Problem Set

Problem 1. Prove that $(p \implies q) \lor (q \implies p)$ is a tautology using a truth table.

Solution

$p$	$q$	$p \implies q$	$q \implies p$	disjunction
T	T	T	T	T
T	F	F	T	T
F	T	T	F	T
F	F	T	T	T

All rows give $T$ So it is a tautology.

If you get this wrong, revise: Section 1.1 and Section 1.4.

Problem 2. Convert $(\neg p \lor q) \land (r \lor \neg s)$ to DNF.

Solution

The formula is a conjunction of two clauses. Distribute $\land$ over $\lor$ :

$(\neg p \lor q) \land (r \lor \neg s) = (\neg p \land r) \lor (\neg p \land \neg s) \lor (q \land r) \lor (q \land \neg s)$ .

This is in DNF (disjunction of four terms, each a conjunction of two literals).

If you get this wrong, revise: Section 1.6.

Problem 3. Negate: “For every real number $x$ There exists a real number $y$ such that $y \gt x$ .”

Solution

Original: $\forall x\, \exists y\, (y \gt x)$ .

Negation: $\exists x\, \forall y\, (y \leq x)$ I.e., “there exists a real number $x$ such that every Real number $y$ satisfies $y \leq x$ .”

If you get this wrong, revise: Section 1.2.

Problem 4. Prove that $A \subseteq B$ if and only if $A \cap B^c = \emptyset$ .

Solution

( $\Rightarrow$ ) Assume $A \subseteq B$ . Let $x \in A \cap B^c$ . Then $x \in A$ and $x \notin B$ . But $A \subseteq B$ implies $x \in B$ Contradiction. So $A \cap B^c = \emptyset$ .

( $\Leftarrow$ ) Assume $A \cap B^c = \emptyset$ . Let $x \in A$ . If $x \notin B$ Then $x \in B^c$ So $x \in A \cap B^c = \emptyset$ Contradiction. Hence $x \in B$ Proving $A \subseteq B$ . $\blacksquare$

If you get this wrong, revise: Section 2.1.

Problem 5. Show that the relation $R$ on $\mathbb{Z}$ defined by $a\,R\,b$ iff $a - b$ is even Is an equivalence relation. How many equivalence classes are there?

Solution

Reflexive: $a - a = 0$ Which is even. ✓ Symmetric: If $a - b$ is even, then $b - a = -(a - b)$ is even. ✓ Transitive: If $a - b$ and $b - c$ are even, then $a - c = (a - b) + (b - c)$ is even. ✓

The equivalence classes are $[0] = \\{\mathrm{even\; integers{}\\}}$ and $[1] = \\{\mathrm{odd\; integers{}\\}}$ . There are exactly 2 equivalence classes.

If you get this wrong, revise: Section 2.2.

Problem 6. Let $f : \mathbb{R} \to \mathbb{R}$ be $f(x) = 2x + 1$ and $g : \mathbb{R} \to \mathbb{R}$ Be $g(x) = x^2$ . Find $g \circ f$ and $f \circ g$ . Is $g \circ f$ injective?

Solution

$(g \circ f)(x) = g(f(x)) = g(2x + 1) = (2x + 1)^2 = 4x^2 + 4x + 1$ .

$(f \circ g)(x) = f(g(x)) = f(x^2) = 2x^2 + 1$ .

Note $g \circ f \neq f \circ g$ So composition is not commutative.

$g \circ f$ is not injective: $(g \circ f)(0) = 1$ and $(g \circ f)(-1) = 4(-1)^2 + 4(-1) + 1 = 1$ but $0 \neq -1$ .

If you get this wrong, revise: Section 2.3.

Problem 7. Prove that the set of all infinite binary sequences is uncountable.

Solution

Suppose the set $S$ of all infinite binary sequences is countable, so $S = \\{s_1, s_2, s_3, \ldots\\}$ Where $s_i = (s_{i1}, s_{i2}, s_{i3}, \ldots)$ with each $s_{ij} \in \\{0, 1\\}$ .

Define $t = (t_1, t_2, t_3, \ldots)$ by $t_i = 1 - s_{ii}$ (flip the $i$ -th bit of the $i$ -th Sequence). Then $t \in S$ but $t \neq s_i$ for every $i$ (they differ in position $i$ ). This Contradicts $S = \\{s_1, s_2, \ldots\\}$ . Therefore $S$ is uncountable. $\blacksquare$

If you get this wrong, revise: Section 2.4.

Problem 8. Prove: the product of any two odd integers is odd.

Solution

Let $a = 2m + 1$ and $b = 2n + 1$ be odd. Then $ab = (2m+1)(2n+1) = 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1$ Which is odd. $\blacksquare$

If you get this wrong, revise: Section 3.1.

Problem 9. Prove: if $3n + 2$ is odd, then $n$ is odd.

Solution

By contrapositive: assume $n$ is even, so $n = 2k$ . Then $3n + 2 = 6k + 2 = 2(3k + 1)$ Which is Even. $\blacksquare$

If you get this wrong, revise: Section 3.2.

Problem 10. Prove: $\sqrt{3}$ is irrational.

Solution

Suppose $\sqrt{3} = p/q$ in lowest terms. Then $3q^2 = p^2$ So $3 \mid p^2$ Hence $3 \mid p$ . Write $p = 3r$ . Then $3q^2 = 9r^2$ So $q^2 = 3r^2$ Giving $3 \mid q^2$ and $3 \mid q$ . But then $\gcd(p, q) \geq 3$ Contradicting lowest terms. $\blacksquare$

If you get this wrong, revise: Section 3.3.

Problem 11. Prove by induction: $\sum_{i=0}^{n} 2^i = 2^{n+1} - 1$ for all $n \geq 0$ .

Solution

Base case: $n = 0$ : $2^0 = 1 = 2^1 - 1$ . ✓

Inductive step: Assume $\sum_{i=0}^{k} 2^i = 2^{k+1} - 1$ . Then

\sum_{i=0}^{k+1} 2^i = (2^{k+1} - 1) + 2^{k+1} = 2 \cdot 2^{k+1} - 1 = 2^{k+2} - 1

$\blacksquare$

If you get this wrong, revise: Section 3.4.

Problem 12. Prove by strong induction that every integer $n \geq 2$ is a product of primes.

Solution

Base case: $n = 2$ is prime, hence a product of primes.

Inductive step: Assume every integer in $\\{2, \ldots, k\\}$ is a product of primes ( $k \geq 2$ ). If $k + 1$ is prime, done. If $k + 1$ is composite, then $k + 1 = ab$ where $2 \leq a, b \leq k$ . By the induction hypothesis, both $a$ and $b$ are products of primes, so $k + 1 = ab$ is too. $\blacksquare$

If you get this wrong, revise: Section 3.4 (strong induction).

Problem 13. A committee of 5 is to be chosen from 12 people. How many ways if two specific People must either both serve or neither serves?

Solution

Case 1: Both serve. Choose the remaining 3 from the other 10: $\binom{10}{3} = 120$ . Case 2: Neither serves. Choose all 5 from the other 10: $\binom{10}{5} = 252$ .

Total: $120 + 252 = 372$ .

If you get this wrong, revise: Section 4.2.

Problem 14. How many integers from 1 to 500 are divisible by 3 or 7 but not by 5?

Solution

$|A_3| = 166$ , $|A_7| = 71$ , $|A_3 \cap A_7| = 23$ . Divisible by 3 or 7: $166 + 71 - 23 = 214$ .

$|A_3 \cap A_5| = 33$ , $|A_7 \cap A_5| = 14$ , $|A_3 \cap A_7 \cap A_5| = 4$ . Divisible by 3 or 7 and 5: $33 + 14 - 4 = 43$ .

Divisible by 3 or 7 but not 5: $214 - 43 = 171$ .

If you get this wrong, revise: Section 4.3.

Problem 15. How many solutions does $x_1 + x_2 + x_3 = 15$ have with $x_i \geq 1$ ?

Solution

Substitute $y_i = x_i - 1 \geq 0$ : $y_1 + y_2 + y_3 = 12$ with $y_i \geq 0$ . By stars and bars: $\binom{12 + 3 - 1}{3 - 1} = \binom{14}{2} = 91$ .

If you get this wrong, revise: Section 4.4.

Problem 16. Prove that among any 13 people, at least 2 were born in the same month.

Solution

There are 12 months and 13 people. By the pigeonhole principle, at least one month contains at least $\lceil 13/12 \rceil = 2$ people.

If you get this wrong, revise: Section 4.5.

Problem 17. Verify that $C_3 = 5$ using the Catalan recurrence and the closed form.

Solution

By recurrence: $C_3 = C_0 C_2 + C_1 C_1 + C_2 C_0 = 1 \cdot 2 + 1 \cdot 1 + 2 \cdot 1 = 2 + 1 + 2 = 5$ .

By closed form: $C_3 = \frac{1}{4}\binom{6}{3} = \frac{1}{4} \cdot 20 = 5$ . ✓

If you get this wrong, revise: Section 4.6.

Problem 18. Find the chromatic number of $C_5$ (the 5-cycle) and justify.

Solution

$C_5$ is an odd cycle. It cannot be 2-coloured (an odd cycle requires 3 colours: colour the first Vertex 1, alternate 2 and 1 around, and the last vertex (5th) is adjacent to both the 4th (colour 2) And the 1st (colour 1), so it needs colour 3).

A 3-colouring exists: label vertices $v_1, \ldots, v_5$ ; colour $v_1 = 1$ , $v_2 = 2$ , $v_3 = 1$ $v_4 = 2$ , $v_5 = 3$ .

Therefore $\chi(C_5) = 3$ .

If you get this wrong, revise: Section 5.5.

Problem 19. In the bipartite graph with partitions $X = \\{1, 2, 3\\}$ and $Y = \\{a, b, c, d\\}$ And edges $1$ — $a,b$ ; $2$ — $b,c$ ; $3$ — $c,d$ Verify Hall’s condition and find a matching covering $X$ .

Solution

Neighbourhoods: $N(\\{1\\}) = \\{a, b\\}$ , $N(\\{2\\}) = \\{b, c\\}$ , $N(\\{3\\}) = \\{c, d\\}$ . $N(\\{1, 2\\}) = \\{a, b, c\\}$ , $N(\\{1, 3\\}) = \\{a, b, c, d\\}$ , $N(\\{2, 3\\}) = \\{b, c, d\\}$ . $N(\\{1, 2, 3\\}) = \\{a, b, c, d\\}$ .

All satisfy $|N(S)| \geq |S|$ . A matching: $1$ — $a$ , $2$ — $b$ , $3$ — $c$ .

If you get this wrong, revise: Section 5.7.

Problem 20. Solve $a_n = 3a_{n-1} - 2a_{n-2}$ with $a_0 = 0$ , $a_1 = 1$ .

Solution

Characteristic equation: $r^2 - 3r + 2 = 0$ Giving $(r - 1)(r - 2) = 0$ So $r_1 = 1$ , $r_2 = 2$ .

$a_n = A \cdot 1^n + B \cdot 2^n = A + B \cdot 2^n$ .

$a_0 = A + B = 0 \implies A = -B$ . $a_1 = A + 2B = 1 \implies -B + 2B = B = 1$ .

So $A = -1$ , $B = 1$ Giving $a_n = 2^n - 1$ . $\blacksquare$

If you get this wrong, revise: Section 6.2 and Section 6.3.

Common Pitfalls

Confusing $\Rightarrow$ and $\Leftrightarrow$ . $P \Rightarrow Q$ means “if $P$ then $Q$ ” (sufficient condition). $P \Leftrightarrow Q$ means “if and only if” (necessary and sufficient). Fix: $P \Rightarrow Q$ is equivalent to $\neg Q \Rightarrow \neg P$ (contrapositive), not $Q \Rightarrow P$ (converse).
Wrong induction base case. The base case must be the smallest value for which the statement is claimed to hold. Fix: If the claim starts at $n = 1$ , prove $n = 1$ ; if at $n = 0$ , prove $n = 0$ .
Confusing combinations and permutations with repetition. Standard formulas assume distinct, non-repeated items. Fix: With repetition: $n^r$ (ordered) or $\binom{n+r-1}{r}$ (unordered).

Worked Examples

Example 1: Proof by induction

Problem. Prove that $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ for all $n \geq 1$ .

Solution. Base ( $n = 1$ ): $1 = 1(2)(3)/6 = 1$ . ✓ Inductive step: assume true for $n$ . Then $\sum_{k=1}^{n+1} k^2 = \frac{n(n+1)(2n+1)}{6} + (n+1)^2 = \frac{(n+1)[n(2n+1) + 6(n+1)]}{6} = \frac{(n+1)(2n^2 + 7n + 6)}{6} = \frac{(n+1)(n+2)(2n+3)}{6}$ . ✓

$\blacksquare$

Example 2: Graph theory

Problem. A connected planar graph has 6 vertices and 8 edges. How many faces does it have?

Solution. By Euler’s formula: $V - E + F = 2$ . $6 - 8 + F = 2 \implies F = 4$ .

$\blacksquare$

Summary

Logic: propositions, truth tables, quantifiers; implication, contrapositive, converse.
Proof techniques: direct, contrapositive, contradiction, induction.
Combinatorics: permutations, combinations, inclusion-exclusion, pigeonhole principle.
Graph theory: Euler’s formula ( $V - E + F = 2$ ), trees, planarity, chromatic number.

Intuition

Practice problems consolidate theoretical understanding. Work through problems systematically: understand the problem, identify relevant concepts, design a solution approach, implement or prove, and verify. Start with concrete examples, then generalize. This iterative process builds intuition for recognizing problem types and selecting appropriate techniques from your mathematical toolkit.

Cross-References

Topic	Site	Link
Discrete Mathematics (Overview)	WyattsNotes	View
Theory of Computation	WyattsNotes	View
Abstract Algebra	WyattsNotes	View
Number Theory	WyattsNotes	View
Discrete Mathematics, MIT OCW	MIT	View

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Problem Set

Common Pitfalls

Worked Examples

Example 1: Proof by induction

Example 2: Graph theory

Summary

Intuition

Cross-References

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