Student Number: A/U/T* *Delete where necessary NATIONAL UNIVERSITY OF SINGAPORE FACULTY OF SCIENCE SEMESTER 1 EXAMINATION 2012-2013 MA1100 Fundamental Concepts of Mathematics November/December 2012 Time allowed: 2 hours INSTRUCTIONS TO CANDIDATES 1. Write down your matriculation/student number neatly in the space provided above. This booklet (and only this booklet) will be col- lected at the end of the examination. Do not insert any loose pages in the booklet. 2. This examination paper contains a total of EIGHT (8) questions and comprises NINETEEN (19) printed pages. 3. Answer ALL questions. Write your answers and working in the spaces provided inside the booklet following each question. 4. Total marks for this exam is 100. The marks for each question are indicated at the beginning of the question. 5. Candidates may use calculators. However, they should lay out systematically the various steps in the calculations. Examiner’s Use Only Questions Marks 1 2 3 4 5 6 7 8 Total
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Student Number: A/U/T*
*Delete where necessary
NATIONAL UNIVERSITY OF SINGAPORE
FACULTY OF SCIENCE
SEMESTER 1 EXAMINATION 2012-2013
MA1100 Fundamental Concepts of Mathematics
November/December 2012 Time allowed: 2 hours
INSTRUCTIONS TO CANDIDATES
1. Write down your matriculation/student
number neatly in the space provided above.
This booklet (and only this booklet) will be col-
lected at the end of the examination. Do not insert
any loose pages in the booklet.
2. This examination paper contains a total of EIGHT
(8) questions and comprises NINETEEN (19)
printed pages.
3. Answer ALL questions. Write your answers and
working in the spaces provided inside the booklet
following each question.
4. Total marks for this exam is 100. The marks for
each question are indicated at the beginning of the
question.
5. Candidates may use calculators. However, they
should lay out systematically the various steps in
the calculations.
Examiner’s Use Only
Questions Marks
1
2
3
4
5
6
7
8
Total
PAGE 2 MA1100
Question 1 [10 marks]
(a) Use mathematical induction to prove that 1 + 4n < 2n for all integers n ≥ 5.
(b) A sequence is defined by
a1 = 2, a2 = 4, an+2 = 5an+1 − 6an for all n ≥ 1.
Use a version of mathematical induction to prove that an = 2n for all n ∈ Z+.
Show your working below and on the next page.
(a) Let P (n) be 1 + 4n < 2n.
Basis step P (5) :
1 + 4× 5 = 21 < 25 = 32.
So P (5) is true.
Inductive step P (k)→ P (k + 1) (for k ≥ 5):
Given 1 + 4k < 2k.
Then 1 + 4k + 4 < 2k + 4
This implies 1 + 4(k + 1) < 2k + 2k (since k ≥ 5, so 4 < 2k.)
Hence 1 + 4(k + 1) < 2(2k) = 2k+1.
So P (k)→ P (k + 1) is true for all k ≥ 5.
By principle of mathematical induction, P (n) is true for all n ≥ 5.
(b) Let P (n) be an = 2n.
Basis step P (1) and P (2):
We have a1 = 2 = 21, a2 = 4 = 22.
So P (1) and P (2) are true.
Inductive step P (k − 1) ∧ P (k)→ P (k + 1) (for k ≥ 2):
ak+1 = 5ak − 6ak−1
= 5(2k)− 6(2k−1)
= 2k−1(5× 2− 6)
= 2k−1 × 22 = 2k+1.
By strong principle of mathematical induction, P (n) is true for all n ≥ 1.
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PAGE 3 MA1100
(More working spaces for Question 1)
Continue on page 18-19 if you need more space. Please indicate clearly.
· · · − 4−
PAGE 4 MA1100
Question 2 [15 marks]
(a) Define sets A and B as follows:
A = {n ∈ Z | n = 8r − 3 for some integer r} and
B = {m ∈ Z | m = 4s + 1 for some integer s}.
Prove A ⊆ B using element method.
(b) Let A,B,C be three subsets of a universal set. Use algebra of sets to prove that
(A−B) ∩ (A− C) = A− (B ∪ C).
State the properties (laws) that you use for your steps.
(c) Let An =
{1
n,
2
n, · · · , n− 1
n
}for integer n ≥ 2.
Use element method to prove that
∞⋃n=2
An = {q ∈ Q | 0 < q < 1}.
Show your working below and on the next page.
(a) Let x ∈ A.
Then x = 8r − 3 = 4(2r)− 4 + 1 = 4(2r − 1) + 1.
So x = 4s + 1 with s = 2r − 1 ∈ Z.
So x ∈ B.
Hence A ⊆ B.
(b)
LHS = (A−B) ∩ (A− C)
= (A ∩Bc) ∩ (A ∩ Cc) set difference
= (A ∩ A) ∩ (Bc ∩ Cc) associative/commutative
= A ∩ (B ∪ C)c idempotent/de Morgan
= A− (B ∪ C) set difference
= RHS
· · · − 5−
PAGE 5 MA1100
(More working spaces for Question 2)
(c) First part (⊆):
Let x ∈∞⋃n=2
An.
Then x ∈ An for some integer n ≥ 2.
So x =m
nfor some integer m with 1 ≤ m ≤ n− 1.
Therefore x ∈ Q and 0 < x < 1.
i.e., x ∈ {q ∈ Q | 0 < q < 1}.
So we conclude∞⋃n=2
An ⊆ {q ∈ Q | 0 < q < 1}.
Second part (⊇):
Let x ∈ {q ∈ Q | 0 < q < 1}.
Then we can write x =m
nfor some integers m,n ∈ Z+ with 0 < m < n. (This is
possible for n ≥ 2.)
So x ∈ An for some integer n ≥ 2.
i.e.,x ∈∞⋃n=2
An.
So we conclude {q ∈ Q | 0 < q < 1} ⊆∞⋃n=2
An.
Continue on page 18-19 if you need more space. Please indicate clearly.
· · · − 6−
PAGE 6 MA1100
Question 3 [10 marks]
Define a function f : R− {0} → R by the formula f(x) =x + 3
x.
(i) Prove that f is one-to-one (injective).
(ii) Replace the codomain of f with a subset of R so that f becomes a bijection. Justify
your answer.
(iii) Write down the inverse function of the bijection in part (ii).
Show your working below and on the next page.
(i) Let x, y ∈ R− {0} such that f(x) = f(y):
⇒ x + 3
x=
y + 3
y⇒ xy + 3y = xy + 3x
⇒ 3y = 3x
⇒ x = y
So f is one-to-one.
(ii) Replace the codomain by R− {1}.
It is sufficient to show f : R−{0} → R−{1} is onto (in order that f is a bijection).
Let k ∈ R− {1}. Want to show there exists x ∈ R− {0} such that f(x) = k.
Take x =3
k − 1. Note that this represents a non-zero real number. Then
f
(3
k − 1
)=
3k−1 + 3
3k−1
=3k
3= k.
So f is an onto function, and hence a bijection (in view of part (i)).
(iii) f−1 : R− {1} → R− {0} with f−1(x) =3
x− 1.
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PAGE 7 MA1100
(More working spaces for Question 3)
Continue on page 18-19 if you need more space. Please indicate clearly.
· · · − 8−
PAGE 8 MA1100
Question 4 [15 marks]
(a) Let a and b be integers such that a ≡ 5 mod 7 and b ≡ 4 mod 7.
(i) Find integers s, t where 0 ≤ s, t < 7 such that a + b ≡ s mod 7 and
ab ≡ t mod 7.
(ii) Is there an integer c such that ac ≡ 2 mod 14? Justify your answer.
(b) Use congruence modulo to compute the remainder of 18199 when it is divided by 65.
(c) Let a, b and n > 1 be integers. Prove that if m > 1 is a divisor of n and a ≡ b
mod n, then a ≡ b mod m.
Show your working below and on the next page.
(a) (i) a + b ≡ 5 + 4 ≡ 2 mod 7. (i.e. s = 2)
ab ≡ 5× 4 ≡ 6 mod 7. (i.e. t = 6)
(ii) Yes, we can take c = 6.
a ≡ 5 mod 7 ⇒ 3a ≡ 15 ≡ 1 mod 7
⇒ 2(3a) ≡ 2(1) mod 2(7)
⇒ 6a ≡ 2 mod 14
(b) First we note that 182 = 324 ≡ −1 mod 65.
We shall write 199 = 2× 99 + 1.
So 18199 = 182×99+1 = (182)99 × 18 ≡ (−1)99 × 18 ≡ −18 ≡ 47 mod 65.
So the remainder of 18199 is 47 when divided by 65.
(c) We are given m | n (1).
We also have a ≡ b mod n which implies n | (a− b) (2).
By (1) and (2), and the transitivity of divisibility, we have m | (a− b).
By definition of congruence, this implies a ≡ b mod m.
· · · − 9−
PAGE 9 MA1100
(More working spaces for Question 4)
Continue on page 18-19 if you need more space. Please indicate clearly.
· · · − 10−
PAGE 10 MA1100
Question 5 [15 marks]
(a) Let A = {1, 2, 3, 4}. Let R be the relation on A given by a R b if and only if a + b
is even.
(i) Express R as an ordered pair representation. (List all the ordered pairs in R.)
(ii) Show that R is an equivalence relation.
(iii) Find the distinct equivalence classes of R.
(iv) Is it possible to remove one ordered pair from R so that the resulting relation
on A is reflexive, symmetric but not transitive? Justify your answer.
(b) Determine whether the following relation S on Z is reflexive, symmetric and tran-