<s>[INST] Proof that $\sqrt6 - \sqrt2 - \sqrt3$ is irrational. I want to prove that: $$\sqrt6 - \sqrt2 - \sqrt3$$ is irrational. I have tried using squares, the $p/q$ definition of rationality and the facts that
1)rational$\times$ irrational=irrational (unless rational=0),
2)rational$+$irrational=irrational.
However, I haven't been able to reach some conclusion. Things seem harder than when you have two square roots. Any help would be appreciated!
 [/INST] Let us assume $\sqrt{6}-\sqrt{3}-\sqrt{2}$ to be a rational number.
Then from the definition of rational no it can be expressed as :
$\sqrt{6}-\sqrt{3}-\sqrt{2}=\frac{p}{q}$, where p and q are co-primes and $q\ne0.$
Squaring both sides we get:
$11-2\sqrt{18}-2\sqrt{12}+2\sqrt{6}=\frac{p^2}{q^2}$
$2\sqrt{6}-6\sqrt{2}-4\sqrt{3}=\frac{p^2-11q^2}{q^2}$
To prove L.H.S is irrational:
According to our assumption,
$ \sqrt{6}-\sqrt{3}-\sqrt{2}$ is a rational no.
$ \implies 2\sqrt{6}-2\sqrt{3}-2\sqrt{2}$ is rational no.
Now, $2\sqrt{6}-6\sqrt{2}-4\sqrt{3}= (2\sqrt{6}-2\sqrt{2}-2\sqrt{3})-(4\sqrt{2}+2\sqrt{3})$
You can easily prove that $(4\sqrt{2}+2\sqrt{3})$ is an irrational no.
So, $2\sqrt{6}-6\sqrt{2}-4\sqrt{3}= (2\sqrt{6}-2\sqrt{2}-2\sqrt{3})-(4\sqrt{2}+2\sqrt{3})$= a rational no.(from assumption)- an irrational no. 
Now the L.H.S is an irrational no whereas the R.H.S is a rational no. which clearly is a contradiction.
So this contradicts our assumption that $\sqrt{6}-\sqrt{3}-\sqrt{2}$ is a rational no.
Hence it follows  $\sqrt{6}-\sqrt{3}-\sqrt{2}$ is irrational.
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