Non-Conditional Statements Exploration

An integer a is odd if and only if a³ is odd. Proof. Suppose that a is odd. Then a = (a)________________. for some integer n³, and a³ = (2n+1)³ = 8n³+12n² +6n+1 = 2(4n³ +6n² +3n)+1. This shows that a³ is twice an integer, plus 1, so a³ is odd. Thus, we’ve proved that if a is (b)______________ then a³ is (d)__________________. Conversely, we need to show that if a³ is odd, then a is odd. For this we employ (e)_________________________. Suppose a is not odd. Thus, a is even, so a = 2n for some integer n. Then a³ = (2n)³ = 8n³ = 2(4n³) is even (not odd).

Prove the following statement. This exercise is cumulative, covering all techniques addressed in this module. "There is a prime number between 90 and 100."