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If $u$ and $v$ are rationals such that $\sqrt{u}, \sqrt{v}, $ and $\sqrt{uv}$ are all irrational, show that there are no rational $a, b, $ and $c$ such that $a+b\sqrt{u}+c\sqrt{v} =\sqrt{uv} $.

This is a generalization of my answer to this question:

Proof by contradiction: finding integers that satisfy $a+b\sqrt{2}+c\sqrt{3}=\sqrt{6}$.

I will post a solution in two days if there are no posted solutions.

marty cohen
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    I don't see what are you looking for: a simpler proof? That's because you can take exactly the one you posted with $u,v$ replacing $2,3$. – Crostul Nov 27 '16 at 23:11
  • That is what I did. But I thought that this generalization was interesting. – marty cohen Nov 27 '16 at 23:12

4 Answers4

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Note that $$a+b\sqrt{u}+c\sqrt{v} +\sqrt{uv}=(\sqrt u+c)(\sqrt v+b)+(a-bc).$$
Now if there is a $t\in\Bbb{Q}\setminus\{0\}$ such that $\sqrt u+c=t$ and $\sqrt v+b=\dfrac{bc-a}{t},$ then $\sqrt u,\sqrt v\in\Bbb{Q}$ which contradicts to your assumption.
Hence $(\sqrt u+c)(\sqrt v+b)\notin\Bbb{Q}\,\,\,\,\,\forall b,c\in\Bbb{Q}.$
There is a very nice generalization of this argument.

Community
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Bumblebee
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Lemma:  $\sqrt{u}$ and $\sqrt{v}$ must be linearly independent over the rationals.

Else $a \sqrt{u} + b \sqrt{v} \in \mathbb{Q}$ with $a,b \in \mathbb{Q} \setminus \{0\}$ $\implies$ $(a \sqrt{u} + b \sqrt{v})^2 = a^2 u + b^2 v + 2ab \sqrt{uv} \in \mathbb{Q}$ $\implies$ $\sqrt{uv} \in \mathbb{Q}$, contrary to the hypothesis.

Proof:  

$$ \sqrt{uv}=a+b\sqrt{u}+c\sqrt{v} \quad\quad \Big| \cdot \sqrt{u} $$ $$ \begin{align} u \sqrt{v} & =a\sqrt{u}+bu+c\sqrt{uv} \\ & = a\sqrt{u}+bu+c(a+b\sqrt{u}+c\sqrt{v}) \\ & = (a+bc)\sqrt{u} + c^2 \sqrt{v} + ac + bu \end{align} $$

By the linear independence proved in the lemma, the coefficients of $\sqrt{v}$ must match i.e. $\,u = c^2\,$ but that contradicts the irrationality of $\sqrt{u}$.

dxiv
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Here's a sketch of a proof (without looking at yours).

Suppose there's some such combination. Square both sides to get an expression involving $\sqrt{uv}$. Isolate that term to get an expression of $\sqrt{uv}$ as a rational combination of $\sqrt{u}$, $\sqrt{v}$ and $1$.

Set this equal to the supposed expression for $\sqrt{uv}$ that you started with, and move everything to one side. You end up with a rational combination of $1, \sqrt{u}, \sqrt{v}$ that's zero. Write this in the form $$ A = B \sqrt{u} + C\sqrt{v} $$ and square it to get $$ A^2 = B^2 u + C^2 v + 2BC \sqrt{uv} $$ from which we see that $\sqrt{uv}$ is written as a linear combination of rationals, which is a contradiction.

Yeah, you have to worry about various coefficients not being zero. That's why I said it was a sketch. :)

John Hughes
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Let $r = \sqrt{u}, s = \sqrt{v}\implies a + br + cs = rs\implies s = \dfrac{a+br}{r-c}\implies s^2 \in \mathbb{Q}\implies r \in \mathbb{Q}$ since $r^2 \in \mathbb{Q}$, contradiction to $r \in \mathbb{Q^c}$. Thus there are no $a,b,c$ which "satisfies" the given condition.

DeepSea
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    Where have you used the part of the hypothesis that says $\sqrt{uv}$ is irrational? – Barry Cipra Nov 27 '16 at 23:52
  • The proof I employed does not require the use of the information that you pointed out. – DeepSea Nov 27 '16 at 23:55
  • Good catch. Working out the algebra, I get that $s^2(r^2+c^2)-a^2-b^2r^2 = 2r(ab+cs^2)$ so that $ab+cs^2=0$ and $s^2(r^2+c^2)-a^2-b^2r^2=0$ which looks harder than my original proof. – marty cohen Nov 28 '16 at 00:05
  • @DeepSea Without that condition $2+\sqrt{2}-\sqrt{2}=\sqrt{4}$. – dxiv Nov 28 '16 at 00:06