Is that an algebraic number?

[TheOogy]

is \sqrt{2}+\sqrt{5} an algebraic number?
i used 2 and 5 arbitrarily, try any integers (as long as they are not the same integer, in which case it is algebraic)
I tried finding a polynomial with rational coefficients that zeros at this value, but haven't found any.

[Hurkyl]

If it was, its powers would span a finite dimensional vector space over Q.

[matt grime]

http://www.dpmms.cam.ac.uk/~wtg10/galois.html

is a useful link to expand on Hurkyl's idea.

Or.

Define x to be that expression above, what is x^2? what is x^2 - 2 - 5?

[TheOogy]

thanks matt grime!

[HallsofIvy]

Yes, of course it is. If x= \sqrt{2}+ \sqrt{5} then x- \sqrt{2}= \sqrt{5} and (x- \sqrt{2})^2= x^2- 2\sqrt{2}x+ 2= 5. Then x^2- 3= 2\sqrt{2}x so (x^2- 3)^2= x^4- 6x^2+ 9= 8. \sqrt{2}+ \sqrt{5} satisfies the polynomial equation x^4- 6x^2+ 1= 0 and so is algebraic.

[TheOogy]

HallsofIvy,

(\sqrt{5}+\sqrt{2})^4-6(\sqrt{5}+\sqrt{2})^2+1 = 98.596



i got a different result, for any \sqrt{a}, \sqrt{b}
just use (\sqrt{a}+ \sqrt{b})*(\sqrt{a}- \sqrt{b})*(-\sqrt{a}+ \sqrt{b})*(-\sqrt{a}- \sqrt{b}) and expand
i haven't read the whole article, just the start and deducted this (without proof) by factoring the polynomial they gave for 2 and 3

[HallsofIvy]

Thanks for the correction. Here's my mistake:
instead of (x^2- 3)^2= x^4- 6x^2+ 9= 8 is should have
(x^2- 3)^2= x^4- 6x^2+ 9= 8x^2. I dropped the "x" in "2\sqrt{2}x" when I squared.

With that correction, we get x^4- 14x^2+ 9= 0 and this time I checked, with a calculator, that \sqrt{2}+ \sqrt{5} satisfies that equation.

Since \sqrt{2}+ \sqrt{5} satisfies x^4- 14x^2+ 9= 0, it is algebraic.

[JasonRox]

If you can prove that the algebraic numbers form a group additively, then you are done.

[camilus]

jason, what do you mean "form a group additively"? I dont get it, do you mean some sort of commutative property? Although I doubt it..

[Moo Of Doom]

It means if you add or subtract algebraic numbers from each other, you get an algebraic number.

[camilus]

Thats a great question. I was working on a similar question, whether e+pi was transcendental.

[CRGreathouse]

Thats a great question. I was working on a similar question, whether e+pi was transcendental.

Hah! good luck.

[JasonRox]

jason, what do you mean "form a group additively"? I dont get it, do you mean some sort of commutative property? Although I doubt it..

I never said anything about commutativity (even though in this case there is).

[JasonRox]

Thats a great question. I was working on a similar question, whether e+pi was transcendental.

Haha, yeah like CRGreathouse said, good luck.

This question is way beyond the calibre of question compared to the one in the OP.

[Office_Shredder]

The algebraic numbers form a field even, but that's a little tricky to prove (specifically if a and b are algebraic numbers, then a*b is too)