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!
 A: Let us assume that
$$\sqrt6-\sqrt2-\sqrt3=r\in\mathbb Q.$$
From this we get
\begin{align*}
\sqrt6-r&=\sqrt2+\sqrt3\\
(\sqrt6-r)^2&=(\sqrt2+\sqrt3)^2\\
6+r^2-2r\sqrt6&=5+2\sqrt6\\
1+r^2&=2(1+r)\sqrt6
\end{align*}
Since $\sqrt6\notin\mathbb Q$, the last equality can be true only if $1+r=1+r^2=0$
The equality $r+1=0$ is satisfied only for $r=-1$. But for $r=-1$ we get $1+r^2=2\ne0$.
So we get a contradiction.
A: Suppose $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is rational.  
Then, $(\sqrt{3}-1)(\sqrt{2}-1)=\sqrt{6}-\sqrt{2}-\sqrt{3}+1$ is a rational number, say $r\in\mathbb{Q}$.
That is, $\sqrt{3}-1=\frac{r}{\sqrt{2}-1}=r(\sqrt{2}+1)$.
Thus, $\sqrt{3}-r\sqrt{2}=r+1\in\mathbb{Q}$.  
Clearly, $r\neq -1$, whence $\sqrt{3}-r\sqrt{2}\neq 0$.
Now, $\sqrt{3}+r\sqrt{2}=\frac{3-2r^2}{\sqrt{3}-r\sqrt{2}}=\frac{3-2r^2}{r+1}\in\mathbb{Q}$.  
What happens if both $\sqrt{3}-r\sqrt{2}$ and $\sqrt{3}+r\sqrt{2}$ are rational numbers?
This line of reasoning shows that $a\sqrt{pq}+b\sqrt{p}+c\sqrt{q}$ is irrational if $a,b,c\in\mathbb{Q}$ with $a\neq0$ and $p,q\in\mathbb{N}\setminus\{1\}$ are such that $p$ and $q$ are distinct and square-free.
A: Suppose that your expression equals a rational number $r$. Treat this equation as a linear combination, with rational coefficients, in the "unknowns" $\sqrt2,\sqrt3,$ and $\sqrt6$ equating to a rational number. Multiply through by $\sqrt2$ and by $\sqrt3$ to obtain two more such equations. We need to check that this system of equations is nonsingular, say by noting that the determinant $1+2r-r^2$ of the coefficients cannot be zero since $\sqrt2$ is irrational. Solving this system then gives a rational value (in particular) for $\sqrt2$, which we know to be impossible.
A: I think the direct method is fairly simple, really. Not sure it adds anything to the other answers at this point, but just in case.
NOTE:  a commenter helpfully pointed out an algebraic error in the original version of this solution.  This error had no significant impact on the solution, and it has now been corrected.
Suppose $\sqrt 6 -\sqrt 2 - \sqrt 3$ were rational.  Square to see that
$$6 + 2 + 3- 4\sqrt 3 - 6\sqrt 2 + 2\sqrt 6 \;\; \in \mathbb Q$$
Which implies that:
$$ \sqrt 6 - 2\sqrt 3 - 3\sqrt 2 \;\; \in \mathbb Q$$
Subtracting this from the original expression we see that 
$$2 \sqrt 2 + \sqrt3 \;\; \in \mathbb Q$$
Square again and simplify to deduce that $$\sqrt 6 \in \mathbb Q$$
which is false, giving us the contradiction we sought.
A: The brute-force method, for when no clever argument such as in the other answer applies, would be something like:


*

*Let $X=\sqrt{6}-\sqrt2-\sqrt3$.

*Calculate $1$, $X$, $X^2$, $X^3$, $X^4$ as rational linear combinations of $1$, $\sqrt2$, $\sqrt3$ and $\sqrt 6$.

*Because the expressions for these 5 powers of $X$ lie in a 4-dimensional vector space over $\mathbb Q$, they must have a nontrivial linear relation, that is, a degree-4 polynomial with rational coefficients that has $X$ as a root. Find such a polynomial using linear algebra.

*Appply the rational root theorem to see if the polynomial has any rational roots. If not, then $X$ cannot be rational.
A: $X=(1-\sqrt{3}-\sqrt{2})^2 = 6 + 2(\sqrt6 - \sqrt2 -\sqrt3)$
We suppose $(1-\sqrt{3}-\sqrt{2})^2$ is rational. 
$((1-\sqrt2)-\sqrt3)^2((1-\sqrt2)+\sqrt3)^2 = 8$
$Y=(1-\sqrt2+\sqrt3)^2$
since $X$ is rational according to our assumption
We do have $Y$ is rational also
$X+Y=  12- 4\sqrt2$
We do have contradiction, so $X$ is irrational, the irrationality of $X$ implies the irrationality of $\sqrt6-\sqrt2-\sqrt3$
A: 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.
