# Is there a way to show that sqrt(2)+sqrt(3)+sqrt(5) is algebraic over Q?

1. Jun 28, 2009

### ~Death~

Hi,

i cant figure out this problem in my algebra book

I thought it was interesting that this thing was rational though...

anyone know how to start it?

2. Jun 28, 2009

### ~Death~

I think i figured it out i might have been confused on the difference between algebraic and rational

idk how to show that algebraic numbers are closed under + though

3. Jun 28, 2009

### snipez90

Erg, so there was a tiny treatment of algebraic numbers in Spivak's calculus text, but basically I have forgotten that chapter. But I think there was a trivial exercise where you show that if a > 0 is algebraic, then sqrt(a) is also algebraic. Also, intuitively, it seems true that the sum of algebraic numbers is also algebraic. On the other hand, there is probably an easier way.

4. Jun 28, 2009

### g_edgar

In this particular case it is enough to find a polynomial (with integer coefficients) that has this as a root. It will be a polynomial of degree 8.

Start with equation $$x = \sqrt{2}+\sqrt{3}+\sqrt{5}$$ then eliminate the radicals ... isolate a radical on one side, then square. It takes three times to get rid of all three radicals, you end up with $$x$$ squared three times so you have $$x^8$$ in your polynomial.

I get
$$-40\,x^6+576-960\,x^2+352\,x^4+x^8 = 0$$

5. Jun 28, 2009

### Hurkyl

Staff Emeritus
The easiest way I can think is in terms of vector spaces over Q. (Meaning that you can only use rational numbers as scalars, rather than arbitrary real or complex numbers)
Theorem: A complex number a is algebraic over Q if and only if its powers 1, a, a2, a3, ... are linearly dependent.​
Can you prove this theorem? Can you see how to leverage linear algebra to prove the result you want?