MATH 135 Midterm F19
Question 1
Let A and B be statement variables.
(a) Complete the truth table below.
A
T
T
F
F
B
T
F
T
F
¬B
A ∧ (¬B)
B⇔A
(A ∧ (¬B)) ∨ (B ⇔ A)
B⇒A
(b) Determine whether (A ∧ (¬B)) ∨ (B ⇔ A) is logically equivalent to B ⇒ A. Circle
the correct answer. No further justification is needed.
Equivalent
Not Equivalent
Solution (a)
A
T
T
F
F
B
T
F
T
F
¬B
F
T
F
T
A ∧ (¬B)
F
T
F
F
B⇔A
T
F
F
T
(A ∧ (¬B)) ∨ (B ⇔ A)
T
T
F
T
B⇒A
T
T
F
T
(b) Equivalent
Question 2
Let x, y ∈ R. Consider the implication S:
If xy > 6, then x > 2 and y > 3.
(a) State the hypothesis of S.
(b) State the conclusion of S.
(c) State the converse of S.
(d) State the contrapositive of S.
(e) State the negation of S in a form that does not contain an implication.
(f) Indicate clearly whether the given implication S is true or false for all x, y ∈ R. Then
prove or disprove the statement.
Circle the correct answer: True
False
Solution
(a) xy > 6
(b) x > 2 and y > 3
(c) If x > 2 and y > 3, then xy > 6
(d) If x ≤ 2 or y ≤ 3, then xy ≤ 6.
(e) The negation of A ⇒ B is A ∧ (¬B). Hence, the negation of S is (xy > 6) and (x ≤ 2
1
or y ≤ 3).
(f) False. For example, let x = 4 and y = 2. Then xy = 8 > 6, but y ≤ 3.
Question 3
Given a variable x, let P (x) denote the open sentence x ≥ 0, and let Q(x) denote the open
sentence x < 0. Determine if the following statements are true or false. Circle the correct
answers. No justification is needed.
(a) (∀x ∈ R, P (x)) ∨ (∀x ∈ R, Q(x))
True
False
(b) ∀x ∈ R, (P (x) ∨ Q(x))
True
False
(c) (∀x ∈ N, P (x)) ∨ (∀x ∈ N, Q(x))
True
False
(d) ∀x ∈ N, (P (x) ∨ Q(x))
True
False
Solution (a) False
(b) True
(c) True
(d) True
Question 4
For each of the following statements indicate clearly whether the statement is true or false
and then prove or disprove the statement.
(a) ∀x ∈ R, ∃y ∈ R, x + 2y = 0.
Circle the correct answer: True
False
(b) ∃y ∈ R, ∀x ∈ R, x + 2y = 0.
Circle the correct answer: True
False
Solution (a) True. Let x ∈ R. Take y =
−x
2
x + 2y = x + 2 ·
∈ R. Then we have
−x 2
= x − x = 0.
(b) False. We will prove that the negation is true. The negation states that ∀y ∈ R,
∃x ∈ R such that x + 2y 6= 0. Let y ∈ R. Take x = −2y + 1. Then
x + 2y = (−2y + 1) + 2y = 1 6= 0.
Since the negation is true, the original statement is false.
Question 5
Let a1 , a2 , a3 , · · · , be a sequence of positive integers defined by
a1 = 1,
Prove that for all n ∈ N,
a2 = 5,
am = am−1 + 2am−2 ,
an = 2n + (−1)n .
Solution Let P (n) be the statement that
an = 2n + (−1)n .
2
if m ≥ 3.
Base Cases
If n = 1,
LHS = a1 = 1
and
RHS = 21 + (−1)1 = 1.
Since LHS=RHS, P (1) is true.
If n = 2,
LHS = a2 = 5
and
RHS = 22 + (−1)2 = 5.
Since LHS=RHS, P (2) is true.
Inductive Step Let k ∈ N with k ≥ 2 be arbitrary. Suppose that P (1), · · · , P (k) are true,
i.e.,
ar = 2r + (−1)r for 1 ≤ r ≤ k.
Consider P (k + 1) with k ≥ 2. We have
ak+1 = ak + 2ak−1
(by definition since k + 1 ≥ 3)
= (2k + (−1)k ) + 2(2k−1 + (−1)k−1 )
k
k
k
(by inductive hypothesis)
k−1
= 2 + (−1) + 2 + 2 · (−1)
= 2k (1 + 1) + (−1)k−1 (−1 + 2)
= 2k+1 + (−1)k−1
= 2k+1 + (−1)k+1
(since (−1)k−1 = (−1)k+1 ).
Thus P (k + 1) is true. By POSI, P (n) is true for all n ∈ N.
Question 6
n X
n
(a) Let n ∈ N. Evaluate the sum
2i .
n−i
i=0
x22
x2
2
x
14
(b) Determine the coefficient of
in the expansion of
+
.
n
= ni . By taking a = 1 and b = 2, by BT2, we have
Solution (a) Note that n−i
n n X
X
n
n i
i
2 =
2 = (1 + 2)n = 3n .
n−i
i
i=0
i=0
(b) By BT2, we have
14 14 i X
2 14 X 14
14 i 28−3i
2 14−i 2
x +
=
(x )
=
2x
.
x
i
x
i
2
i=0
i=0
If 28 − 3i = 22, then 3i = 6 and i = 2. Hence the coefficient of x22 is
14 2
2 = 91 · 4 = 364.
2
Question 7
Let a, b ∈ Z with a ≥ 2. Prove that if a 6= 13, then a - (3b + 1) or 3a - (7b − 2).
3
Solution By the contrapositive law, it suffices to prove that if a|(3b + 1) and 3a|(7b − 2),
then a = 13. Since a|3a and 3a|(7b − 2), by TD, we have a|(7b − 2). Since a|(3b + 1) and
a|(7b − 2), by DIC,
a| 7(3b + 1) − 3(7b − 2)
=⇒ a|(21b + 7 − 21b + 6) =⇒ a|13.
Note that the only positive divisors of 13 are 1 and 13. Since a|13 and a ≥ 2, we have
a = 13.
Question 8
(a) Prove that for all n ∈ N, n2 + n + 1 is odd.
(b) Let d, n ∈ N. Prove that if d | (n2 + n + 1) and d | (n2 + n + 3), then d = 1.
Solution (a) Let n ∈ N. We consider two cases: n is even or n is odd.
(1) If n is even, say n = 2k for some k ∈ N, then
n2 + n + 1 = (2k)2 + (2k) + 1 = 4k 2 + 2k + 1 = 2(2k 2 + k) + 1,
which is odd.
(2) If n is odd, say n = 2k + 1 for some k ∈ N, then
n2 + n + 1 = (2k + 1)2 + (2k + 1) + 1 = 4k 2 + 4k + 1 + 2k + 1 + 1 = 2(2k 2 + 3k + 1) + 1,
which is odd.
(b) If d|(n2 + n + 1) and d|(n2 + n + 3), by DIC, we have d|((n2 + n + 3) − (n2 + n + 1)),
i.e., d|2. Since d|2 and d ∈ N, we have d = 1 or d = 2. Since d|(n2 + n + 1) and n2 + n + 1
is odd by (a), d 6= 2. It follows that d = 1.
Question 9
Let a, b be non-negative integers. Prove that 3a = 8b if and only if a = b = 0.
Solution (⇐) If a = b = 0, then 30 = 1 = 80 and hence 3a = 8b .
(⇒) Suppose that 3a = 8b . We prove this implication by contradiction. Suppose that
there exist a ≥ 1 or b ≥ 1 such that 3a = 8b . If a ≥ 1, since 3a = 8b , we have b ≥ 1. If
b ≥ 1, since 3a = 8b , we have a ≥ 1. Hence, we have a ≥ 1 and b ≥ 1. Since 3 is odd and
a ≥ 1, the number 3a is odd. Also, since 8 is even and b ≥ 1, the number 8b is even. It
follows that 3a 6= 8b which leads to a contradiction. Hence, such a ≥ 1 or b ≥ 1 does not
exist and we have a = 0 and b = 0.
Question 10
Prove that for all n ∈ N,
n
X
1
1
≤2− .
2
i
n
i=1
Solution Let P (n) denote the statement
n
X
1
1
≤2− .
2
i
n
i=1
4
Base Case If n = 1,
LHS =
1
X
i2 = 1
and
RHS = 2 −
i=1
1
= 1.
1
Since LHS ≤ RHS, P (1) is true.
Inductive Step Let k ∈ N be arbitrary. Suppose that P (k) is true, i.e.,
k
X
1
1
≤2− .
2
i
k
i=1
Consider P (k + 1). We have
k+1
k
X
X
1
1
1
1
1
=
+
≤2− +
i2
i2 (k + 1)2
k (k + 1)2
i=1
(by inductive hypothesis).
i=1
Note that
1
1
1
−k 2 − 2k − 2 + k + k 2 + k
−1
−
+
=
=
≤ 0.
2
2
(k + 1)
k k+1
k(k + 1)
k(k + 1)2
Hence, we have
1
1
1
− ≤−
.
2
(k + 1)
k
k+1
It follows that
k+1
X
1
1
1
1
≤2− +
≤2−
.
i2
k (k + 1)2
k+1
i=1
Thus P (k + 1) is true. By POMI, P (n) is true for all n ∈ N.
5