MAT3702 Assignment ONE 2024

MAT3702 ASSIGNMENT 1 2024

Question 1

𝑓(𝑥) = 2𝑥 + 3

𝑔(𝑥) = 𝑥 2 + 1

(𝑓𝑔)(𝑥) = 𝑓(𝑔(𝑥))

(𝑓𝑔)(𝑥) = 𝑓(𝑥 2 + 1)

(𝑓𝑔)(𝑥) = 2(𝑥 2 + 1) + 3

(𝑓𝑔)(𝑥) = 2𝑥 2 + 2 + 3

(𝑓𝑔)(𝑥) = 2𝑥 2 + 5


(𝑔𝑓)(𝑥) = 𝑔(𝑓(𝑥))

(𝑔𝑓)(𝑥) = 𝑔(2𝑥 + 3)

(𝑔𝑓)(𝑥) = (2𝑥 + 3)2 + 1

(𝑔𝑓)(𝑥) = 4𝑥 2 + 12𝑥 + 9 + 1

(𝑔𝑓)(𝑥) = 4𝑥 2 + 12𝑥 + 10


Therefore, (𝑓𝑔)(𝑥) ≠ (𝑔𝑓)(𝑥)

Question 2

~ is an equivalence relation on set 𝐴 with 𝑎, 𝑏 ∈ 𝐴

To prove that 𝑎~𝑏 ⇔ [𝑎] = [𝑏]

(i) ⇒:

Suppose that
𝑎~𝑏

Let
𝑥 ∈ [𝑎]

⇒ 𝑥~𝑎

So, 𝑥~𝑎 and 𝑎~𝑏

Now, since ~ is an equivalence relation, then ~ is transitive

, 𝑥~𝑎 and 𝑎~𝑏

⇒ 𝑥~𝑏

⇒ 𝑥 ∈ [𝑏]

So, 𝑎~𝑏 ⇒ [𝑎] = [𝑏]


(ii) ⇐:

Suppose that
[𝑎] = [𝑏]

Note that 𝑎~𝑎 because equivalence relations are reflexive.

⇒ 𝑎 ∈ [𝑎]

⇒ 𝑎 ∈ [𝑏] because [𝑎] = [𝑏]

⇒ 𝑎 ∈ [𝑏]

⇒ 𝑎~𝑏

So, [𝑎] = [𝑏] ⇒ 𝑎~𝑏

By (i) and (ii), 𝑎~𝑏 ⇔ [𝑎] = [𝑏]

Question 3

1 0 𝑖 0 0 1 0 𝑖
1=[ ], 𝑖=[ ], 𝑗=[ ], 𝑘=[ ]
0 1 0 −𝑖 −1 0 𝑖 0

𝑖 0 𝑖 0
𝑖2 = [ ][ ]
0 −𝑖 0 −𝑖
2 2
𝑖2 = [ 𝑖 + 0 𝑖 × 0 − 0𝑖 ]
0𝑖 − 𝑖 × 0 02 + 𝑖 2

−1 0
𝑖2 = [ ]
0 −1

1 0
𝑖2 = − [ ]
0 1

𝑖 2 = −1


0 1 0 1
𝑗2 = [ ][ ]
−1 0 −1 0

0×0−1×1 0×1+1×0
𝑗2 = [ ]
−1 × 0 − 1 × 0 −1 × 1 + 0 × 0