Proof that sqrt(6)-sqrt(2)-sqrt(3) is irrational

[abruzzi]

I want to prove that $$\sqrt 6 - \sqrt 2- \sqrt 3$$ is irrational.

I already know that $$\sqrt 2+\sqrt 3$$ is irrational (by squaring it). I would like a proof that doesn't use a polynomial and the rational root theorem.

Thanks.

 

[Hurkyl]

I want to proof that $$\sqrt 6 - \sqrt 2- \sqrt 3$$ is irrational.

I already know that $$\sqrt 2+\sqrt 3$$ is irrational (by squaring it). I would like a proof that doesn't use a polynomial and the rational root theorem.

Thanks.

Is $\mahtbb{Q}(\sqrt{2}, \sqrt{3})$ Galois over Q? If so, something is rational if and only if it is equal to all of its conjugates. (The same is true if the field isn't galois... it's just that in that case, its conjugates would live in other fields)

 

[abruzzi]

Sorry, but I have no idea about groups, rings, fields, etc. I am looking for a basic proof.

For example the proof that I have that $$\sqrt 2 + \sqrt 3$$ is irrational is that, supposing it is rational, its square should also be rational. But $$(\sqrt 2 + \sqrt 3)^2 = 5+2\sqrt6$$ is irrational because $$\sqrt 6$$ is irrational.

I would like to come up with a similar proof for $$\sqrt 6 - (\sqrt 2 + \sqrt 3)$$

 

[tiny-tim]

II already know that $$\sqrt 2+\sqrt 3$$ is irrational (by squaring it).

Hi abruzzi!

How about squaring $$n+\sqrt 2+\sqrt 3$$, for some whole number n?

 

[abruzzi]

By taking n=-1 and squaring we get $$2\sqrt 6 - 2\sqrt 3 - 2\sqrt 2 + 6 = 6 + 2(\sqrt 6 -\sqrt 3-\sqrt 2)$$

But from this I cannot conclude anything - since knowing that number a is irrational doesn't mean that a^2 is (for example $$a=\sqrt 2$$).

Or am I not on the right path?

 

[tiny-tim]

oops!

ooh, you're right!

ok, let's try this:

rational + irrational = irrational.

rational x irrational = irrational.

Suppose √6 - √3 - √2 is rational.

Then (1 +√3)(1 + √2) is irrational, because it is (1 + 2√6) - (√6 - √3 - √2).

But (1 - √3)(1 - √2) is rational, because it is 1 + (√6 - √3 - √2).

So the product (1 +√3)(1 + √2)(1 - √3)(1 - √2) is irrational.

But it isn't - it's 2.

 

[mathwonk]

sqrt2 + sqrt3 = sqrt 6 + r, with r rational,

implies, by squaring both sides, that 1 + r^2 = (2-2r)sqrt6. which equates a rational and an irrational.

 

[tiny-tim]

… mathwonk is cool …

ooh, mathwonk, that's much better! ​

 

[abruzzi]

That was exactly what I was looking for, thanks!

 

[lurflurf]

Much nicer than mine
let x=√6-√3-√2
x³-3x²-15x-3=4√2
->x is irrational
or
let y=√323-√19-√17
y³-19y²-393y+4883=72√17
->y is irrational
mathwonks works for that too
√323=(297+y²)/(2-2y)
->y is irrational
just two was of writing m√ab
one in Z[√a,√b]
one in the field of fractions
the fraction sure give nice numbers though

 

[Alex48674]

Well I think if you can prove that anyone of the terms is irrational, which should be easy, and if the number subtracted by it will be irrational, unless the number you subtract by is the same number or follows the same irrational pattern, which would be quite hard as it is impossible to find a pattern, thus if you were to subtract 2 irrational numbers, there should be an infinite probability that they will be irrational and rational =\.

for example you have an irrational number

.342526524525352325...n where n are the rest of the terms
-.1412413431413...n then the 2 n's would cause repeating zeros, thus being rational.

Just bringing up irrat-irrat

But there is no way to prove that at some point the rest of the digits would be the same (aka n), yet as it goes on for ever it could happen. Very weird little thing infinity =P, I'm in very low level maths, and I just wanted to contribute so this is all I could think of =], I'm sure there is a clever way to prove it though.