Direct Proof involving Rational Numbers

Question

I have been trying to figure out the following proof: "If a is rational and b is rational, then a + b is rational."

I am new to proofs, and still trying to familiarize myself with the process. I know that definitions of things are very important to proofs, and I know that a rational number is any 2 integers that can be written as a fraction. But I am not sure how to express this idea in the proof.

Answer 1

$a\in\mathbb{Q}\iff \exists p,q\in\mathbb{Z}:a=\frac{p}{q},\quad q\neq 0$

Then $a,b\in\mathbb{Q}\implies$ we can write $a=\frac{p}{q}$ and $b=\frac{m}{n}$ with the above properties for $p,q$ and $m,n$ respectively.

Then $a+b =\frac{p}{q}+\frac{m}{n} = \frac{pn+mq}{qn}$

Now since $p,q,m,n\in\mathbb{Z}$, we also have that $pn+mq\in\mathbb{Z}$ and $qn\in\mathbb{Z}$. Therefore $a+b\in\mathbb{Q}$ as required.

Answer 2

"I know that definitions of things are very important to proofs, and I know that a rational number is any 2 integers that can be written as a fraction. But I am not sure how to express this idea in the proof."

Then that means you have to prove that if $\frac ab$ is a rational number and $\frac cd$ is a rational number then $\frac ab + \frac cd = \frac ef$ for some rational number that is $\frac ef$ for some two integers $e,f$.

So

$\frac ab + \frac cd = \frac {ad}{bd} + \frac {cb}{bd} = \frac {ad+cb}{bd}$.

$ad+cb= e$ is an integer. $f= bd$ is an integer. So $\frac ef = \frac {ad+cb}{bd} = \frac ab + \frac cd$ is a rational number. That's it.

"I know that a rational number is any 2 integers that can be written as a fraction"

Well... nitpick. The rational is the fraction; not the two integers. But I know what you meant.