ASSIGNMENT 01
Closing Date: 30 April 2026
Total Marks: 100
UNIQUE ASSIGNMENT NUMBER: 197154
All questions solved with full working. If need assignment or exam help any module
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AT2613 Assignment 01 Complete Solutions
, Question 1
Given statement:
𝐴:If 𝑛2 is odd, then 𝑛 is odd.
(1.1) Contrapositive
The contrapositive of 𝑃 ⇒ 𝑄is ¬𝑄 ⇒ ¬𝑃.
• 𝑃: 𝑛2 is odd
• 𝑄: 𝑛is odd
Thus:
¬𝑄: 𝑛 is even, ¬𝑃: 𝑛2 is even
Contrapositive:
If 𝑛 is even, then 𝑛2 is even.
Proof:
Let 𝑛 = 2𝑘, where 𝑘 ∈ ℤ.
𝑛2 = (2𝑘)2 = 4𝑘 2 = 2(2𝑘 2 )
Thus, 𝑛2 is even.
Conclusion:
The contrapositive is true, hence the original statement is true.
(1.2) Converse
The converse is:
If 𝑛 is odd, then 𝑛2 is odd.
Proof:
Let 𝑛 = 2𝑘 + 1, 𝑘 ∈ ℤ.
𝑛2 = (2𝑘 + 1)2 = 4𝑘 2 + 4𝑘 + 1 = 2(2𝑘 2 + 2𝑘) + 1