Question

Show that if n is an integer and \(\left( {{n^3} + 5} \right)\)is odd, then n is even using

a) a proof by contraposition.

b) a proof by contradiction.

Short answer

If \(n\)is an integer and \(\left( {{n^3} + 5} \right)\)is odd, then \(n\)has to be even is true by both the methods of contraposition and contradiction.

Step-by-step solution

Introduction for part (a)

(a)

“If \(n\)is an integer and\(\left( {{n^3} + 5} \right)\)is odd, then\(n\)has to be even.”

Contraposition statement

Contraposition is \( - q \to - p\)when the statement is\(p \to q\).

In the above statement,

\(p\): \(n\)is an integer and\(\left( {{n^3} + 5} \right)\)is to be odd

\( - p\): \(n\)is an integer and\(\left( {{n^3} + 5} \right)\)is to be even

\(q\): \(n\)is even

\( - q\): \(n\)is odd

Contraposition statement: If \(n\)is odd, then \(n\) is an integer and\(\left( {{n^3} + 5} \right)\)has to be even.

Prove of part (a) using contraposition

Assume that \(n\)is odd, then by definition of odd numbers \(n = 2k + 1\)for any integer\(k\).

Consider

\({n^3} + 5 = {\left( {2k + 1} \right)^3} + 5\)

\(\begin{aligned}{l} = 8{k^3} + 12{k^2} + 6k + 1 + 5\\ = 2\left( {4{k^3} + 6{k^2} + 3k + 3} \right)\\ = 2p\end{aligned}\)

Taking \(p = 4{k^3} + 6{k^2} + 3k + 3\)

Thus, \(\left( {{n^3} + 5} \right)\)is even.

Hence, the contraposition statement is,

“If \(n\)is odd, then \(n\) is an integer and\(\left( {{n^3} + 5} \right)\)has to be even.”

Therefore, the original statement is,

“If \(n\)is an integer and\(\left( {{n^3} + 5} \right)\)is odd, then \(n\)has to be even.”

Introduction for part (b)

Consider the statement,

“If \(n\)is an integer and\(\left( {{n^3} + 5} \right)\)is odd, then \(n\)has to be even.”

The purpose is to show this statement true using contradiction method of proving.

Prove of part (b) using contradiction

Suppose that \(\left( {{n^3} + 5} \right)\)is odd and \(n\)is odd.

Using the result that product of two odd numbers is odd twice to\({n^3}\).

It gives that \({n^3}\)is odd.

But, \(\left( {{n^3} + 5} \right) - {n^3} = 5\)is an even number as the subtraction of two odd numbers is even.

That is, \(\left( {{n^3} + 5} \right) - {n^3} = 5\)is even.

This is not true as 5 is odd number.

Thus, the supposition is wrong.

Therefore, if\(n\)is an integer and \(\left( {{n^3} + 5} \right)\)is odd, then \(n\)has to be even.