NUS MA1100

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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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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(More working spaces for Question 1)

Continue on page 18-19 if you need more space. Please indicate clearly.

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(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}.


(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

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(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.


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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).


(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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(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.


(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.

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(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-

sitive? Justify your answers.

S = {(a, b) ∈ Z× Z | (a− b)2 < 9}.


(a) (i) R = {(1, 1), (1, 3), (2, 2), (2, 4), (3, 1), (3, 3), (4, 2), (4, 4)}.

(ii) Reflexive: For all x ∈ A, x + x = 2x is even. So (x, x) ∈ R.

Symmetric: For all x, y ∈ A, x+y is even implies y+x is even. In other words,

(x, y) ∈ R→ (y, x) ∈ R.

Transitive: For all x, y, z ∈ A, if x+y and y+z are even, then (x+y)+(y+z) =

x + 2y + z is even. Since 2y is even, so x + z is also even. Hence (x, z) ∈ R

We conclude that R is an equivalent relation.

(iii) The distinct equivalent classes are [1] = {1, 3} and [2] = {2, 4}.

(iv) No. If we remove (a, a) for any a ∈ A, the resulting relation will not be

reflexive.

If we remove (a, b) for any two distinct a, b ∈ A, then we will have (b, a) in the

resulting relation but not (a, b), and hence the relation is not symmetric.

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(b) S is reflexive:

For all a ∈ Z, (a− a)2 = 0 < 9→ (a, a) ∈ S.

S is symmetric:

For all a, b ∈ Z, (a− b)2 = (b− a)2.

So (a− b)2 < 9→ (b− a)2 < 9.

So (a, b) ∈ S → (b, a) ∈ S.

S is not transitive:

For example, a = 1, b = 3, c = 5:

Then (a− b)2 = 4 < 9 so (a, b) ∈ S.

(b− c)2 = 4 < 9 so (b, c) ∈ S.

(a− c)2 = 16 > 9 so (a, c) 6∈ S.


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Determine whether the following sets are countable or uncountable. Justify your answers.

(a) A = {n ∈ Z | n is a square free number}.

(b) B = {x ∈ R | x =√n for some integer n }.

(c) C = [0, 1]−{

1,1

2,1

3, . . . ,

1

n, . . .

}.


(a) A is countable.

A is a subset of Z which is countable. So A is countable.

(b) B is countable.

We define a function f : B → Znonneg by

f(x) = x2.

For any x =√n ∈ B, f(

√n) = n, which is a non-negative integer. So f is well-

defined.

For any x, y ∈ B, x =√n and y =

√m for some non-negative integers n,m. If

f(x) = f(y), then n = m which implies√n =

√m. So x = y, and hence f is

injective.

For any n ∈ Znonneg, let x =√n ∈ B. Then f(x) = f(

√n) = n. So f is surjective.

Since f is a bijection and Znonneg is contable, so B is countable.

(c) C is uncountable.

Note that C ∪{

1, 12, 13, . . . , 1

n, . . .

}= [0, 1].

Suppose C is countable.

Since{

1, 12, 13, . . . , 1

n, . . .

}is countable (being a subset of Q+), then [0, 1] is the union

of two countable sets, which is countable. This is contradiction to the fact that [0, 1]

is uncountable.

So we conclude that C is uncountable.

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Let S be the set of all finite strings in 0’s and 1’s where the left most digit is 1.

(a) Define a function g : S → Z as follows:

for each string s in S,

g(s) = the number of 1’s in s minus the number of 0’s in s.

(i) What is g(101011)? g(100100)?

(ii) Is g one-to-one? Prove or give a counterexample.

(iii) Is g onto? Prove or give a counterexample.

(b) Define a function h : S → Z+ as follows:

for each string s = anan−1 . . . a1a0 in S,

h(s) = an × 2n + an−1 × 2n−1 + · · ·+ a1 × 2 + a0.

Is h a bijection? Justify your answer. (You may use any result proven in class.)


(a) (i) g(101011) = 4− 2 = 2.

g(100100) = 2− 4 = −2.

(ii) No. For (110), (101) ∈ S, we have g(110) = g(101) = 2− 1 = 1.

(iii) Yes. For positive integer n, take the string s = 11 . . . 1 ∈ S (repeated n times).

Then g(s) = n− 0 = n.

For n = 0, take the string s = 10 ∈ S. Then g(s) = 1− 1 = 0.

For negative integer n, take the string s = 10 . . . 0 ∈ S (with 1− n 0’s). Then

g(s) = 1− (1− n) = n.

(b) Yes. h is a bijection.

Surjective: We use the result that

Every positive integer m can be written as a sum of distinct powers of 2.

So m = an × 2n + an−1 × 2n−1 + · · ·+ a1 × 2 + a0 for some n ≥ 0 where ai = 0 or 1

with an = 1.

Then s = anan−1 . . . a1a0 is an element in S and h(s) = m.

So h is surjective.

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(b) (cont.)

Injective: This is the same as showing, for any m ∈ Z+, we can only write m as the

sum of distincts power of 2 in exactly one way.

Proof by strong PMI:

Let P (m): There is only one way to express m as a sum of distinct powers of 2.

Basis step P (1):

Note that for n ≥ 1, the sum an × 2n + an−1 × 2n−1 + · · ·+ a1 × 2 + a0 > 1.

Therefore the only way to express 1 as sum of power of 2 is a0 = 1.

Inductive step P (1) ∧ P (2) ∧ . . . ∧ P (k)→ P (k + 1).

If k + 1 is a power of 2, say 2h, then:

(i) if n < h, then an×2n +an−1×2n−1 + · · ·+a1×2 +a0 < 2h, (and hence 6= k+ 1);

(ii) if n > h, then an×2n +an−1×2n−1 + · · ·+a1×2+a0 > 2h , (and hence 6= k+1);

(iii) if n = h, then an × 2n + an−1 × 2n−1 + · · ·+ a1 × 2 + a0 > 2h if some ai 6= 0 for

some i < n , (and hence 6= k + 1).

Hence the only way to express k + 1 = 2h as a distinct powers of 2 is 2h itself.

If k+1 is not a power of 2, let 2h be the largest power of 2 smaller than k+1. Then

k + 1 = 2h + m with m < 2h < k + 1.

By induction hypothesis, P (m) is true. So there is a unique way to write m as sum

of power of 2, say

m = 2t + at−1 × 2t−1 + · · ·+ a1 × 2 + a0

with t < h.

Since 2h is the largest power of 2 smaller than k + 1, so 2h is unique.

Hence k + 1 can be uniquely written as 2h + 2t + at−1 × 2t−1 + · · ·+ a1 × 2 + a0.

By strong PMI, we have proven that, for all m ≥ 1, m can be written as the sum

of distincts power of 2 in exactly one way.

We can then conclude that the function h is injective.


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Let pn be the nth prime number, i.e. p1 = 2, p2 = 3, p3 = 5 . . . etc.

(a) Is it true that p1p2 · · · pn ≡ 2 mod 4 for all n ∈ Z+? Justify your answer.

(b) Prove that pn < 22n for all n ∈ Z+.


(a) Yes. p1 = 2 and pi is odd for all i ≥ 2.

So p1p2 · · · pn is even and p2 · · · pn is odd for all n ∈ Z+.

Hence p1p2 · · · pn ≡ 0 or 2 mod 4 for all n ∈ Z+.

Suppose p1p2 · · · pn ≡ 0 mod 4 for some n. Then p1p2 · · · pn = 4k for some integer

k and p2 · · · pn = 2k which contradicts that p2 · · · pn is odd.

So we conclude that p1p2 · · · pn ≡ 2 mod 4.

(b) We shall prove by strong mathematical induction.

Let P (n) : pn < 22n .

Basis step: Let n = 1. We have p1 = 2 < 221 . So P (1) is true.

Inductive step: Assume P (1), P (2), . . . , P (k) are true. i.e. pi < 22i for all i =

1, 2, . . . , k. We can also write this inequality as pi + 1 ≤ 22i .

Let N = p1p2 · · · pk + 1. Then by hypothesis,

N = p1p2 · · · pk + 1

< (p1 + 1)(p2 + 1) · · · (pk + 1)

≤ (221)(222) · · · (22k)

= 2(21+22+···+2k)

= 2(2k+1−2)

< 22k+1

Let p be a prime factor of N . Then p 6= pi for 1 ≤ i ≤ k. Otherwise, p | N and

p | (N − 1). This implies p | 1 which is impossible.

So p is a prime bigger than pk. i.e. pk+1 ≤ p ≤ N < 22k+1.

This proves that if P (1), P (2), . . . , P (k) are true, then P (k + 1) is true.

Hence by strong mathematical induction, we have proven that pn < 22n for all

positive integers n.

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(Additional working spaces for ALL questions - indicate your question numbers clearly.)

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(Additional working spaces for ALL questions - indicate your question numbers clearly.)

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