Prove that $b \sqrt{3} - a\sqrt{2} \notin \mathbb{Q}, a,b \in \mathbb{Z}$ Basically I want to show that $b \sqrt{3} - a\sqrt{2} \notin \mathbb{Q}, a,b \in \mathbb{Z}$ for my proof to show that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$.
I'm trying to do it via contradiction.
Suppose there exists $p,q \in \mathbb{Q}$ such that $p + q\sqrt{2} =\sqrt{3}$. Let $ q = \frac{a}{b}, p = \frac{c}{d}, a,b,c,d \in \mathbb{Z}$.  Then $\frac{cb}{d} = b\sqrt{3} - a\sqrt{2}$
I'm not sure where to go from here, any help would be appreciated.
 A: Some comments:

*

*I'm assuming that you've seen the proof that $\sqrt{2}$ is irrational and that you can generalize it to show that $\sqrt{3}$ is irrational.

*The given claim is incorrect when $a,b=0$.

*It's a little easier to show that $\sqrt{3}\not\in\mathbb{Q}(\sqrt{2})$ than to show the claimed fact.

Suppose that $b\sqrt{3}-a\sqrt{2}$ were rational.  Let $x$ be this rational.  Then $b\sqrt{3}-a\sqrt{2}=x$.  By rearranging, we have $b\sqrt{3}=x+a\sqrt{2}$.  Squaring both sides gives $3b^2=x^2+2a^2+2ax\sqrt{2}$.

 It must be that $ax=0$ since otherwise we could solve for $\sqrt{2}$ and conclude that $\sqrt{2}$ is rational.  Similarly, we can't have $a=0$ since then we could conclude $\sqrt{3}$ would be rational, unless $b=0$.


 This leaves the condition $x=0$.  But then we have $3b^2=2a^2$.  Now, by following the style of the proof that $2$ is irrational, you can assume that you divide out powers of $2$ from both $a$ and $b$ so that one of them is odd (or they're both $0$).  Since $2\mid 3b^2$, it follows that $2\mid b$.  But then $4\mid 2a^2$ so $2\mid a^2$ and so $a\mid b$.

Therefore, we have a contradiction, unless $a$ and $b$ are both zero.
A: Let assume you know already that $\sqrt{2}$ and $\sqrt{3}$ are irrational and set $x=b\sqrt{3}-a\sqrt{2}$.
If $a=0$ then $x=b\sqrt{3}\notin\mathbb Q$ (unless $b=0$ too) and you are done.
If $a\neq 0$, let assume $x\in\mathbb Q$ then $y=x+2a\sqrt{2}\notin\mathbb Q$ and therefore $xy\notin\mathbb Q$.
But $xy=3b^2-2a^2\in\mathbb Z$ contradiction.
