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 $$r+s$$ is irrational.

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

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

Then $r=(a/b)–s$.  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.  But that contradicts the hypothesis that r is irrational.

Since we get a contra