There are some similar questions, but not exactly this one.
First:
As, Zero is a rational number. So, by using the counterexample
$a=0$ and $b=x$ ($x$ being an irrational number) we get $ab = 0$. Thus it is not true that $ab$ necessarily has to be irrational.
Second:
However, I have learned a technique that I have used to proof similar but slightly different combination ($a$ is rational and $b$ is irrational, is $a+b$ irrational?) of questions.
With this technique (Proof by contradiction):
Given $a$ is rational and $b$ is irrational. Let, $ab$ be rational.
Suppose, $a= \frac{m}{n}$ and $ab= z =\frac{p}{q}$ , where $m,n,p,q$ are intgers and $n,q \neq 0$
$b=x$ where $x$ is an irrational number.
$$a \cdot b=z$$
$$\frac{m}{n} \cdot x=\frac{p}{q}$$
$$x=\frac{p}{q} \cdot \frac{n}{m}$$
Now, as $p,q,m,n$ are all integers, therefore $x$ is an integer, therefore rational, which contradicts our starting assumption that $x$ is an irrational number. Therefore, $ab$ is irrational.
Now, I think the First conclusion that $ab$ is not necessarily irrational is correct. The second proof is definitely wrong (I think), but I am new to proofs and don't know where it is wrong. What did I miss and not consider in the second approach?
Best Answer
If $k = m/n$ is rational and $j = p/q\ne 0$ is rational, then $k/j = mq/np$ is rational (and if $j = 0$ then $k/j$ is not irrational; it is simply undefined and meaningless and not a number or anything at all).
So if $ab$ is rational. And $a$ is rational. And $a \ne 0$ then than $ab/a = b$ is rational.
So the only way. $a$ can rational and $b$ be irrational but somehow $ab $ is rational is if $ab/a$ is not defined. That only happens if $a$ is zero. That is the one and only counter example.
As for $a +b$.... Note if $k = m/n$ and $j = p/q$ are rational, then $k \pm j = \frac {mq \pm pn}{nq}$ is rational.
So if $a$ is rational and $a + b$ is rational, then $(b+a) - a = b$ must also be rational.
So if $a$ is rational and $b$ is irrational there is no way possible for $a + b$ to be rational. There are absolutely no counterexamples.