How can I show that $a\sqrt 6 + b \sqrt 3 + c \sqrt 2$ is irrational given that at least one of $b, c$ is nonzero? I've seen the argument when $a = b = c = 1$, but in the general case the algebra is a lot more tedious (I need to avoid division by $0$, for example). Here's what I have so far:

Suppose that $a\sqrt6 + b\sqrt3 + c\sqrt2 \in \mathbb{Q}$. If $a = 0$, then $b\sqrt3 + c\sqrt2 \in \mathbb{Q}$ (note that $b\sqrt3 + c\sqrt2 \neq 0$, otherwise $b\sqrt3 = -c\sqrt2 \implies \sqrt6 = -2c/b$ which is a contradiction since $\sqrt6$ is irrational) so that

\begin{align*} b\sqrt3 - c\sqrt2 = \frac{3b - 2c}{b\sqrt3 + c\sqrt2} \in \mathbb{Q} \end{align*}


\begin{align*} (b\sqrt3 + c\sqrt2) + (b\sqrt3 - c\sqrt2) = 2b\sqrt3 \in \mathbb{Q} \end{align*}

which is a contradiction if $b \neq 0$. (Note that if $b = 0$, then by assumption $c \neq 0$ and we can use a similar argument to arrive at a contradiction)

Therefore, assume that $a \neq 0$. Then, it follows that

\begin{equation*} r = (a\sqrt3 + c)\left(\sqrt2 + \frac{b}{a}\right) = (a\sqrt6 + b \sqrt3 + c\sqrt2) + \frac{cb}{a} \in \mathbb{Q} \end{equation*}

since the sum of two rationals is rational. Then,

\begin{align*} a\sqrt3 + c &= \frac{r}{\sqrt2 + \frac{b}{a}} \\&= \frac{r}{\sqrt2 + \frac{b}{a}} \cdot \frac{\sqrt2 - \frac{b}{a}}{\sqrt2 - \frac{b}{a}} \\&= \frac{r}{2 - \frac{b^2}{a^2}}\cdot\left(\sqrt2 - \frac{b}{a}\right) \end{align*}

and so

\begin{align*} a\sqrt3 - \frac{r\sqrt2}{2 - \frac{b^2}{a^2}}&= \frac{-rb}{a\left(2 - \frac{b^2}{a^2}\right)}-c \in \mathbb{Q} \end{align*}

Note that $a\sqrt3 - \frac{r\sqrt2}{2 - \frac{b^2}{a^2}} \neq 0$, otherwise $a\sqrt6 = \frac{2r}{2 - \frac{b^2}{a^2}} \in \mathbb{Q}$. On the other hand,

\begin{align*} a\sqrt3 + \frac{r\sqrt2}{2 - \frac{b^2}{a^2}}&= \frac{3a^2 - \frac{2r^2}{\left(2 - \frac{b^2}{a^2}\right)^2}}{a\sqrt3 - \frac{r\sqrt2}{2 - \frac{b^2}{a^2}}}\\&= \frac{3a^2 - \frac{2r^2}{\left(2 - \frac{b^2}{a^2}\right)^2}}{\frac{-rb}{2a - \frac{b^2}{a}}-c}\in \mathbb{Q} \end{align*}

However, if $a\sqrt3 + \frac{r\sqrt2}{2 - \frac{b^2}{a^2}} \in \mathbb{Q}$ and $a\sqrt3 - \frac{r\sqrt2}{2 - \frac{b^2}{a^2}} \in \mathbb{Q}$, then \begin{align*} \left(a\sqrt3 + \frac{r\sqrt2}{2 - \frac{b^2}{a^2}}\right) + \left(a\sqrt3 - \frac{r\sqrt2}{2 - \frac{b^2}{a^2}}\right) = 2a\sqrt{3} \in \mathbb{Q} \end{align*}

This is a contradiction, since we assumed $a \neq 0$.

As you see the argument is very tedious (and perhaps there are even holes / mistakes, in which case please let me know). Is there a better (cleaner) way to prove the statement

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    $\begingroup$ what are a, b and c? (rationals? natural?) $\endgroup$ – user160823 Sep 23 '16 at 6:29
  • $\begingroup$ In the first line of proof, I mention that they are rationals. Perhaps I should have mentioned in title $\endgroup$ – b_pcakes Sep 23 '16 at 6:31

Let $q = a\sqrt{6}+b\sqrt{3}+c\sqrt{2}$ with the premise that $a, b, c, q \in \mathbb{Q}$ ,then if $q = 0 \implies (-a\sqrt{6})^2 = (b\sqrt{3}+c\sqrt{2})^2\implies 6a^2 = 3b^2 + 2bc\sqrt{6} + 2c^2\implies \sqrt{6} \in \mathbb{Q}$, a contradiction unless $b = 0 $ or $c = 0$ which in turn yields a much simpler case and a contradiction is easily obtained. If $q \neq 0$, then normalize the equation we can assume from now on $q = 1$ and repeat the steps above we have $\sqrt{6}$ is rational or $a = -bc$. The former can't happen and if the latter does, then we have: $ 1 = -bc\sqrt{6} + b\sqrt{3} + c\sqrt{2}\implies (b\sqrt{3} - 1)(1-c\sqrt{2}) = 0\implies \sqrt{3} = \dfrac{1}{b}$ or $\sqrt{2} = \dfrac{1}{c}$ which are rationals, and we have a contradiction again. Done.

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  • $\begingroup$ Duh. Thanks haha $\endgroup$ – b_pcakes Sep 23 '16 at 6:39
  • $\begingroup$ I end up with $q^2 + 6a - 3b^2 + 2c^2 = 2\sqrt6 (bc+qa)$, but why do we know that $(bc+qa) \neq 0$? $\endgroup$ – b_pcakes Sep 23 '16 at 6:46
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    $\begingroup$ @SimonZhu: I think your LHS should be $q^2+6a^2-3b^2-2c^2$. But your question is still valid, and means that the proof is incomplete. Perhaps you should unaccept it? $\endgroup$ – TonyK Sep 23 '16 at 15:46
  • $\begingroup$ @SimonZhu: See my edited post. $\endgroup$ – DeepSea Sep 23 '16 at 20:38
  • $\begingroup$ Very clear, thanks $\endgroup$ – b_pcakes Sep 24 '16 at 6:44

... because $a\sqrt 3-\frac{r\sqrt 2}{2-\frac {b^2}{a^2}}\ne 0$ (cf. the $a=0$ case).

Here's a more systematicapproach:

Let $F$ be a subfield of $\Bbb R$, $\alpha\in\Bbb R$ with $\alpha\notin F$, $\alpha^2\in F$. Then $E:=\{\,a+b\alpha\mid a,b\in E\,\}$ is a subfield of $\Bbb R$, is the smallest field containing $E$ and $\alpha$, and $a+b\alpha\mapsto a-b\alpha$ is a field automorphism of $E$ leaving precisely $F$ fixed.

To show that $E$ is a field, the only interesting parts are that $(a+b\alpha)(c+d\alpha)=(ac+bd\alpha^2)+(ad+bc)\alpha\in E$ and that $(a+b\alpha)\frac{a-b\alpha}{a^2-b^2\alpha^2}=1$, where $a^2-b^2\alpha^2\in F$ is non-zero because $a+b\alpha$ and $a-b\alpha$ are non-zero. The rest of the claims are quite straightforward.

Applying this twice, we find $\{\,a+b\sqrt 2+c\sqrt 3+d\sqrt 6\mid a,b,c,d\in\Bbb Q\,\}$ is a field with some interesting automorphisms: one given by $\sqrt 2\mapsto-\sqrt 2$, $\sqrt 3\mapsto \sqrt 3$, $\sqrt 6\mapsto -\sqrt 6$, another given by $\sqrt 2\mapsto\sqrt 2$, $\sqrt 3\mapsto -\sqrt 3$, $\sqrt 6\mapsto -\sqrt 6$, and yet another given by $\sqrt 2\mapsto-\sqrt 2$, $\sqrt 3\mapsto -\sqrt 3$, $\sqrt 6\mapsto \sqrt 6$; these leave the subfields $\{\,a+b\sqrt 3\mid a,b\in\Bbb Q\,\}$, $\{\,a+b\sqrt 2\mid a,b\in\Bbb Q\,\}$, and $\{\,a+b\sqrt 6\mid a,b\in\Bbb Q\,\}$, respectively, fixed. According to the lemma, the elements fixed under all these (or even just under the first two) are precisely the rationals. Now assume $a\sqrt 2+b\sqrt 3+c\sqrt 6$ is rational. Then $$a\sqrt 2+b\sqrt 3+c\sqrt 6=-a\sqrt 2+b\sqrt 3-c\sqrt 6=a\sqrt 2-b\sqrt 3-c\sqrt 6 $$ and hence the differences are zero, in particular $$ 2\sqrt 2(a+c\sqrt 3)=0,\qquad 2\sqrt 3(b+c\sqrt 2)=0$$ Per irrationality of $\sqrt 3$ and $\sqrt 2$, we obtain $a=b=c=0$.

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  • $\begingroup$ Thanks, I've actually already figured it out and changed the question. However, the main purpose of the question was to seek a cleaner argument. $\endgroup$ – b_pcakes Sep 23 '16 at 6:32
  • $\begingroup$ OP wants a very nice solution which uses only elementary algebra and not using higher level techniques... $\endgroup$ – DeepSea Sep 23 '16 at 20:11

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