Is that an algebraic number?

**TheOogy** · May31-09, 06:56 AM

is $LaTeX Code: \\sqrt{2}+\\sqrt{5}$ 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** · May31-09, 07:04 AM

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

**matt grime** · May31-09, 08:22 AM

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** · May31-09, 12:22 PM

thanks matt grime!

**HallsofIvy** · May31-09, 01:51 PM

Yes, of course it is. If $LaTeX Code: x= \\sqrt{2}+ \\sqrt{5}$ then $LaTeX Code: x- \\sqrt{2}= \\sqrt{5}$ and $LaTeX Code: (x- \\sqrt{2})^2= x^2- 2\\sqrt{2}x+ 2= 5$ . Then $LaTeX Code: x^2- 3= 2\\sqrt{2}x$ so

. $LaTeX Code: \\sqrt{2}+ \\sqrt{5}$ satisfies the polynomial equation

and so is algebraic.

**TheOogy** · May31-09, 02:40 PM

HallsofIvy,
$LaTeX Code: <BR> (\\sqrt{5}+\\sqrt{2})^4-6(\\sqrt{5}+\\sqrt{2})^2+1 = 98.596<BR>$

i got a different result, for any $LaTeX Code: \\sqrt{a}, \\sqrt{b}$
just use $LaTeX Code: (\\sqrt{a}+ \\sqrt{b})*(\\sqrt{a}- \\sqrt{b})*(-\\sqrt{a}+ \\sqrt{b})*(-\\sqrt{a}- \\sqrt{b})$ 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** · Jun2-09, 02:01 PM

Thanks for the correction. Here's my mistake:
instead of

is should have

LaTeX Code: (x^2- 3)^2= x^4- 6x^2+ 9= 8x^2

. I dropped the "x" in " $LaTeX Code: 2\\sqrt{2}x$ " when I squared.

With that correction, we get

and this time I checked, with a calculator, that $LaTeX Code: \\sqrt{2}+ \\sqrt{5}$ satisfies that equation.

Since $LaTeX Code: \\sqrt{2}+ \\sqrt{5}$ satisfies

, it is algebraic.

**JasonRox** · Jun9-09, 12:06 AM

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

**camilus** · Jun9-09, 04:55 PM

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** · Jun9-09, 04:59 PM

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

**camilus** · Jun9-09, 05:17 PM

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

**CRGreathouse** · Jun10-09, 12:24 AM

Originally Posted by camilus

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

Hah! good luck.

**JasonRox** · Jun10-09, 11:43 AM

Originally Posted 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).

**JasonRox** · Jun10-09, 11:44 AM

Originally Posted 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.

**Office_Shredder** · Jun10-09, 02:15 PM

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)