Give a proof by contradiction of the theorem
show that “If 3n + 2 is odd, then n is odd.”
Solution:
Let
p : “3n + 2 is odd”
q: “n is odd.”
¬ p: “3n + 2 is even”
¬ q: “n is even.”
To construct a proof by contradiction,
we have to prove that
¬ q → ¬ p is true i.e " If n is even 3n + 2 , then 3n + 2 is even"
Because n is even, there is an integer k such
that n = 2k.
This implies that
3n + 2 = 3(2k) + 2 = 6k + 2 = 2(3k + 1) is even which is contradiction where 3n + 2 is odd.
therefore, ¬ q → ¬ p is true that implies that q → p is true
so, If 3n + 2 is odd, then n is odd
