# Logic Integers Sets Functions Prime Numbers & Relations Questions

Description

Topics: Logic, integers, sets and functions, prime numbers, relations, real and complex

numbers, greatest common divisor and modular arithmetic, infimum and supremum, sequences,

limits of sequences, functions and limits of functions, continuity, groups.

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lOMoARcPSD|5093418
MT2116 Abstract mathematics
Examiners’ commentaries 2017
MT2116 Abstract mathematics
Important note
This commentary reflects the examination and assessment arrangements for this course in the
academic year 2016–17. The format and structure of the examination may change in future years,
and any such changes will be publicised on the virtual learning environment (VLE).
references
Unless otherwise stated, all cross-references will be to the latest version of the subject guide (2011).
You should always attempt to use the most recent edition of any Essential reading textbook, even if
the commentary and/or online reading list and/or subject guide refer to an earlier edition. If
none are available, please use the contents list and index of the new edition to find the relevant
section.
Comments on specific questions – Zone A
Candidates should answer SIX of the following EIGHT questions: THREE from Section A and
from Section A and your best THREE answers from Section B will count towards the final mark.
All questions carry equal marks.
Section A
Answer any three questions from this section.
Question 1
(a) Let p, q and r be statements. Prove, using a truth table or otherwise, that
[(p ∨ r) ∨ (q ∨ r)] ∧ ¬ [(p ∨ r) ∧ (q ∨ r)] =⇒ ¬r.
This question builds on the material of Chapter 1 of the subject guide.
Approaching the question
Let S1 denote the statement (p ∨ r) ∨ (q ∨ r), let S2 be ¬ [(p ∨ r) ∧ (q ∨ r)] and let S be
S1 ∧ S2 . (So, S is the left-hand side of the implication given in the question.) Then we have
the following truth table.
4
lOMoARcPSD|5093418
Examiners’ commentaries 2017
p
T
T
T
F
T
F
F
F
q
T
T
F
T
F
T
F
F
r
T
F
T
T
F
F
T
F
p∨r
T
T
T
T
T
F
T
F
q∨r
T
T
T
T
F
T
T
F
S1
T
T
T
T
T
T
T
F
(p ∨ r) ∧ (q ∨ r)
T
T
T
T
F
F
T
F
S2
F
F
F
F
T
T
F
T
S
F
F
F
F
T
T
F
T
¬r
F
T
F
F
T
T
F
T
We need to verify that S =⇒ ¬r. What that means is that we need to verify that whenever
S is true, so too is ¬r. This can be seen to be the case from the final two columns of the
table.
(b) Let S be the following statement:
If an integer, n, is even, then its cube, n3 , is even.
Prove that S is true.
Write down the converse of S.
Write down the contrapositive of the converse of S.
Approaching the question
If n is even, then n = 2k for some integer k, and so n3 = 8k 3 = 2(4k 3 ) is even.
The converse is: ‘If n3 is even, then n is even.’
The contrapositive of the converse if: ‘If n is not even, then n3 is not even,’ That is, it is ‘If
n is odd, then n3 is odd.’
The contrapositive of the converse is true: for, if n is odd, then n = 2k + 1 for some integer
k, and so:
n3 = (2k + 1)3 = 8k 3 + 12k 2 + 6k + 1 = 2(4k 3 + 6k 2 + 3k) + 1
which is odd.
Since the contrapositive of the converse is logically equivalent to the converse, it follows that
the converse is true.
Question 2
(a) Use the method of induction to prove that the following statement is true for all
natural numbers n:
n
X
r(r + 2)(r + 4) =
r=1
1
4
n(n + 1)(n + 4)(n + 5).
Proof by induction is discussed in Chapter 3 of the subject guide.
Approaching the question
The base case n = 1 is easily seen to be true since both sides of the expression are 15 in this
case. Suppose the result is true for a positive integer n, so that:
n
X
r=1
r(r + 2)(r + 4) =
1
n(n + 1)(n + 4)(n + 5).
4
5
lOMoARcPSD|5093418
MT2116 Abstract mathematics
Then:
n+1
X
r(r + 2)(r + 4)
=
(n + 1)(n + 3)(n + 5) +
n
X
r(r + 2)(r + 4)
r=1
r=1
=
=
=
=
=
1
(n + 1)(n + 3)(n + 5) + n(n + 1)(n + 4)(n + 5)
4
1
(n + 1)(n + 5) [4(n + 3) + n(n + 4)]
4
1
(n + 1)(n + 5)(n2 + 8n + 12)
4
1
(n + 1)(n + 5)(n + 2)(n + 6)
4
1
(n + 1)((n + 1) + 1)((n + 1) + 4)((n + 1) + 5)
4
which establishes that the result is true for n + 1. By induction, therefore, the result holds
for all n.
(b) The sequence x1 , x2 , . . . is defined as follows:
x1 = 1, x2 = 2 and, for n ≥ 3, xn = 2xn−1 + xn−2 − 2.
Prove by induction that: xn is even if and only if n is even.
Approaching the question
Since x1 = 1 and x2 = 2, the statement is true for n = 1, 2. Suppose it is true for all n ≤ k.
Suppose first that k + 1 is even, so k is odd. Then xk is odd and xk−1 is even. Therefore,
xk+1 = 2xk + xk−1 + 2 is even. If k + 1 is odd, then xk is even and xk−1 is odd. In this case,
xk+1 = 2xk + xk−1 + 2 is odd. So the result holds.
(c) Let X = R and let Y be the interval (−1, 1). Prove that the function f : X → Y
given by
x
f (x) =
1 + |x|
is a bijection from X to Y .
Functions and their properties are dealt with in Chapter 4 of the subject guide.
Approaching the question
First, we show f in injective. Suppose f (x) = f (y). Then:
y
x
=
1 + |x|
1 + |y|
so:
x(1 + |y|) = y(1 + |x|)
which means:
x + x|y| = y + y|x|.
Now:
x
y
=
1 + |x|
1 + |y|
implies that x, y have the same sign (both are at most, or at least, 0). So x|y| = y|x|. We
conclude (after cancellation) that x = y.
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lOMoARcPSD|5093418
Examiners’ commentaries 2017
Now we show f is surjective. Given y ∈ (−1, 1), we want x such that:
x
= y.
1 + |x|
If y ≥ 0, we want x ≥ 0, so we solve:
x
=y
1+x
(b) i. State the Intermediate Value Theorem.
ii. Let f : [0, ∞) → R be a decreasing continuous function √
such that f (0) = 1.
Show that there is exactly one x > 0 such that f (x) = x.
See Chapter 11 of the subject guide for relevant background material.
Approaching the question
i. Intermediate Value Theorem (Theorem 11.6 in the subject guide): suppose g is a
continuous function on the closed interval [a, b], and z lies between g(a) and g(b). Then
there is some c ∈ [a, b] with g(c) = z.
10
lOMoARcPSD|5093418
Examiners’ commentaries 2017

ii. We apply the Intermediate Value Theorem with g(x) =√ x − f (x) and [a, b] = [0, 1].
Function g is continuous at every point x ≥ 0 because x and −f (x) are continuous. We
have g(0) = −f (0) = −1, and g(1) = 1 − f (1) ≥ 1 − f (0) = 0 because f is decreasing.
By

the Intermediate Value Theorem there is x such that g(x) = 0, that is, f (x) = x.

Suppose that x1 < x2 are two solutions. Then f (x1 ) = x1 < x2 = f (x2 ), which contradicts the fact that f is decreasing. Question 6 (a) Define what it means to say that a sequence (an ) converges to 1. Reading for this question See Chapter 10 of the subject guide for material related to this question. Approaching the question A sequence (an ) is convergent to 1 if, for every ǫ > 0, there exists N such that for all n > N
we have |an − 1| < ǫ. (b) Suppose that a sequence (an ) converges to 1. i. Show, directly from the definition, that the sequence (a2n ) converges to 1. Let bn = max{an , a2n }, for n ∈ N. Here max{x, y} = x for x ≥ y, and max{x, y} = y for x < y. ii. Show that (bn ) converges to 1. Approaching the question i. Given ǫ > 0, we pick N such that for n > N we have |an − 1| < min{1, ǫ/3}. Then: |a2n − 1| = |an − 1||an + 1| < ǫ ǫ (|an − 1| + 2) < · 3 = ǫ. 3 3 ii. Given ǫ > 0, we pick N1 such that for n > N1 we have |an − 1| < ǫ, and we choose N2 such that for n > N2 we have |a2n − 1| < ǫ. Take N = max{N1 , N2 }. For n > N , we have an < 1 + ǫ and a2n < 1 + ǫ. Hence, bn < 1 + ǫ as well. We also have bn ≥ an > 1 − ǫ. So, for n > N , we have |bn − 1| < ǫ. (c) For each of the following sequences, find the limit of the sequence or show that the sequence does not converge. Justify your answers. (You may use any results from the course, provided they are clearly stated.) √ √ √ i. (cn ), where cn = n( n + 1 − n − 1); ii. (dn ), where dn = (−1)n 21/n . Approaching the question i. Observe that: √ n+1− √ √ √ √ √ ( n + 1 − n − 1)( n + 1 + n − 1) n + 1 − (n − 1) √ √ √ n−1= =√ . n+1+ n−1 n+1+ n−1 Hence: cn = √ √ n + 1 − (n − 1) 2 n 2 p √ √ n√ =√ =p n+1+ n−1 n+1+ n−1 1 + 1/n + 1 − 1/n and, by the algebra of limits (Theorem 10.5 in the subject guide): lim cn = lim p n→∞ n→∞ 2 1 + 1/n + p 1 − 1/n =√ Downloaded by lex lexus (localfocalpoint@gmail.com) 2 √ = 1. 1+0+ 1−0 11 lOMoARcPSD|5093418 MT2116 Abstract mathematics ii. We know that 21/n → 20 = 1 as n → ∞. So, using part a) with ǫ = 1/2, for some N , we have that 21/n > 1 − ǫ = 1/2 for n > N .
ii. Let ℓ = inf (A ∪ B). We show that ℓ is a lower bound on A. Hence, by the second
property of infimum, inf A ≥ ℓ. Indeed, if a ∈ A, then a ∈ A ∪ B, so a ≥ ℓ.
iii. Suppose that A minorizes B, let t = inf A, and take any x ∈ A ∪ B. Either x ∈ A, in
which case x ≥ t because t is a lower bound on A, or x ∈ B, in which case there is some
a ∈ A such that x ≥ a ≥ t (because t is a lower bound on A). Thus, t is a lower bound
for A ∪ B.
Consequently, by the second property of infimum, inf (A ∪ B) ≥ t = inf A, and we have
equality by i.
iv. This is false. Consider A = (0, 1), B = [0, 1). Then inf (A ∪ B) = inf A = 0, but there is
no a ∈ A such that a ≤ 0 ∈ B, so A cannot minorize B.
v. This is true. Take any b ∈ B. As t is a lower bound for B, but t 6∈ B, we have b > t.
Now, as t is the infimum of A, b is not a lower bound for A, and so there is some a ∈ A
with a < b. Hence A minorizes B. (b) i. State the Intermediate Value Theorem. ii. Let f : [0, ∞) → R be a decreasing continuous function such that f (0) = 1. Show that there is exactly one x > 0 such that f (x) = x2 .
See Chapter 11 of the subject guide for relevant background material.
Approaching the question
i. Intermediate Value Theorem (Theorem 11.6 in the subject guide): suppose g is a
continuous function on the closed interval [a, b], and z lies between g(a) and g(b). Then
there is some c ∈ [a, b] with g(c) = z.
ii. We apply the Intermediate Value Theorem with g(x) = x2 − f (x) and [a, b] = [0, 1].
Function g is continuous at every point x ≥ 0 because x2 and −f (x) are continuous. We
have g(0) = −f (0) = −1, and g(1) = 1 − f (1) ≥ 1 − f (0) = 0 because f is decreasing. By
the Intermediate Value Theorem, there is x such that g(x) = 0, that is, f (x) = x2 .
Suppose that x1 < x2 are two solutions. Then f (x1 ) = x21 < x22 = f (x2 ), which contradicts the fact that f is decreasing. 21 Downloaded by lex lexus (localfocalpoint@gmail.com) lOMoARcPSD|5093418 MT2116 Abstract mathematics Question 6 (a) Define what it means to say that a sequence (an ) converges to 1. Reading for this question See Chapter 10 of the subject guide for material related to this question. Approaching the question A sequence (an ) is convergent to 1 if, for every ǫ > 0, there exists N such that for all n > N
we have |an − 1| < ǫ. (b) Suppose that a sequence (an ) converges to 1. i. Show, directly from the definition, that the sequence (a2n ) converges to 1. Let bn = min{an , a2n }, for n ∈ N. Here min{x, y} = y for x ≥ y, and min{x, y} = x for x < y. ii. Show that (bn ) converges to 1. Approaching the question i. Given ǫ > 0, we pick N such that for n > N we have |an − 1| < min{1, ǫ/3}. Then: |a2n − 1| = |an − 1||an + 1| < ǫ ǫ (|an − 1| + 2) < · 3 = ǫ. 3 3 ii. Given ǫ > 0, we pick N1 such that for n > N1 we have |an − 1| < ǫ, and we choose N2 such that for n > N2 we have |a2n − 1| < ǫ. Take N = max{N1 , N2 }. For n > N , we have an > 1 − ǫ and a2n > 1 − ǫ. Hence,
bn > 1 − ǫ as well. We also have bn ≤ an < 1 + ǫ. So, for n > N , we have |bn − 1| < ǫ. (c) For each of the following sequences, find the limit of the sequence or show that the sequence does not converge. Justify your answers. (You may use any results from the course, provided they are clearly stated.) √ √ i. (cn ), where cn = n( n2 + 1 − n2 − 1); ii. (dn ), where dn = cos(πn)21/n . Approaching the question i. Observe that: p n2 + 1 − Hence: p n2 − 1 = √ √ √ √ ( n2 + 1 − n2 − 1)( n2 + 1 + n2 − 1) n2 + 1 − (n2 − 1) √ √ √ =√ . n2 + 1 + n2 − 1 n2 + 1 + n2 − 1 2n 2 n2 + 1 − (n2 − 1) p √ √ =√ =p cn = n √ 2 n2 + 1 + n2 − 1 n2 + 1 + n2 − 1 1 + 1/n + 1 − 1/n2 and, by the algebra of limits (Theorem 10.5 in the subject guide): lim cn = lim p n→∞ n→∞ 2 1 + 1/n2 + p 1 − 1/n2 =√ 2 √ = 1. 1+0+ 1−0 ii. We know that 21/n → 20 = 1 as n → ∞. So, using part a) with ǫ = 1/2, for some N , we have that 21/n > 1 − ǫ = 1/2 for n > N .
Hence, for even n > N , we have cos(πn) = 1 and dn > 1/2 and, for odd n > N , we have
cos(πn) = −1 and dn < −1/2. This implies that (dn ) does not converge. 22 Downloaded by lex lexus (localfocalpoint@gmail.com) lOMoARcPSD|5093418 Examiners’ commentaries 2017 Question 7 (a) Let (G, ⋆) be a group, and let a ∈ G. Give the definition of am , where m ∈ Z. Define the order of the element a of G. Reading for this question See Chapters 12, 13 and 14 of the subject guide for relevant reading for this question. Approaching the question m −1 −m 0 If m > 0, then am = a

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