Question

Prove that the following is true for all positive integers n: n is odd if and only if n2 is odd.

Answer 1

The contrapositive of an If-Then Statement
If p, then q is If ~q, then ~p

The conditional statement is logically equivalent to its contrapositive.

So,

p→q ≡ ~q→ ~p which means if the conditional stamement is true, its contrapositive is also true.

The even number can be expressed as the product of 2 and an integer, that is, 2k for some integer k.

Example,

10= 2(5)

12= 2(6)

The contrapositive of the theorem that we want to prove. If the contrapositive is proven true, the original statement must be also true.

If n^2 is odd, then n is odd    is If n is even, then n^2 is even.

Assume that n is even that means we can express it as n = 2k.

So,

n²=(2k)²=4k²

Now factor out 2 from RHS,

n²=2(2k²) (If k is an integer 2k²is also an integer)

So, we can write it as, where t is an integer and 2t is clearly the algebraic form of an even number.

n²=2t

The contrapositive to be true makes the original conditional statement true. So, we have that if n² is odd, then n is odd.