There are no rationals $r, s$ such that $\sqrt{3} = r + s\sqrt{2}$. Our professor gave us lengthy and a bit of a complex proof. I tried to construct my own, more simple proof. Could you please verificate it ?  Suppose there are $r,s$ such that$\sqrt{3} = r + s\sqrt{2}$. Then $\sqrt{3}-r= s\sqrt{2} \to \frac{\sqrt{3}}{\sqrt{2}} - \frac{r}{\sqrt{2}} = s$. $\frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2} = \frac{1}{2} * \sqrt{6}$  Clearly this number is irrational $\frac{r}{\sqrt{2}} = \frac{r\sqrt{2}}{2} = \frac{r}{2} * \sqrt{2}$ Again this is irrational. The only way to substract irrationals and receive an integer is if they are equal. Clearly, this is impossible in our case. Hence, this is contradiction.  Actually could you please explain why difference of irrationals is equal to irrational or $0$. UPD: I now realize my proof was wrong. I constructed a new one. $\sqrt{3} - r = s\sqrt{2}$. Squaring this equation we get $3-2r\sqrt{3} + r^2 = 2s^2 \to 3-2s^2+r^2=2r\sqrt{3}$ Clearly, $2r\in Q $ and so RHS is irrational. Suppose $r=0$. Then  $3-2s^2=0 \to s=\sqrt{\frac{3}{2}}$ Contradiction, since s is rational. Suppose $r\neq 0$ Then there is contradiction since LHS is rational and RHS is irrational. Therefore there does not exist such rational r, s.
Possible Duplicate:          Consider the field extension $\mathbb{Q}(\sqrt2)$. I want to show that $\sqrt5 \notin \mathbb{Q}(\sqrt2)$. If this were not the case, then we could write $\sqrt5 = a + b\sqrt2$ for $a,b\in\mathbb{Q}$. However, I do not see the contradiction here. Is there a better/easier way to prove this?
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