Irrationality of $(a_1+\sqrt{b_1})(a_2+\sqrt{b_2})$ Sorry, for a rather silly question.
Suppose $a_1$, $b_1$, $a_2$, $b_2$ are integers, all different from zero, while $b_1$ and $b_2$ are co-prime positive integers, neither being a complete square.
Is there an elementary proof that $(a_1+\sqrt{b_1})(a_2+\sqrt{b_2})$ is irrational? Or maybe it's just wrong?
Multiplying gives three distinct surds $n_1\sqrt{b_1}$, $n_2\sqrt{b_2}$, $n_3\sqrt{b_1\;b_2}$, so it doesn't seem to help, while taking a power gives again a product of two irrationals.
How about more general case $(a_1+\sqrt[m_1]{b_1})(a_2+\sqrt[m_2]{b_2})$ ?

$(a + \sqrt b)(c + \sqrt d) = k \in \mathbb Q$ would mean
$\sqrt{b} = \frac k{c+\sqrt d} - a$
$b = ( \frac k{c+\sqrt d} - a)^2 \in \mathbb Q$ which can probably be proven false.  
Indeed $( \frac k{c+\sqrt d} - a)^2 = $
$\frac {k(c - \sqrt d)}{c^2 - d} -a)^2 =$
$(m\sqrt d - n)^2$ where $m = \frac k{c^2-d}\in \mathbb Q$ and $n =\frac {kc}{c^2 -d} -a \in \mathbb Q$.
And $(m\sqrt d -n)^2= m^2d -n^2 - 2nm\sqrt d$ which is not rational.
Unless $n$ or $m$ is $0$.
As $k\ne 0$, $m \ne 0$.  $n = 0$ if $a = \frac {kc}{c^2-d}$.
Hmmm.... 
