# Is that an algebraic number?

1. May 31, 2009

### 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.

2. May 31, 2009

### Hurkyl

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

3. May 31, 2009

### matt grime

Last edited: May 31, 2009
4. May 31, 2009

### TheOogy

thanks matt grime!

5. May 31, 2009

### 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.

6. May 31, 2009

### 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

7. Jun 2, 2009

### 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.

8. Jun 8, 2009

### JasonRox

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

9. Jun 9, 2009

### 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..

10. Jun 9, 2009

### Moo Of Doom

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

11. Jun 9, 2009

### camilus

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

12. Jun 9, 2009

### CRGreathouse

Hah! good luck.

13. Jun 10, 2009

### JasonRox

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

14. Jun 10, 2009

### JasonRox

Haha, yeah like CRGreathouse said, good luck.

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

15. Jun 10, 2009

### Office_Shredder

Staff Emeritus
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)