In order to prove that the sum of a rational number and an irrational number is irrational, we will use a proof by contraction.

What is a proof by contradiction?

Suppose we want to prove that a math statement is true. The first thing to do is to assume that the math statement is false. Then you need to show that this will lead to a contradiction. If it leads to a contradiction, then the statement must be true.

Assume that the the sum of a rational number and an irrational number is not irrational, but rational.

rational number + irrational number = rational

Let x / y represent the rational number, let a represent the irrational number, and let m / n represent the sum that is "rational"

x / y + a = m / n

Isolate a by subtracting x / y from each side of the equation.

x / y - x / y + a = m / n - x / y

0 + a = m / n - x / y

a = m / n - x / y

Write m / n - x / y with the same denominator.

a = m / n - x / y

The common denominator is ny

Multiply m / n and x / y by ny / ny

a = (m / n)(ny / ny) - (x / y)(ny / ny)

a = (m × ny / (n^{2}y) - (x × ny / (ny^{2})

a = (my / (ny) - (xn / (ny)

a = (my - xn) / (ny)

If x / y and m / n are rational numbers, then we know that x, y, m, and n are integers. This means my, xn, and ny are integers. This will also mean that my - xn is an integer.

Therefore, the expression a = (my - xn) / (ny) is a rational number since it is made by dividing two integers.

This is a contradiction since a is an irrational number as stated above. a cannot be rational and irrational at the same.

Therefore, we can conclude that the sum of a rational number and an irrational number is indeed irrational.