Irrationality of $(a_1+\sqrt{b_1})(a_2+\sqrt{b_2})$

Question

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 $(a_1+\sqrt[m_1]{b_1})(a_2+\sqrt[m_2]{b_2})$ ?

Answer 1

Suppose $(a+\sqrt{b})(c+\sqrt{d})=ac+a\sqrt d+c\sqrt b+\sqrt{bd}=p\in\Bbb Q.$

Then \begin{align} da^2+bc^2+bd+2(ac\sqrt{bd}+ad\sqrt b+bc\sqrt d) &=(p-ac)^2 \\ ac\sqrt{bd}+ad\sqrt b+bc\sqrt d &=q:=\frac{(p-ac)^2-da^2-bc^2-bd}{2} \\ \sqrt{bd}(ac+a\sqrt d+c\sqrt b)=\sqrt{bd}(p-\sqrt{bd}) &=q \\ \sqrt{bd}=\frac{q+bd}{p}\in \Bbb Q\end{align}which is impossible given that $b$ and $d$ are coprime.

Answer 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....

Answer 3

$\sqrt a = r + s\sqrt b\ $ follows from $\, c + \sqrt{a} = \dfrac{e}{d+\sqrt b}\,$ by rationalizing denominator. Now apply

Theorem $ $ If $\,r,s\in\Bbb Q\,$ and $\,\sqrt a = r + s\sqrt b\,$ then $\sqrt a,\sqrt b\not\in \Bbb Q\Rightarrow \sqrt{ab}\in\Bbb Q$

Proof $\, $ Squaring $\Rightarrow a = r^2\!+s^2b + 2rs\, \sqrt b\ $ so $\ rs = 0\,$ (else solving for $\sqrt b\,$ $\,\Rightarrow \sqrt b\in \Bbb Q)$.

$s\neq 0\,$ (else $\,a = r^2,\,$ contra $\,\sqrt a \not\in \Bbb Q)\,$ so $\,r=0\,$ so $\,\sqrt a = s\sqrt b\,$ $\overset{\large\times\sqrt b}\Longrightarrow\sqrt{ab} = sb\in \Bbb Q$.

Note $ $ The proof can be reformulated in field theoretic language as the Lemma below (which inductively generalizes to any number of sqrts and also $n$'th roots).

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Lemma $\rm\ \ [K(\sqrt{a},\sqrt{b}) : K] = 4\ $ if $\rm\ \sqrt{a},\ \sqrt{b},\ \sqrt{a\,b}\ $ all are not in $\rm\,K\,$ and $\rm\, 2 \ne 0\,$ in the field $\rm\,K.$

Proof $\ $ Let $\rm\ L = K(\sqrt{b}).\,$ Then $\rm\, [L:K] = 2\,$ via $\rm\,\sqrt{b} \not\in K,\,$ so it suffices to prove $\rm\, [L(\sqrt{a}):L] = 2.\,$ It fails only if $\rm\,\sqrt{a} \in L = K(\sqrt{b})\, $ and then $\rm\, \sqrt{a}\ =\ r + s\, \sqrt{b}\ $ for $\rm\ r,s\in K.\,$ But that's impossible,

since squaring $\Rightarrow \rm(1)\!:\ \ a\ =\ r^2 + b\ s^2 + 2\,r\,s\ \sqrt{b},\, $ which contradicts hypotheses as follows:

$\rm\qquad\qquad rs \ne 0\ \ \Rightarrow\ \ \sqrt{b}\ \in\ K\ \ $ by solving $(1)$ for $\rm\sqrt{b}\,,\,$ using $\rm\,2 \ne 0$

$\rm\qquad\qquad\ s = 0\ \ \Rightarrow\ \ \ \sqrt{a}\ \in\ K\ \ $ via $\rm\ \sqrt{a}\ =\ r+s\,\sqrt b = r \in K$

$\rm\qquad\qquad\ r = 0\ \ \Rightarrow\ \ \sqrt{a\,b}\in K\ \ $ via $\rm\ \sqrt{a}\ =\ s\, \sqrt{b},\ \ $times $\rm\,\sqrt{b}\quad$