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This question already has an answer here:

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!

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marked as duplicate by Watson, T. Bongers, Cesareo, Jaideep Khare, user10354138 Nov 27 '18 at 5:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Hint: Squaring the expression was a good start. Suppose that the original was rational, what does the squared form tell you? $\endgroup$ – lulu Jul 23 '15 at 3:06
  • $\begingroup$ The answer below solved my problem but what about the squaring approach? Personally it didn't seem to get me anywhere since I couldnt make the L.H.S. to be irrational. $\endgroup$ – Jezulas Jul 23 '15 at 3:29
  • $\begingroup$ Subtraction is not associative. $1-(0-1)=2$ and $(1-0)-1=0$. Of course in either interpretation of your question you get an irrational number. But it's always good to be accurate. $\endgroup$ – Asaf Karagila Jul 23 '15 at 4:41
  • $\begingroup$ A more generic answer would be that a linear combination of square roots of integers with integer coefficients is always an algebraic integer. And an algebraic integer is rational if and only if it is an integer. Therefore questions of this type are under this or this umbrella question for more. Basically the answers say that you never get rational numbers from such combos except trivially, when all those integers have integer square roots. Questions linked to those two give plenty of examples. $\endgroup$ – Jyrki Lahtonen Jul 23 '15 at 9:43
  • $\begingroup$ @Jezulas. I'll write it out and post it below. I see that other people have thoroughly answered the question, though some of the answers look a little complex. I think it's not so hard really... Anyway, I'll write it up. $\endgroup$ – lulu Jul 23 '15 at 10:49
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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.

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  • $\begingroup$ Thank you! This one would have probably taken me a lot of time. $\endgroup$ – Jezulas Jul 23 '15 at 3:26
  • $\begingroup$ There is another path from $\sqrt{3} - r\sqrt{2} = r+1 \in \mathbb{Q}$, using the fact that $\sqrt{3} \not \in \mathbb{Q}[\sqrt{2}]$. $\endgroup$ – Carl Mummert Jul 23 '15 at 13:34
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    $\begingroup$ Well, I was avoiding more advanced stuff. $\endgroup$ – Batominovski Jul 23 '15 at 14:06
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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.

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  • $\begingroup$ Thank you! This one is fairly simple indeed but I somehow managed to miss it. $\endgroup$ – Jezulas Jul 23 '15 at 12:36
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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.

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    $\begingroup$ In particular, $X$ satisfies $X^4-22X^2-48X-23=0$ (found by asking WolframAlpha for the minimal polynomial of $\sqrt{6}-\sqrt{2}-\sqrt{3}$ after giving up doing it by hand). Appealing to the rational root theorem, all rational roots must be integer divisors of $23$, but neither $23$, $0$, nor $-23$ are roots (alternatively, put $p/q$ into the polynomial and show it can never be a root). $\endgroup$ – Kyle Miller Jul 23 '15 at 3:47
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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.

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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.

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$X=(1-\sqrt{3}-\sqrt{2})^2 = 6 + 2(\sqrt6 - \sqrt2 -\sqrt3)$

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 $\sqrt6-\sqrt2-\sqrt3$

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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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    $\begingroup$ How do you know the L.H.S. is irrational? That seems more or less equivalent to the original question. $\endgroup$ – lulu Jul 23 '15 at 3:21
  • $\begingroup$ That is pretty much the same conclusion I came to after squaring. I could not prove that the L.H.S. was irrational. $\endgroup$ – Jezulas Jul 23 '15 at 3:24
  • $\begingroup$ While it's true that the left-hand side is irrational, assuming it is at this point is wishful. Try squaring it again; you'll probably get a polynomial on the R.H.S. which is very similar to the one I posted under Henning's answer. $\endgroup$ – Kyle Miller Jul 23 '15 at 3:50
  • $\begingroup$ @lulu How do i know the L.H.S. is irrational? I had done my edits. Hope you can agree now that the L.H.S is irrational. $\endgroup$ – yasir Jul 23 '15 at 9:33

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