MAT3702
Assignment 2
Due: 15 August 2025
, MAT3702 Assignment 2 (2025)
Question 1
Let R be a ring with a, b, c ∈ R, and let 0R ∈ R be the zero element of R.
(i) If a + b = a + c, then show that b = c
Step 1: Recall that under addition, (R, +) is an abelian group. Hence, every
element has an additive inverse.
Step 2: Given a + b = a + c, we add the additive inverse of a, denoted
−a, to both sides:
(−a) + (a + b) = (−a) + (a + c)
Step 3: Using associativity:
((−a) + a) + b = ((−a) + a) + c
Step 4: Since (−a) + a = 0R , we have:
0R + b = 0R + c
Step 5: Using the identity property of addition:
b=c
Final Answer: b = c, as required.
(ii) Show that a · 0R = 0R = 0R · a
Step 1: Note that in any ring, the distributive law holds: for all a, b, c ∈ R,
a(b + c) = ab + ac and (b + c)a = ba + ca
Also, recall that 0R + 0R = 0R .
Step 2: Compute:
a · 0R = a · (0R + 0R ) = a · 0R + a · 0R
Step 3: Subtract a · 0R from both sides:
a · 0R − a · 0R = (a · 0R + a · 0R ) − a · 0R
1
Assignment 2
Due: 15 August 2025
, MAT3702 Assignment 2 (2025)
Question 1
Let R be a ring with a, b, c ∈ R, and let 0R ∈ R be the zero element of R.
(i) If a + b = a + c, then show that b = c
Step 1: Recall that under addition, (R, +) is an abelian group. Hence, every
element has an additive inverse.
Step 2: Given a + b = a + c, we add the additive inverse of a, denoted
−a, to both sides:
(−a) + (a + b) = (−a) + (a + c)
Step 3: Using associativity:
((−a) + a) + b = ((−a) + a) + c
Step 4: Since (−a) + a = 0R , we have:
0R + b = 0R + c
Step 5: Using the identity property of addition:
b=c
Final Answer: b = c, as required.
(ii) Show that a · 0R = 0R = 0R · a
Step 1: Note that in any ring, the distributive law holds: for all a, b, c ∈ R,
a(b + c) = ab + ac and (b + c)a = ba + ca
Also, recall that 0R + 0R = 0R .
Step 2: Compute:
a · 0R = a · (0R + 0R ) = a · 0R + a · 0R
Step 3: Subtract a · 0R from both sides:
a · 0R − a · 0R = (a · 0R + a · 0R ) − a · 0R
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