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Section A: Long Answer Questions

Attempt any TWO questions.

3 questions·10 marks each
1long10 marks

Define proposition. Construct the truth table for the compound proposition ((p \rightarrow q) \leftrightarrow (\neg q \rightarrow \neg p)) and determine whether it is a tautology.

Proposition

A proposition (or statement) is a declarative sentence that is either true or false, but not both. For example, "2+3=52 + 3 = 5" is a true proposition, while "77 is even" is a false proposition. Sentences that are questions, commands, or have no definite truth value are not propositions.

Truth Table for (pq)(¬q¬p)(p \rightarrow q) \leftrightarrow (\neg q \rightarrow \neg p)

The implication pqp \rightarrow q is false only when pp is true and qq is false. ¬q¬p\neg q \rightarrow \neg p is the contrapositive of pqp \rightarrow q.

ppqq¬p\neg p¬q\neg qpqp \rightarrow q¬q¬p\neg q \rightarrow \neg p(pq)(¬q¬p)(p \rightarrow q) \leftrightarrow (\neg q \rightarrow \neg p)
TTFFTTT
TFFTFFT
FTTFTTT
FFTTTTT

Conclusion

The last column is true in every row. Therefore the compound proposition (pq)(¬q¬p)(p \rightarrow q) \leftrightarrow (\neg q \rightarrow \neg p) is a tautology. This confirms the logical equivalence (pq)(¬q¬p)(p \rightarrow q) \equiv (\neg q \rightarrow \neg p), i.e. an implication is logically equivalent to its contrapositive.

logictruth-table
2long10 marks

What is mathematical induction? Using the principle of mathematical induction, prove that (1 + 2 + 3 + \dots + n = n(n+1)/2) for all positive integers n.

Mathematical Induction

Mathematical induction is a proof technique used to prove that a statement P(n)P(n) holds for all positive integers nn. It consists of two steps:

  1. Basis step: Show that P(1)P(1) (or the smallest value) is true.
  2. Inductive step: Assume P(k)P(k) is true (the inductive hypothesis) and use it to prove that P(k+1)P(k+1) is true.

If both steps hold, then by the principle of mathematical induction P(n)P(n) is true for all n1n \geq 1.

Proof that 1+2+3++n=n(n+1)21 + 2 + 3 + \dots + n = \dfrac{n(n+1)}{2}

Let P(n):  1+2++n=n(n+1)2P(n): \; 1 + 2 + \dots + n = \dfrac{n(n+1)}{2}.

Basis step (n=1n = 1):

LHS=1,RHS=1(1+1)2=22=1.\text{LHS} = 1, \qquad \text{RHS} = \frac{1(1+1)}{2} = \frac{2}{2} = 1.

Since LHS = RHS, P(1)P(1) is true.

Inductive step: Assume P(k)P(k) holds for some positive integer kk:

1+2++k=k(k+1)2.(inductive hypothesis)1 + 2 + \dots + k = \frac{k(k+1)}{2}. \qquad \text{(inductive hypothesis)}

We must show P(k+1)P(k+1) holds, i.e. 1+2++k+(k+1)=(k+1)(k+2)21 + 2 + \dots + k + (k+1) = \dfrac{(k+1)(k+2)}{2}.

Adding (k+1)(k+1) to both sides of the hypothesis:

1+2++k+(k+1)=k(k+1)2+(k+1)=(k+1)(k2+1)=(k+1)k+22=(k+1)(k+2)2.1 + 2 + \dots + k + (k+1) = \frac{k(k+1)}{2} + (k+1) = (k+1)\left(\frac{k}{2} + 1\right) = (k+1)\cdot\frac{k+2}{2} = \frac{(k+1)(k+2)}{2}.

This is exactly P(k+1)P(k+1).

Conclusion

The basis step and inductive step both hold, so by the principle of mathematical induction,

1+2+3++n=n(n+1)2for all positive integers n.1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2} \quad \text{for all positive integers } n. \qquad \blacksquare
inductionproof
3long10 marks

Define a graph. Explain the different types of graphs with examples. Describe the adjacency matrix and incidence matrix representation of a graph.

Definition of a Graph

A graph G=(V,E)G = (V, E) consists of a non-empty set of vertices (nodes) VV and a set of edges EE, where each edge connects a pair of vertices. Graphs are used to model pairwise relationships between objects.

Types of Graphs (with examples)

  • Null/Empty graph: A graph with vertices but no edges. Example: V={a,b,c},E=V = \{a,b,c\}, E = \emptyset.
  • Simple graph: No self-loops and no parallel (multiple) edges between the same pair of vertices.
  • Multigraph: Allows multiple (parallel) edges between the same pair of vertices.
  • Directed graph (digraph): Each edge has a direction, represented as an ordered pair (u,v)(u,v). Example: a one-way road network.
  • Undirected graph: Edges have no direction; {u,v}\{u,v\} is unordered.
  • Weighted graph: Each edge is assigned a weight/cost. Example: a road map with distances.
  • Complete graph KnK_n: Every pair of distinct vertices is joined by an edge; it has n(n1)2\dfrac{n(n-1)}{2} edges.
  • Bipartite graph: Vertices split into two disjoint sets with edges only between the sets. Example: K2,3K_{2,3}.
  • Regular graph: Every vertex has the same degree.

Adjacency Matrix

For a graph with nn vertices, the adjacency matrix AA is an n×nn \times n matrix where

Aij={1if there is an edge between vi and vj0otherwise.A_{ij} = \begin{cases} 1 & \text{if there is an edge between } v_i \text{ and } v_j \\ 0 & \text{otherwise.}\end{cases}

For an undirected graph AA is symmetric. Example: for a triangle with vertices {1,2,3}\{1,2,3\} and edges {1-2,2-3,1-3}\{1\text{-}2, 2\text{-}3, 1\text{-}3\}:

A=[011101110].A = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}.

Incidence Matrix

For a graph with nn vertices and mm edges, the incidence matrix MM is an n×mn \times m matrix where

Mij={1if vertex vi is incident on edge ej0otherwise.M_{ij} = \begin{cases} 1 & \text{if vertex } v_i \text{ is incident on edge } e_j \\ 0 & \text{otherwise.}\end{cases}

Each column has exactly two 1's (the endpoints of that edge). For the triangle above with edges e1=1-2e_1 = 1\text{-}2, e2=2-3e_2 = 2\text{-}3, e3=1-3e_3 = 1\text{-}3:

M=[101110011].M = \begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix}.
graphrepresentation
B

Section B: Short Answer Questions

Attempt any EIGHT questions.

9 questions·5 marks each
4short5 marks

Convert the decimal number 245 into binary, octal and hexadecimal.

Convert 24510245_{10} to binary, octal and hexadecimal by repeated division.

Binary (divide by 2):

245÷2=122245 \div 2 = 122 r 11; 122÷2=61122 \div 2 = 61 r 00; 61÷2=3061 \div 2 = 30 r 11; 30÷2=1530 \div 2 = 15 r 00; 15÷2=715 \div 2 = 7 r 11; 7÷2=37 \div 2 = 3 r 11; 3÷2=13 \div 2 = 1 r 11; 1÷2=01 \div 2 = 0 r 11.

Reading remainders bottom-up: 24510=111101012245_{10} = 11110101_2.

Octal (divide by 8):

245÷8=30245 \div 8 = 30 r 55; 30÷8=330 \div 8 = 3 r 66; 3÷8=03 \div 8 = 0 r 33.

So 24510=3658245_{10} = 365_8. (Check: group the binary as 011110101=365011\,110\,101 = 3\,6\,5.)

Hexadecimal (divide by 16):

245÷16=15245 \div 16 = 15 r 55; 15÷16=015 \div 16 = 0 r 15=F15 = \text{F}.

So 24510=F516245_{10} = \text{F5}_{16}. (Check: group the binary as 11110101=F51111\,0101 = \text{F}\,5.)

Result: 24510=111101012=3658=F516245_{10} = 11110101_2 = 365_8 = \text{F5}_{16}.

number-system
5short5 marks

Define conjunction, disjunction and negation with truth tables.

Let pp and qq be propositions.

Conjunction (pqp \wedge q): "pp AND qq". It is true only when both pp and qq are true.

Disjunction (pqp \vee q): "pp OR qq" (inclusive). It is true when at least one of pp, qq is true.

Negation (¬p\neg p): "NOT pp". It reverses the truth value of pp.

ppqqpqp \wedge qpqp \vee q¬p\neg p
TTTTF
TFFTF
FTFTT
FFFFT
logic
6short5 marks

What is a bijective function? Give an example.

Bijective Function

A function f:ABf: A \rightarrow B is bijective (a one-to-one correspondence) if it is both:

  • Injective (one-to-one): distinct elements of AA map to distinct elements of BB, i.e. f(x1)=f(x2)    x1=x2f(x_1) = f(x_2) \implies x_1 = x_2.
  • Surjective (onto): every element of BB is the image of some element of AA, i.e. range =B= B.

A bijective function has an inverse f1:BAf^{-1}: B \rightarrow A.

Example

Let f:RRf: \mathbb{R} \rightarrow \mathbb{R} be defined by f(x)=2x+3f(x) = 2x + 3.

  • Injective: If 2x1+3=2x2+32x_1 + 3 = 2x_2 + 3 then x1=x2x_1 = x_2.
  • Surjective: For any yRy \in \mathbb{R}, choose x=y32x = \dfrac{y-3}{2}; then f(x)=yf(x) = y.

Hence ff is bijective, with inverse f1(y)=y32f^{-1}(y) = \dfrac{y-3}{2}.

functions
7short5 marks

State the principle of inclusion and exclusion.

Principle of Inclusion and Exclusion

For two finite sets AA and BB:

AB=A+BAB.|A \cup B| = |A| + |B| - |A \cap B|.

For three finite sets AA, BB, CC:

ABC=A+B+CABBCAC+ABC.|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |A \cap C| + |A \cap B \cap C|.

In general, the cardinality of a union of nn sets is obtained by adding the sizes of single sets, subtracting the sizes of all pairwise intersections, adding back the triple intersections, and so on, alternating signs:

i=1nAi=iAii<jAiAj+i<j<kAiAjAk+(1)n+1A1An.\left| \bigcup_{i=1}^{n} A_i \right| = \sum_i |A_i| - \sum_{i<j} |A_i \cap A_j| + \sum_{i<j<k} |A_i \cap A_j \cap A_k| - \dots + (-1)^{n+1}|A_1 \cap \dots \cap A_n|.

The principle corrects for over-counting of elements that lie in more than one set.

sets
8short5 marks

Explain the difference between a walk, path and circuit in a graph.

In a graph, these terms describe sequences of vertices and edges:

  • Walk: A finite alternating sequence of vertices and edges v0,e1,v1,e2,,ek,vkv_0, e_1, v_1, e_2, \dots, e_k, v_k where each edge eie_i joins vi1v_{i-1} and viv_i. Vertices and edges may be repeated. Its length is the number of edges.

  • Path: A walk in which no vertex is repeated (and hence no edge is repeated). It is an open walk (start vertex \neq end vertex).

  • Circuit (closed trail): A closed walk that starts and ends at the same vertex (v0=vkv_0 = v_k) and in which no edge is repeated (vertices other than the start/end may repeat).

Summary table:

TypeRepeated vertices?Repeated edges?Open/Closed
Walkallowedallowedeither
Pathnot allowednot allowedopen
Circuitallowed (except start=end)not allowedclosed

(A cycle is a closed path: a circuit with no repeated vertices except the start = end.)

graph
9short5 marks

Define equivalence relation with an example.

Equivalence Relation

A relation RR on a set AA is an equivalence relation if it is:

  1. Reflexive: aRaaRa for all aAa \in A.
  2. Symmetric: if aRbaRb then bRabRa.
  3. Transitive: if aRbaRb and bRcbRc then aRcaRc.

An equivalence relation partitions the set AA into disjoint equivalence classes.

Example

Define the relation "congruent modulo 3" on the set of integers Z\mathbb{Z}: aRb    3(ab)a \, R \, b \iff 3 \mid (a - b).

  • Reflexive: 3(aa)=03 \mid (a - a) = 0, so aRaaRa.
  • Symmetric: if 3(ab)3 \mid (a-b) then 3(ba)3 \mid (b-a).
  • Transitive: if 3(ab)3 \mid (a-b) and 3(bc)3 \mid (b-c), then 3(ac)3 \mid (a-c).

Hence it is an equivalence relation, partitioning Z\mathbb{Z} into three classes {[0],[1],[2]}\{[0], [1], [2]\} (remainders mod 3).

relations
10short5 marks

Solve the recurrence relation (a_n = 2a_{n-1}), (a_0 = 3).

Solving an=2an1,  a0=3a_n = 2a_{n-1},\; a_0 = 3

This is a first-order linear homogeneous recurrence (a geometric progression with common ratio 22). Expanding:

an=2an1=22an2==2na0.a_n = 2a_{n-1} = 2^2 a_{n-2} = \dots = 2^n a_0.

Substituting a0=3a_0 = 3:

an=32n.\boxed{a_n = 3 \cdot 2^n}.

Verification: a0=320=3a_0 = 3\cdot 2^0 = 3 ✓, a1=32=6=2a0a_1 = 3\cdot 2 = 6 = 2a_0 ✓, a2=12=2a1a_2 = 12 = 2a_1 ✓.

recurrence
11short5 marks

What is a planar graph? State Euler's formula for planar graphs.

Planar Graph

A planar graph is a graph that can be drawn (embedded) in the plane so that no two of its edges cross each other except at common vertices. Such a drawing is called a plane graph. For example, K4K_4 is planar, whereas K5K_5 and K3,3K_{3,3} are not planar (by Kuratowski's theorem).

Euler's Formula

For a connected planar graph with VV vertices, EE edges and FF faces (regions, including the unbounded outer face):

VE+F=2.V - E + F = 2.

For a connected simple planar graph with V3V \geq 3, this also gives the bound E3V6E \leq 3V - 6.

Example: A cube graph has V=8V = 8, E=12E = 12, F=6F = 6: 812+6=28 - 12 + 6 = 2 ✓.

graph
12short5 marks

Find the number of ways to select 3 students from a group of 10.

Selecting 3 Students from 10

Since order does not matter, this is a combination problem, (nr)\binom{n}{r} with n=10n = 10, r=3r = 3:

(103)=10!3!(103)!=10×9×83×2×1=7206=120.\binom{10}{3} = \frac{10!}{3!\,(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120.

Result: There are 120\boxed{120} ways to select 3 students from a group of 10.

counting

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