Prove that $\sqrt 2 +\sqrt 3$ is irrational. [duplicate]

As the rationals are closed under addition, if you know $\sqrt 2 + \sqrt 3$ is rational and that $\sqrt 3 - \sqrt 2$ is rational, their sum $2 \sqrt 3 $ is rational, then divide by $2$

Added: we can even make it explicit. If $\sqrt 2+\sqrt 3=\frac ab, \sqrt 3-\sqrt 2=\frac ba$ and $\sqrt 3=\frac 12 (\frac ab + \frac ba)$

It's badly phrased. Its phrasing makes it appear that it's saying that if $\sqrt{3}-\sqrt{2}$ is rational, then $\sqrt{3}$ must be rational. But it actually means that if BOTH $\sqrt{3}-\sqrt{2}$ and $\sqrt{3}+\sqrt{2}$ are rational, then so is $\sqrt{3}$. That's because if they're both rational, then their sum is rational. Their sum is $2\sqrt{3}$. It's easy to see that if that's rational, then so is $\sqrt{3}$.

if $a,b$ are rational, so is $a+b$...

As $\sqrt{3}-\sqrt{2}$ and $\sqrt{3}+\sqrt{2}$ are rational, so is $\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}=2\sqrt{3}$

if $a,ab$ are rational, so is $b$...

As $2,2\sqrt{3}$ are rational so is $\sqrt{3}$...

here we have used two statements

if $a,b$ are rational, so is $a+b$...

if $a,ab$ are rational, so is $b$...

convince your self that these results can be seen easily.. If not,

for first ststement:

As you can see if $a=\frac{p}{q},b=\frac{r}{s}$ then $a+b=\frac{ps+qr}{qs}$

and for second statement

suppose $a=\frac{p}{q}$ and $ab=\frac{r}{s}$ then $b=\frac{qr}{ps}$

P.S : I like your idea $(\sqrt 3 +\sqrt 2)(\sqrt 3 -\sqrt 2)=1$ to prove irrationality... :)That is the reason I have tried to help you... all the best