CS 1050A: Constructing Proofs
Solutions to Problem Set 5
Problem 1
1. f(n) is O(g(n)) and h(n) is O(g(n))
⇒f(n)≤c1·g(n)∀n>N1and h(n)≤c2·g(n)∀n>N2
⇒f(n) + h(n)≤(c1+c2)·g(n)∀n > max(N1, N2)
⇒(f(n) + h(n) is O(g(n)).
2. f(n) is O(g(n)) and h(n) is O(g(n))
⇒f(n)≤c1·g(n)∀n>N1and h(n)≤c2·g(n)∀n>N2
⇒f(n)·h(n)≤(c1·c2)·g(n)2∀n > max(N1, N2)
⇒(f(n)·h(n) is O(g(n)2).
Problem 2
1. a|cand b|d
⇒c=x1·aand d=x2·bfor some x1, x2∈Z
⇒c·d=x1x2ab ⇒ab|cd
2. Let gcd(a, b) = g1and gcd(c, d) = g2
g1|aand a|c⇒g1|c
g1|band b|d⇒g1|d
g2=s·c+t·dfor some s, t, ∈Z
⇒g1|g2
Problem 3
1. Set of Divisors of 150 = {1,2,3,5,6,10,15,25,30,50,75,150}
Set of Divisors of 42 = {1,2,3,6,7,14,21,42}
Therefore, gcd(150,42) = 6
2. 150 = 21·31·52
42 = 21·31·71
Therefore gcd(150,42) = 2min(1,1) ·3min(1,1) ·5min(2,0) ·7min(0,1) = 6
3. 150 = 3 ·42 + 24
42 = 1 ·24 + 18
24 = 1 ·18 + 6
18 = 3 ·6+0
Therefore gcd(150,42) = 6
24 = 150 −3(42)
18 = 42 −24 = 42 −150 + 3(42) = 4(42) −150
6 = 24 −18 = 150 −3(42) −4(42) + 150 = 2(150) + (−7)42
Therefore s= 2 and t=−7
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