Now examine the sum of a rational number, y, and an irrational number, x. The rational number y can be written as y = a/b, where a and b are integers and b ≠ 0. Leave the irrational number x as x because it can’t be written as the ratio of two integers. Let’s look at a proof by contradiction. In other words, we’re trying to show that x + y is equal to a rational number instead of an irrational number. Let the sum equal m/n, where m and n are integers and n ≠ 0. The process for rewriting the sum for x is shown.

Statement l Reason

x + a/b = m/n l substitution

x + a/b - a/b = m/n - a/b l subtraction property of equality

x = m/n -a/b l

x = (b/b) (m/n) -( n/n) (a/b) l Create common denominators.
x = bm/bn - an/bn l

x = bm-an/bn l Simplify

Based on what we established about the classification of x and using the closure of integers, what does the equation tell you about the type of number x must be for the sum to be rational? What conclusion can you now make about the result of adding a rational and an irrational number?