Answer to Question #350542 in Discrete Mathematics for Kenetha

Question

Using proof by contraposition, show that if n is an integer and 5 added to its cube is odd then n is even. Before showing your solution, rewrite the statement to the proper form of a conditional statement then assign variables to the simple propositions. Show also the contrapositive form of the simple propositions before proceeding to your solution.

Answer 1

Assign variables to the simple propositions:

"A(n) = "n^3 + 5 \\text{ is odd}"; \\\\\nB(n) = "n \\text{ is even}"."

Then we need to show that

$A(n) \implies B(n).$

By contraposition it is equivalent to

$\lnot B(n) \implies \lnot A(n)$

(contrapositive form), that is "if n is odd then n³ + 5 is even".

Let n be odd. Then n³ is also odd (because if n doesn't have 2 as a divisor, then n³ doesn't as well). Hence n³ + 5 is even as the sum of two odd numbers. Q. E. D.