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

The purpose of this question is to prove that $n$ is a positive and even integer if and only if $7n + 4$ is also even.

Even numbers can be equally divided into two pairs or groups and are completely divisible by two. For instance, $2, 4, 6, 8$, and so on are said to be even numbers, which can be divided into equal groups. This type of pairing cannot be made for numbers such as $5, 7, 9$, or $11$. As a result, $5, 7, 9$, or $11$ are not even numbers. The sum and difference of any two even numbers is also an even number. The product of two even numbers is even in addition to being divisible by $4$. The even number leaves a remainder of $0$ when it is divisible by $2$.

Odd numbers are those that simply cannot be equally divided by two. For instance, $1, 3, 5, 7$, and so on are odd integers. An odd number leaves a remainder of $1$ when divided by $2$. Odd numbers are the inverse notion of even numbers. Odd numbers cannot be grouped into pairs.  More generally, all numbers other than multiples of $2$ are odd.

Suppose that $n$ is even then by definition, there exists an integer $k$ such that $n=2k$. Substituting this in $7n + 4$:

$7(2k)+4$

$=14k+4$

$=2(7k+2)$

Hence, an integer $m=7k+2$ can be found such that $7n+4=2m$. Or to put it in another way, $7n+4$ is an even number.

Now to prove that if $7n+4$ is an even number then $n$ is even. For this, suppose that $n$ is odd, and then by definition, there exists an integer $k$ such that $n=2k+1$. Substituting this in $7n + 4$:

$7(2k+1)+4$

$=14k+7+4$

$=14k+10+1$

$=2(7k+5)+1$

Hence, an integer $m=7k+5$ can be found such that $7n+4=2m+1$. Or to put it in another way, $7n+4$ is an odd number which is a contradiction. Thus, the contradiction arises due to the wrong supposition and hence $n$ is an even number.

## Example

Prove that the difference between two odd numbers is an even number.

### Solution

Suppose that $p$ and $q$ are two odd numbers, then by definition:

$p=2k_1+1$  and  $q=2k_2+1$, where $k_1$ and $k_2$ belongs to the set of integers.

Now,  $p-q=2k_1+1-(2k_2+1)$

$p-q=2k_1-2k_2$

$p-q=2(k_1-k_2)$

which will leave a remainder of $0$ when divided by $2$, and hence it is proved that the difference between two odd numbers is an even number.