Question

Prove that if n is a positive integer, then \(n\) is even if and only if \(7n + 4\)is even.

Short answer

If n is a positive integer, and then n will be even if and only if\(\left( {7n + 4} \right)\)is even.

Step-by-step solution

Introduction

The purpose is to show that if n is a positive integer, and then n is even if and only if \(7n + 4\)is even.

To prove \(p \leftrightarrow q\) it is enough to prove that \(\left( {p \to q} \right)\)and\(\left( {q \to p} \right)\).

Here, the statements are as follows:

p: n is even

q: \(\left( {7n + 4} \right)\)is even.

Proof of \(\left( {p \to q} \right)\) )part

Firstly prove \(\left( {p \to q} \right)\)part that is, if n is even then \(\left( {7n + 4} \right)\) is even

Suppose that n is an even number.

Then, \(n = 2k\)for any integer k.

Now,

\(7n + 4 = 7\left( {2k} \right) + 4\)

\(\begin{array}{l} = 14k + 4\\ = 2\left( {7k + 2} \right)\\ = 2\left( m \right)\end{array}\)

For \(m = 7k + 2\)

Therefore, \(\left( {7n + 4} \right)\)is even. …………… (1)

Prove of \(\left( {q \to p} \right)\)

Now, prove the second part ,\(\left( {q \to p} \right)\).

This means, if \(\left( {7n + 4} \right)\)is even then n is an even.

To prove this, use contrapositive result.

That is,

Thus, assume that n is not even means n is odd.

This implies, \(n = 2k + 1\)for some integer k.

Then,

\(7n + 4 = 7\left( {2k + 1} \right) + 4\)

\(\begin{array}{l} = 14k + 7 + 4\\ = 14k + 11\\ = 14k + 10 + 1\\ = 2\left( {7k + 5} \right) + 1\\ = 2\left( m \right) + 1\end{array}\)

For,\(m = 7k + 5\).

Hence,\(\left( {7n + 4} \right)\)is odd.

Thus, proved that \( - p \to - q\)

This is equivalent to\(\left( {q \to p} \right)\).

Therefore, if n is a positive integer, and then n will be even if and only if \(\left( {7n + 4} \right)\)is even.