Proof that $\sqrt6 - \sqrt2 - \sqrt3$ is irrational.

Question

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!

Answer 1

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.

Answer 2

The brute-force method, for when no clever argument such as in the other answer applies, would be something like:

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

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

3. 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.

4. Appply the rational root theorem to see if the polynomial has any rational roots. If not, then $X$ cannot be rational.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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.