There are no rationals $r, s$ such that $\sqrt{3} = r + s\sqrt{2}$.

Question

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.

Answer 1

The proof is wrong .

For example $2+\sqrt{3}$ and $-2+\sqrt{3}$ are irrtionlas and give an integer that is not zero on subtraction

Answer 2

On the other hand, it can be proved like this:

if

$\sqrt 3 = r + s\sqrt 2; \; r, s \in \Bbb Q, \tag 1$

then

$3 = (r + s\sqrt 2)^2 = r^2 + 2rs\sqrt 2 + 2s^2, \tag 2$

whence

$2rs \sqrt 2 = 3 - r^2 - 2s^2 \in \Bbb Q; \tag 3$

note that

$rs \ne 0, \tag 3$

for if

$s = 0, \tag 4$

then

$\sqrt 3 = r \in \Bbb Q, \tag 5$

contradicting the irrationality of $\sqrt 3$; if

$r = 0, \tag 6$

we have

$\sqrt 3 = s \sqrt 2, \tag 7$

whence

$3 = 2s^2; \tag 8$

writing

$s = \dfrac{p}{q}, \; p, q \in \Bbb Z, \; \gcd(p, q) = 1, \tag 9$

we find

$3 = 2\dfrac{p^2}{q^2}, \tag{10}$

or

$3q^2 = 2p^2; \tag{11}$

then

$2 \mid 3q^2 \Longrightarrow 2 \mid q^2 \Longrightarrow 2 \mid q \Longrightarrow 4 \mid q^2 \Longrightarrow 4 \mid 2p^2 \Longrightarrow 2 \mid p^2 \Longrightarrow 2 \mid p; \tag{12}$

but $2 \mid p$ and $2 \mid q$ contradicts

$\gcd(p, q) = 1; \tag{13}$

thus no such $p$, $q$ exist and hence

$r \ne 0; \tag{14}$

but now

$rs \ne 0, \tag{15}$

whence from (2),

$\sqrt 2 = \dfrac{3 - r^2 - 2s^2}{2rs} \in \Bbb Q, \tag{16}$

impossible. Thus (1) is forbidden as well. $OE\Delta$.

Note Added in Edit, Wednesday 15 October 2020, 4:28 PM PST: I add here a few remarks on our OP Rustem Sadykov's "new proof", as requested in the comments to this answer.

First, however, I would like to point out that the assertion

"difference of irrationals is equal to irrational or $0$."

is false; for if $r \in \Bbb Q$ and $\alpha \in \Bbb R \setminus \Bbb Q$, then $r + \alpha \in \Bbb R \setminus \Bbb Q$ but $(r + \alpha) - \alpha = r$, a rational.

Having resolved this issue, we turn to details of the "new proof", which I believe to be essentially correct, and not entirely dissimilar to mine. Both start by squaring $\sqrt 3 = r + 2\sqrt 2$ or its equivalent $\sqrt 3 - r = s\sqrt 2$; both branch into cases contingent upon whether $r = 0$ or not; in both approaches, the case $r = 0$ leads to considering $s^2 = 3/2$ in one form or another, this being the substance of ca. (7)-(14) in my approach. (I provided details in order to avoid taking $s = \sqrt 3 / \sqrt 2 \notin \Bbb Q$ by assumption.) The case $r \ne 0$ leads in each proof to a rational expression in $r$, $s$ taking an irrational value, a contradiction. Hopefully these observations will clarify the relationship 'twixt these two similar, though not identical, approaches to this problem. End of Note.