if $a$ and $b$ are irrational and $a \neq b$, then is $ab$ necessarily irrational? $\sqrt{2}\sqrt{2}$ is rational so it is not the case $\forall$ m,n $\in$ 
$\Bbb{R} - \Bbb{Q}$ , mn $\in$ $\Bbb{R} - \Bbb{Q}$. What about if m $\neq$ n? Is there a case where m $\neq$ n and $mn$ is rational?
 A: Counter example for first case, and an example for second case: $a = \sqrt{2}, b = \dfrac{1}{\sqrt{2}}$. $a \neq b$, yet $ab = 1 \in \mathbb{Q}$. For the other case, take $a = \sqrt{2}-1, b = \sqrt{2}$, $a \neq b$, and $ab = 2-\sqrt{2} \notin \mathbb{Q}$.
A: It's trivial that
\begin{eqnarray}
&& a\in\mathbb{Q}\,,\,b\in\mathbb{Q}\,\Rightarrow ab\in\mathbb{Q}\\
&& a\in\mathbb{Q}\,,\,b\in\mathbb{Q}^c\,\Rightarrow ab\in\mathbb{Q}^c\,\,\,,\,a\neq0\\
\end{eqnarray}
but with other cases:
\begin{eqnarray}
&& \sqrt{2}\in\mathbb{Q}^c\,,\,\sqrt{2}\in\mathbb{Q}^c\,\,\,\,\text{but}\,\,\, 2\in\mathbb{Q}\\
&& \sqrt{3}\in\mathbb{Q}^c\,,\,\sqrt{2}\in\mathbb{Q}^c\,\,\,\,\text{but}\,\,\, \sqrt{6}\in\mathbb{Q}^c\\
\end{eqnarray}
More, that's trivial that
\begin{eqnarray}
&& a\in\mathbb{Q}\,,\,b\in\mathbb{Q}\,\Rightarrow a+b\in\mathbb{Q}\\
&& a\in\mathbb{Q}\,,\,b\in\mathbb{Q}^c\,\Rightarrow a+b\in\mathbb{Q}^c\\
\end{eqnarray}
but with other cases:
\begin{eqnarray}
&& \sqrt{2}\in\mathbb{Q}^c\,,\,-\sqrt{2}\in\mathbb{Q}^c\,\,\,\,\text{but}\,\,\, 0\in\mathbb{Q}\\
&& \sqrt{3}\in\mathbb{Q}^c\,,\,\sqrt{2}\in\mathbb{Q}^c\,\,\,\,\text{but}\,\,\, \sqrt{3}+\sqrt{2}\in\mathbb{Q}^c\\
\end{eqnarray}
A: You are asking if $\forall m,n\in\mathbb R, m\ne n\implies mn\in \mathbb Q$.
That would mean that $\forall m\in\mathbb R-\mathbb Q, m\ne\sqrt2\implies m\sqrt2\in \mathbb Q$.
There are many more counterexamples than examples, as $\mathbb R-\mathbb Q$ is uncountable, while $\mathbb Q$ is.
A: Be $q$ a rational number other than $0$. Be $a$ an irrational number with $a^2\ne q$. Then $b=\frac{q}{a}$ is an irrational number that is not equal to $a$, but $ab=q$ is rational by construction.
So the product of two different irrational numbers is not always irrational. Indeed, the number of counterexamples is uncountable (as for any given $q$ almost all irrational numbers can be chosen as $a$).
Interestingly this means there are as many pairs of irrational numbers whose product is rational as there are products of irrational numbers whose product is irrational, despite there being more irrational numbers than rational numbers.
