<s>[INST] Show that $a = b = c = 0$ for $a\sqrt{2} + b = c\sqrt{3}$ is This is the following question:
Suppose that $a, b, c$ are integers such that
$a\sqrt{2} + b = c\sqrt{3}$
(i) By squaring both sides of the equation, show that $a = b = c = 0$
The answer says that you put the equation into the following form:
if $ab \neq 0$
$\sqrt{2} = \frac{3c^2 − 2a^2 − b^2}{2ab}$
is rational — a contradiction and so a = 0 or b = 0.
Why would $a$ or $b$ be 0? (I get that you cannot express an irrational number as the quotient of two rational numbers).
 [/INST] If $a,b \in \mathbb{Z}$, then
$$
2a^{2} + b^{2} + 2ab\sqrt{2} = 3c^{2},
$$
and then
\begin{align}
(*)\ \ \ \  ab\sqrt{2} = \frac{3c^{2}-2a^{2}-b^{2}}{2}  .
\end{align}
The number $\sqrt{2}$ is irrational;
so $ab \neq 0$ leads to a contradiction,
and hence $ab = 0$.
We claim that $a=b=c = 0$; without loss of generality,
let $a = 0$ and let $b \neq 0$.
Then from $(*)$ we have
$|b| = \sqrt{3}|c|$,
which,
by the fact that $\sqrt{3}$ is irrational,
shows that $bc \neq 0$ is absurd;
hence $bc = 0$.
But either $b \neq 0$ or $c \neq 0$ also gives absurdity,
so $b=c=0$.
</s>