# Real number is Algebraic

1. Sep 14, 2010

### kathrynag

1. The problem statement, all variables and given/known data
A real number x$$\in$$R is called algebraic if there exists integers $$a_{0}$$$$x^{n}$$+$$a_{n1}$$$$x^{n1}$$+.....+$$a_{1}$$x+$$a_{0}$$=0.
Show that $$\sqrt{2}$$,$$\sqrt[3]{2}$$, and 3+$$\sqrt{2}$$ are algebraic.
Fix n$$\in$$N and let $$A_{n}$$ be the algebraic numbers obtained as roots of polynomials with integer coefficients that have degree n. Using the fact that every polynomial has a finite number of roots, show that $$A_{n}$$ is countable.

2. Relevant equations

3. The attempt at a solution
Completely confused on this one.

2. Sep 14, 2010

### kathrynag

I think it's the notation that's getting me.

3. Sep 14, 2010

### Dick

What part are you confused about? Surely you can show sqrt(2) is algebraic?

4. Sep 14, 2010

### kathrynag

It's the whole notation. I'm not sure how to apply the a's and x's to sqrt(2).

5. Sep 14, 2010

### Dick

It just means show sqrt(2) is the solution of polynomial equation with integer coefficients. I'll get you started. sqrt(2) is a root of 1*x^2-2=0. That's all the a_n*x^n+... is supposed to mean. How about (2)^(1/3)?

6. Sep 14, 2010

### kathrynag

x^3-2=0
Well for 3+sqrt(2)
x^2-(3+sqrt(2))=0

7. Sep 14, 2010

### Dick

x^3-2=0 is good. That's a start. As for x^2-(3+sqrt(2))=0, (3+sqrt(2)) doesn't solve that equation AND (3+sqrt(2)) isn't an integer. You need an equation with integer coefficients. You have to work a little harder on this one.

8. Sep 14, 2010

### kathrynag

I'm assuming it is of the ax^2+bx+c=0, but I'm having trouble coming up with a, b, and c.

9. Sep 14, 2010

### kathrynag

x^2-6x+7

10. Sep 14, 2010

### Dick

11. Sep 14, 2010

### Dick

Yes. Though x^2-6x+7=0 is the equation, right? So 3+sqrt(2) is algebraic.

12. Sep 14, 2010

### kathrynag

x^2=9+6sqrt(2)+2
x^2-6sqrt(2)=11
x^4-12sqrt(2)+72=121

13. Sep 14, 2010

### Dick

You already figured out that one equation is x^2-6x+7=0, didn't you? That's fine. You can turn that what you just wrote into a eighth degree equation for 3+sqrt(2) but you don't need it. You've already got one.