Kenneth Rosen Edition 7 Exercise 1.7 Question 17 (Page No. 91)

Question

Show that if $n$ is an integer and $n^3+5$ is odd, then $n$ is even using.

1. a proof by contraposition.
2. a proof by contradiction.

Answer 1

Proof by Contraposition:

To prove a statement of the form “If A, then B,” do the following:

1. Form the contrapositive. In particular, negate A and negate B.
2. Prove directly that ¬B implies ¬A.

Hence to prove “ if n is an integer and n^3+5 is odd, then n is even”, we will prove “If n is odd, then n^3 + 5 is even”.

If “n” is odd, it can be written in the form :

• n = 2k +1,
• therefore, n^3+5 = (2k+1)^3 + 5 =  = (8k^3 + 1 + 12k^2 + 6k) + 5  = 8k^3 + 12k^2 + 6k + 6 which is equal to 2(4k^3 + 6k^2 + 3k + 3), which is in the form of even integer

Proof by Contradiction:

Assume that “n”  is odd.

Since n is odd, the product of odd numbers is odd(n X n X n). So we can say that n^3 is odd. Also n^ 3 + 5 is odd(from question)

Now if we subtract 5 from n^3 + 5(an odd number) we are getting n^3(an odd number from our assumption). But on subtracting an odd number from another odd number, we should be getting an even number. Thus, our assumption was wrong and it is a contradiction.