Is \sqrt{2}+\sqrt{5} an Algebraic Number?

[TheOogy]

is [tex]\sqrt{2}+\sqrt{5}[/tex] 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 [itex]x= \sqrt{2}+ \sqrt{5}[/itex] then [itex]x- \sqrt{2}= \sqrt{5}[/itex] and [itex](x- \sqrt{2})^2= x^2- 2\sqrt{2}x+ 2= 5[/itex]. Then [itex]x^2- 3= 2\sqrt{2}x[/itex] so [itex](x^2- 3)^2= x^4- 6x^2+ 9= 8[/itex]. [itex]\sqrt{2}+ \sqrt{5}[/itex] satisfies the polynomial equation [itex]x^4- 6x^2+ 1= 0[/itex] and so is algebraic.

 

[TheOogy]

HallsofIvy,
[itex]
(\sqrt{5}+\sqrt{2})^4-6(\sqrt{5}+\sqrt{2})^2+1 = 98.596
[/itex]


i got a different result, for any [itex]\sqrt{a}, \sqrt{b}[/itex]
just use [itex](\sqrt{a}+ \sqrt{b})*(\sqrt{a}- \sqrt{b})*(-\sqrt{a}+ \sqrt{b})*(-\sqrt{a}- \sqrt{b})[/itex] 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 [itex](x^2- 3)^2= x^4- 6x^2+ 9= 8[/itex] is should have
[itex](x^2- 3)^2= x^4- 6x^2+ 9= 8x^2[/itex]. I dropped the "x" in "[itex]2\sqrt{2}x[/itex]" when I squared.

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

Since [itex]\sqrt{2}+ \sqrt{5}[/itex] satisfies [itex]x^4- 14x^2+ 9= 0[/itex], 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 don't 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 don't 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)