## Is that an algebraic number?

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.

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 Recognitions: Gold Member Science Advisor Staff Emeritus If it was, its powers would span a finite dimensional vector space over Q.
 Recognitions: Homework Help Science Advisor 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?

## Is that an algebraic number?

thanks matt grime!

 Recognitions: Gold Member Science Advisor Staff Emeritus 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.
 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
 Recognitions: Gold Member Science Advisor Staff Emeritus 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.
 Recognitions: Gold Member Homework Help If you can prove that the algebraic numbers form a group additively, then you are done.
 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..
 It means if you add or subtract algebraic numbers from each other, you get an algebraic number.
 Thats a great question. I was working on a similar question, whether e+pi was transcendental.

Recognitions:
Homework Help
 Quote by camilus Thats a great question. I was working on a similar question, whether e+pi was transcendental.
Hah! good luck.

Recognitions:
Gold Member
Homework Help
 Quote by 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..
I never said anything about commutativity (even though in this case there is).

Recognitions:
Gold Member
Homework Help
 Quote by camilus 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.

 Blog Entries: 1 Recognitions: Homework Help 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)