 + (0 + 0)′ = 0 + 010 + 0′ = 0 + 0 + 1 = 1. α(1, 1 , 1) = 11 + 11′1 + (1 + 1)′ = 1 + 101 + 1′ = 1 + 0 + 0 = 1. [Note: 1 + x = 1 for any x.]

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  1. (4 points) Find a Boolean expression corresponding to the circuit Solution: The Boolean expression is (xy)′ + (y + z).

  2. (4 points) Construct a circuit to implement xy′ + x′y. Solution:

  3. (4 points) Let S = { 1 , 2 , 3 } and T = { 1 , 2 , 3 , 4 , 5 , 6 }. Let ρ be a relation from S to T such that xρy means 3 | (x + y). Find ρ as a subset of S × T. Solution: If x = 1, then when y = 2, 5, we know 3 | (1 + y). If x = 2, then when y = 1, 4, we know 3 | (2 + y). If x = 3, then when y = 3, 6, we know 3 | (3 + y). Thus ρ = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6)}. [This relation is simple. You can directly write down the solution without any steps.]

  4. (8 points) Let S = { 1 , 2 , 3 }. Let ρ be a relation on S such that xρy means x ≤ 2 y.

(a) Find ρ and ρ− 1 as subsets of S × S. Solution: If x = 1, then from 1 ≤ 2 y, we have y = 1, 2 , 3. If x = 2, then from 2 ≤ 2 y, we have x = 1, 2 , 3. If x = 3, then from 3 ≤ 2 y, we have x = 2, 3. Thus

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Math 1271 3071 Assignment 3 Solutions

Course: Discrete Mathematics  (Mathematics 1271)

23 Documents
Students shared 23 documents in this course

University: Lakehead University

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Math 1271 / 3071: Assignment 3 Solutions
Show all your work with all necessary steps.
1. (4 points) Express the number (12.34)5in base 10 as m
nwhere m, n are positive integers such
that m, n do not have common divisor larger than 1, that is, gcd(m, n) = 1.
Solution: Let x= (12.34)5. Then (100)5x= (1234.34)5. (100)5x−x= (1234.34)5−(12.34)5=
(1222)5.25x−x= 187, x= 187/24.
You can also use geometric series to find the number. (12.34)5= 5 + 2 + 3/5 + 4/52+ 3/53+
4/44+... = 7 + 3/5·(1 + 1/25 + 1/252+...)+4/25 ·(1 + 1/25 + 1/252+...) = 7 + 3/4·1/(1 −
1/25) + 4/25 ·1/(1 −1/25) = 7 + 5/8+1/6 = 187/24.
2. (4 points) Let xnbe a sequence of real numbers defined as x1= 1.5 and xn+1 = 1 + 1
xnfor n≥1.
Use the Principle of Mathematical Induction to show that 1 < xn<2 for all n≥1.
Solution: Let P(n):1< xn<2.
Basis step: x1= 1.5, 1 < x1<2. So P(1) is true.
Inductive step: For any n≥1, assume P(n) is true (or you can use strong induction, P(1), ..., P (n)
are all true), that is, 1 < xn<2. Consider xn+1 = 1 + 1
xn. We know 1 <1+1/2<1+1/xn<
1+1/1 = 2. That is, 1 < xn+1 <2. So P(n+ 1) is true.
By mathematical induction, 1 < xn<2 for all n≥1.
3. (4 points) Use mathematical induction to show that 1+ 3 + 5 + · · · + (2n−1) = n2for all positive
integers n.
Solution: We want to show P(n) : 1 + 3 + 5 + · · · + (2n−1) = n2for all n≥1.
Basis step: P(1): 1 = 12is true.
Inductive step: Assume P(n) is true, that is, 1 + 3 + ... + (2n−1) = n2for every n≥1. Now
1+3+...+(2n−1)+(2n+1) = n2+(2n+1) = (n+1)2. Thus P(n+1) is true. By mathematical
induction, 1 + 3 + 5 + · · · + (2n−1) = n2for all n≥1.
4. (4 points) Show that in any Boolean algebra B,x+x0y=x+yfor any x, y ∈B. Indicate which
law is used for each step.
Solution:
x+x0y= (x+x0)(x+y) (Distributive Law or B3(a))
= 1(x+y) (Complement Law or B5(a))
=x+y(Identity Law or B4(b)).
5. (6 points) The Boolean expression α(x, y, z) = xy +xy0z+ (y+z)0. Evaluate
(a) α(0,0,0).
(b) α(1,1,1).
Solution: α(0,0,0) = 00 + 0000 + (0 + 0)0= 0 + 010 + 00= 0 + 0 + 1 = 1.
α(1,1,1) = 11 + 1101 + (1 + 1)0= 1 + 101 + 10= 1 + 0 + 0 = 1.[Note: 1 + x= 1 for any x.]
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