# Solution

Prove that if $a$ is a rational number and $b$ is an irrational number, then $a+b$ is an irrational number.

Let's try to argue indirectly...

Assume that $a+b$ is a rational number. Then we can find integers $m$ and $n$ where $$a+b=\frac{m}{n}$$ Since $a$ is also rational, we can find integers $s$ and $t$ where $$a = \frac{s}{t}$$ Consider what happens when we solve the first equation for $b$: \begin{align}b &= \frac{m}{n} - a\\\\ &= \frac{m}{n} - \frac{s}{t}\\\\ &= \frac{mt}{nt} - \frac{ns}{nt}\\\\ &= \frac{mt-ns}{nt} \end{align} Notice that both $mt-ns$ and $nt$ are integers. We have just written $b$ as the quotient of two integers. But then $b$ must be rational -- which contradicts the given fact that $b$ is irrational.

As such, our assumption must be rejected, and the opposite must be true: $a+b$ is an irrational number.

QED.

We can tighten this argument up substantially if we instead appeal the closure of the rational numbers under subtraction:

Again, assume that $a+b$ is rational. That is to say:

$$a+b = r \quad \textrm{ for some } r \in \mathbb{Q}$$

But then $b = r - a$, so $b$ is a difference of rational numbers and must then be rational itself as $\mathbb{Q}$ is closed under subtraction. We have just contradicted the given irrationality of $b$.

As such, our assumption must be rejected, and the opposite must be true: $a+b$ is an irrational number.

QED.