MAT3702 Assignment 2 2026 | Due 19 August 2026 - Distinction Guaranteed

MAT 3702 Assignment 2 2026
Due Date: 19 August 2026



1. Prove that b | a if and only if (−b) | a.

2. If a | b and b | c, then prove that a | c.

3. Let R, S be rings and consider the following subsets of R × S



Question 1

Prove that b ∣ a if and only if (−b) ∣ a.


We work in the ring of integers Z.
Recall the definition: For integers x, y , we say x ∣ y if there exists an integer k such that y = xk .

(⇒) Assume b ∣ a. Then a = bk for some integer k .
Since bk = (−b)(−k), we have a = (−b)(−k).
Hence (−b) ∣ a.

(⇐) Assume (−b) ∣ a. Then a = (−b)m for some integer m.
But (−b)m = b(−m), so a = b(−m).
Hence b ∣ a.

Thus b ∣ a ⟺ (−b) ∣ a.



Question 2

If a ∣ b and b ∣ c, prove that a ∣ c.
Solution
Assume a ∣ b and b ∣ c. Then there exist integers x, y such that
b = ax and c = by .
Substituting: c = (ax)y = a(xy).
Since xy is an integer, we conclude a ∣ c.

, Question 3




Let R, S be rings. Consider the subsets
R = {(r, 0S ) ∣ r ∈ R} and S = {(0R , s) ∣ s ∈ S},
where 0R , 0S are the zero elements of R, S respectively.

(i) If R = Z3 and S = Z5 , find R and S .
Solution
The elements of Z3 are 03 , 13 , 23 ; the zero element is 03 .
The elements of Z5 are 05 , 15 , 25 , 35 , 45 ; the zero element is 05 .

Thus



R = {(03 , 05 ), (13 , 05 ), (23 , 05 )}.

S = {(03 , 05 ), (03 , 15 ), (03 , 25 ), (03 , 35 ), (03 , 45 )}.