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Prove that $\sqrt{18} - \sqrt{12}-\sqrt{45} + \sqrt{6}$ is irrational

I tried to let $x = \sqrt{18} - \sqrt{12}-\sqrt{45} + \sqrt{6}$, and assume for the sake of contradiction that $x$ is rational.
Then I squared numerous times to attempt to isolate an irrational on one side, and a supposedly rational expression on another side, since we assumed $x$ is rational, to arrive at a contradiction.
However, all of my attempts have resulted in very complicated expressions that I am unable to complete.

Is this the right approach, or are there other methods such as looking at it as a polynomial root?

Prem
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fds-mqdwqmqkdwl
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2 Answers2

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To assist OP & at the risk of downvoting , I will add a very simple answer which the Particular numbers involved here allow (The Duplicates do not seem to allow this method) . . . .

Let $X=\sqrt{18} - \sqrt{12}-\sqrt{45} + \sqrt{6}=3\sqrt{2} - 2\sqrt{3}-3\sqrt{5} + \sqrt{2}\sqrt{3}$

Hence $X - 3\sqrt{2} + 2\sqrt{3} - \sqrt{2}\sqrt{3} = - 3\sqrt{5}$
Squaring will eliminate $\sqrt{5}$ , while retaining $\sqrt{2}$ & $\sqrt{3}$ & $\sqrt{2}\sqrt{3}$

Move the terms around to get :
$\square 1+\square \sqrt{3} = \square \sqrt{2} + \square \sqrt{2}\sqrt{3} = \sqrt{2} ( \square + \square \sqrt{3} )$ , where $\square$ indicates some arbitrary Co-Efficients involving $X$ , $X^2$ & Constants , though not involving radicals.

Square this to eliminate $\sqrt{2}$ , while retaining $\sqrt{3}$ on either side.

Collect all $\sqrt{3}$ terms to make it $\sqrt{3}=\square/\square$ which is a Contradiction.
[[ Checking that it is not $0/0$ is left out here , though it is not very hard ]]

Summary / Overview :

This is a Proof By Contradiction.
Assume that the given $X$ is rational.
What will that tell us about $X^2$ ? That too must be rational.
What about $X^2-6X+8$ ? Rational too. What about $1/(X^2-6X+8)$ ? Rational too.
Basically , we can Add , Subtract , Divide , Multiply (& Square !) rational numbers to get new rational numbers.
What occur when we have $Y=(X^2+12X+32)/(X^2+3X+2)$ ?
That $Y$ must be rational too.
Well , when we have $\sqrt{3}=(X^2+12X+32)/(X^2+3X+2)$ , we will get Contradiction that $\sqrt{3}$ is rational , which is not true.

What will that Contradiction tell us about our original assumption ?
Well , that assumption must be not true : We then know that the given $X$ is not rational.

Prem
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  • Dear friend if I have $x=\sqrt2$ and I want to prove the irrationality of $x$, can I take the square of $x$ to prove this? What theoretical support does squaring your number $X$ have? Greetings. – Piquito Mar 17 '24 at 14:32
  • No , we can not show irrationality of $\sqrt{2}$ using the square of $\sqrt{2}$. It is unrelated to my answer which is not doing that , @Piquito , I have added a "Summary" which will explain more. – Prem Mar 18 '24 at 15:43
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HINT.-You have to prove that $$3\sqrt2-2\sqrt3-3\sqrt5+\sqrt6$$ Show first that $x=\sqrt a+\sqrt b$ is irrational for $a,b$ both integers square-free. It is very known that $x$ is of degree $4$ over $\Bbb Q$ so $x$ is irrational (its minimal polynomial is $x^4-2(a+b)x^2+(a-b)^2=0$.

It follows that, for example, $x=3\sqrt2-2\sqrt3$ and $y=-3\sqrt5+\sqrt6$ are both irrational so its sum, $x+y$, is rational if and only if the sum of its fractional parts is integer. You can verify that this is not so with a little calculation of only two decimal digits.

Piquito
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  • [A] Why should sum be integer when irrational/rational numbers are involved ? [B] Consider $x=0.0000001+\sqrt{2}-\sqrt{3}$ , $y=0.0000001-\sqrt{2}+\sqrt{3}$ , then $x+y$ is rational & not integer. – Prem Mar 14 '24 at 19:08
  • @Prem: Here $x=3\sqrt2-2\sqrt3$ and $y=-3\sqrt5+\sqrt6$ not other numbers (no doubt about irrationality of both $x$ and $y$?). Besides $x=\lfloor x\rfloor+{x}$ and $y=\lfloor y\rfloor+{y}$ so $$x+y=\lfloor x\rfloor+\lfloor y\rfloor+{x}+{y}=\text{ integer }+({x}+{y})$$ – Piquito Mar 15 '24 at 09:21
  • I might be missing something here : I am not sure why your $(x+y)$ might not have fractional value like $0.00004$ when you state "sum is rational if and only if sum of fractional parts is integer" which is not generally true with my $(x+y)$ Example [[ PS : No Downvote from me ]] – Prem Mar 15 '24 at 14:37
  • Here $x$ and $y$ have fixed values and it is clear that I refer to these fixed numbers $x$ and $y$. – Piquito Mar 17 '24 at 04:44