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

[~Death~]

Hi,

i can't 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?

 

[~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

 

[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.

 

[g_edgar]

Hi,

i can't 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?

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

 

[Hurkyl]

Also, intuitively, it seems true that the sum of algebraic numbers is also algebraic. On the other hand, there is probably an easier way.

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?