MTH202 Discrete Mathematics Assignment 1 for Fall 2026 covers propositional logic and sets. The assignment carries 20 marks and the due date is 9th November, 2026. The solution file linked below answers all four questions: conditional statements, Boolean circuit equivalence, logical symbols for inequalities, and finding three sets from their intersections and differences.
What Does the Solution File Cover?
What Are the Converse, Inverse and Contrapositive?
For the statement “If a number is divisible by 6, then it is divisible by 3”, the converse swaps the two parts: “If a number is divisible by 3, then it is divisible by 6.” The inverse negates both parts: “If a number is not divisible by 6, then it is not divisible by 3.” The contrapositive swaps and negates both parts: “If a number is not divisible by 3, then it is not divisible by 6.” Only the contrapositive is logically equivalent to the original statement. The same step by step answer style is used in the STA301 Assignment 1 solution for Fall 2026.
How Are the Two Boolean Circuits Proved Equivalent?
The first circuit feeds three AND gates into one OR gate, giving (P ∧ Q) ∨ (¬P ∧ Q) ∨ (P ∧ ¬Q). The second circuit is a single OR gate, P ∨ Q. The proof uses only the laws of logic, as the question requires. Factoring Q out of the first two terms with the distributive law, then applying the negation and identity laws, reduces the long expression to Q ∨ (P ∧ ¬Q). One more distributive step and the negation law leave Q ∨ P, which is P ∨ Q by the commutative law. Verifying an answer against the given conditions, rather than guessing, is the same approach used in the CS201 Assignment 1 solution.
How Are the Inequalities Written With p, q and r?
With p: x > 1, q: x < 9 and r: x = 1, the inequality x ≤ 1 is simply ¬p, because it is the exact negation of x > 1. The range 1 ≤ x < 9 needs both parts: (p ∨ r) ∧ q, since x must equal 1 or exceed it while staying below 9. The inequality x ≥ 9 is ¬q, the exact negation of x < 9.
What Are the Sets A, B and C?
The element 5 sits in both A ∩ B and A ∩ C, so B ∩ C = {5}. That gives B = (B − C) ∪ (B ∩ C) = {2, 4, 5, 6} and C = (C − B) ∪ (B ∩ C) = {5, 7, 8}. Set A collects its three disjoint parts: A = (A ∩ B) ∪ (A ∩ C) ∪ (A − (B ∪ C)) = {1, 2, 3, 5, 7}. All five given conditions check out against these three sets.
How Do You Download the Solution File?
You can download the complete solution file from Google Drive using the link below. It contains all four questions with short answers, the full laws of logic proof for Question 2, and the set working with a final check for Question 4.
Frequently Asked Questions
Which of the Converse, Inverse and Contrapositive Is Equivalent to the Original?
The contrapositive. For any conditional statement, the contrapositive always has the same truth value as the original, while the converse and inverse are equivalent to each other but not to the original.
Why Does the Circuit Proof Avoid a Truth Table?
Because the question asks for the laws of logic, not a truth table. The distributive, negation, identity and commutative laws transform the long expression into P ∨ Q step by step, which also shows why the two circuits behave identically.
How Do You Rebuild Set B From the Given Information?
Every element of B is either in C or outside it. The elements outside C are given directly as B − C = {2, 4, 6}, and the elements inside C come from B ∩ C = {5}, found through the two intersections with A. Their union is B = {2, 4, 5, 6}.
Last updated: November 2026.
