CS336
Homework Assignment ♯3 Solutions
Section 1.5
20. U = Z, N(x) : x is negative, P (x) : x is positive
a) ∀x∀y[N(x) ∧ N(y) → P (xy)]
]
b) ∀x∀y[P (x) ∧ P (y) → P ( x+y
2
c) ∃x∃y[N(x) ∧ N(y) ∧ ¬N(x − y)]
d) ∀x∀y[|x + y| ≤ |x| + |y|]
36 abd
a) U = all people, P(x): x has lost more than 1000 dollars playing the lottery
Original statement: ¬∃xP (x)
Negation: ∃xP (x)
Negation in English: Someone has lost more than 1000 dollars playing the lottery.
b) C(x, y): x has chatted with y, U = students in this class
Original: ∃x∃y[C(x, y) ∧ ∀z[C(x, z) → y = z]]
Negation: ∀x∀y[¬C(x, y) ∨ ∃z[C(x, z) ∧ y 6= z]]
Negation in English: Every student has either chatted with no one in the class or with at
least two students.
d) U = all students, P(x): x has solved every exercise in this book
Original: ∃xP (x)
Negation: ∀x¬P (x) Negation in English: No student has solved every exercise in this book.
46. ∃x∀y(x ≤ y 2 ). What is the truth value?
a)U = R+ F
b) U = Z T
c) U = R − {0} T
Section 1.6
24. Step 3 does not follow from step 2 - simplification has been used incorrectly. Likewise,
step 5 does not follow from step 2.
Section 1.7
8. Prove: If n is a perfect square, then n + 2 is not a perfect square.
Proof: Assume that n is a perfect square. So n = m2 for some integer m ≥ 0. If n = 0,
then n + 2 = 2 is not a perfect square. So assume n ≥ 1, and so m ≥ 1. Then the smallest
perfect square which is bigger than n is (m + 1)2 , and
(m + 1)2 = m2 + 2m + 1 = n + 2m + 1 ≥ n + 2 + 1 = n + 3 > n + 2.
Therefore m2 = n < n + 2 < (m + 1)2 , so n + 2 is not a perfect square. 10. Use a direct proof to show that the product of two rational numbers is rational.
1
Proof: Let x, y be arbitrary rational numbers. Then by definition of rational number, x = m
n
and y = rs , where m, n, r, s ∈ Z and n and s are non-zero. Then xy = m
( r ) = mr
, where
n s
ns
mr, ns ∈ Z since m, r, n, s ∈ Z, and ns 6= 0 since n and s are non-zero. Therefore xy is
rational. 12. Prove: the product of a nonzero rational number and any irrational number is irrational.
Proof: Let x, y be arbitrary real numbers. Assume that x is a nonzero rational number and
y is irrational. Assume by way of contradiction that xy is rational. So x = m
, where m, n are
n
nr
nonzero integers, and xy = rs , where r, s ∈ Z, and s 6= 0. Since x 6= 0, y = (1/x)(xy) = ms
,
where nr, ms ∈ Z since n, r, m, s ∈ Z, and ms 6= 0 since m 6= 0 and s 6= 0. Thus y is rational.
Contradiction! 18. Prove that if n is an integer and 3n + 2 is even, then n is even, using
a) contraposition
Proof: Let n be an arbitrary integer. Assume that n is odd, i.e., n = 2k + 1 for some integer
k. We must show that 3n + 2 is odd.
Since n = 2k + 1, 3n + 2 = 3(2k + 1) + 2 = 6k + 5 = 2(3k + 2) + 1, where 3k + 2 ∈ Z since
k ∈ Z. So 3n + 2 is odd.
b) contradiction
Proof: Let n be an arbitrary integer. Assume by way of contradiction that 3n + 2 is even
and n is odd. Then n = 2k + 1 for some integer k, and so 3n + 2 = 3(2k + 1) + 2 = 6k + 5 =
2(3k + 2) + 1, where 3k + 2 ∈ Z since k ∈ Z. So 3n + 2 is odd. So 3n + 2 is odd and even.
Contradiction!
Section 5.3
2bc
b)f(0) = 3, f(n+1) = 3f(n)+7 for n = 0, 1, ...
f(1) = 3(3)+7 = 16, f(2) = 3(16)+7 = 55, f(3) = 3(55)+7 = 172
c) f(0) = 3, f(n+1) = f (n)2 − 2f (n) − 2
f(1) = 9-6-2 = 1, f(2) = 1-2-2 = -3, f(3) = 9+6-2 = 13
8.
a) Base Case: a1 = 2
Recursive Step: an+1 = an + 4 for n = 1, 2, ...
b)Base Step: a1 = 0
Recursive Step: an+1 = an + 2(−1)n+1 for n = 1, 2, ...
c) Base Case: a1 = 2
Recursive Step: an+1 = an + 2(n + 1) for n = 1, 2, 3, ....
d) Base step: a1 = 1
Recursive step: an+1 = an + 2n + 1 for n = 1, 2, 3, ...
2
10. Base step: Sm (0) = m
Recursive step: Sm (n + 1) = Sm (n) + 1 for n = 0, 1, ...
12. Prove that f12 + f22 + ... + fn2 = fn fn+1 for all n ≥ 1.
Proof:
Base case(n = 1): f12 = 12 = 1 = 1(1) = f1 f2
Induction hypothesis: Let k ≥ 1 be arbitrary. Assume that f12 + ...fk2 = fk fk+1 .
2
2
So f12 + ... + fk2 + fk+1
= fk fk+1 + fk+1
by IH
= fk+1(fk + fk+1 )
= fk+1(fk+2 ) by definition of the fibonnaci sequence
24.
a) Define the set S by
Base step: 1 ∈ S
Recursive step: If a ∈ S, then a + 2 ∈ S.
b) Define the set S by
Base step: 3 ∈ S
Recursive step: If a ∈ S, then 3a ∈ S.
c) Define the set S by
Base step: x ∈ S, and c ∈ S for any integer constant c
Recursive step: If p ∈ S and q ∈ S, then ps, p + s, p − s ∈ S.
26.
a) (2, 3), (3, 2);
(4, 6), (5, 5), (6, 4);
(6, 9), (7, 8), (8, 7), (9, 6);
(8, 12), (9, 11), (10, 10), (11, 9), (12, 8);
(10, 15), (11, 14), (12, 13), (13, 12), (14, 11), (15, 10)
b) Let P (n) be the statement that 5|a + b whenever (a, b) ∈ S is obtained by n applications
of the recursive step.
Basis Step: P (0) is true, because only (0, 0) ∈ S and 0 + 0 = 5 × 0, 5|0 + 0
IH: Assume that 5|a + b whenever (a, b) ∈ S is obtained by k ≥ 0 or fewer applications of
the recursive step.
Inductive Step: Consider an element (c, d) obtained with k + 1 applications of the recursive
step.
Assume (a, b) is an element obtained with k or fewer applications of the recursive step and
is the one to construct (c, d).
3
Case (c + d) = (a + 2, b + 3):
c + d = (a + 2) + (b + 3) = a + b + 5
= 5t + 5 where t ∈ Z (by IH)
= 5(t + 1)
Case (c + d) = (a + 3, b + 2):
c + d = (a + 3) + (b + 2) = a + b + 5
= 5t + 5 where t ∈ Z (by IH)
= 5(t + 1)
Therefore, in both cases, 5|c + d.
c) Basis step: For (0, 0) ∈ S, then 5|0 + 0
Inductive step: Assume 5|a + b holds for arbitrary (a, b) ∈ S. Then it also holds for all
(a + 2, b + 3) and (a + 3, b + 2) because
(a + 2) + (b + 3) = (a + 3) + (b + 2) = a + b + 5 = 5t + 5 = 5(t + 1), where t ∈ Z.
28a. S = {(a, b)|a ∈ Z+ , b ∈ Z+ , a + b is odd}
Basis step: (1, 2) ∈ S and (2, 1) ∈ S
Recursive step: If (a, b) ∈ S, then (a + 1, b + 1), (a, b + 2), (a + 2, b) ∈ S.
32.
a) Basis step: ones(λ) = 0
Recursive step: ones(sb) = ones(s) + x, where x = 0 when b = ‘0′ and x = 1 when b = ‘1′
b) Let P (t) be ones(st) = ones(s) + ones(t), s, t ∈ Σ∗ , Σ = {‘0′ , ‘1′ }
Base: Show P (λ) is true. That is, ones(sλ) = ones(s) = ones(s) + 0 = ones(s) + ones(λ)
IH: Assume P (t) is true for arbitrary t ∈ Σ∗ . We must show that P (tb) is true, where b ∈ Σ.
Recursive Step: Show ones(stb) = ones(s) + ones(tb).
ones(stb)
= ones(st) + x where x = 0 when b = ‘0′ and x = 1 when b = ‘1′ (by the recursive definition
of ones(w))
= ones(s) + ones(t) + x (by IH)
= ones(s) + ones(tb) (by the recursive definition of ones(w))
Section 3.2
2. a) No
b) Yes
c) No
14. a) No
b) Yes
c) Yes
d) Yes
4
e) Yes
f) Yes
5