# A Simple Proof by Contradiction

Use a proof by contradiction to show that the sum of an irrational number and a rational number is irrational (using the definitions of rational and irrational real numbers; cf p85 of the textbook, or your notes).

Theorem: If $r$ is an irrational number and $s$ is a rational number, then $latex r+s$ is irrational.

Proof: For a proof by contradiction, assume that the hypotheses are true (i.e., thatĀ $r$ is irrational and $s$ isĀ  rational) but that the conclusion is false, i.e., $r+s$ is not irrational.

That meansĀ $r+s$ is rational, and so by the definition of rational numbers, $r + s = a/b$ for integers $a, b$.

Then $r = (a/b) – s$.Ā  There are two cases for $s$:

(i) $s$ is irrational: this contradicts the hypothesis that s is rational.

(ii) $s$ is rational: then $s = c/d$ for integers $c, d$. But then

$$r = (a/b) – s = (a/b) – (c/d) = (ad – bc)/bd$$

meaning r is rational.Ā  But that contradicts the hypothesis that r is irrational.

Since we get a contra