THIS ISNOTHOMEWORK! This is a practice problem from our textbook. Class link: Discrete Math I

Please correct if wrong:

My solution:

1) If n is odd, then n^2 is odd.

2) Suppose that n=2a+1, where a is an arbitrary integer.

3) Then n^2=(2a+1)^2 (Substitution)

<=> (2a+1)(2a+1) (Distributive law)

<=> 4a^2 + 4a + 1 (Distributive law, Associative law, Commutative law)

<=> 2(2a^2 + 2a) + 1 (Distributive law)

4) Since 2(2a^2 + 2a) + 1 = 2k + 1 (where k=2a^2 + 2a , and is therefore an integer) the proposition is true using the definition of odd: 2k + 1