# Prove that the sum of a rational number and an irrational number is irrational

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 m / n

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

x
/ y x / y a = m / n x / y

a = m /x / y

a = m /x / y

Write m /x /with the same denominator.

a = m /x y

The common denominator is ny

Multiply m /and x / y by ny ny

a = (m / n)(ny ny) - (x / y)(ny ny)

a = (× n/ (n2y) - (× ny / (ny2)

a = (m/ (ny) - (xn / (ny)

a = (my - xn) / (ny)

If/ y and m /are rational numbers, then we know that x, y, m, and n are integers. This means myxn, 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.

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