Is Every Real Number Algebraic?

[kathrynag]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

Completely confused on this one.

 

[kathrynag]

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

 

[Dick]

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

 

[kathrynag]

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

 

[Dick]

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

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)?

 

[kathrynag]

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

 

[Dick]

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

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.

 

[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.

 

[kathrynag]

x^2-6x+7

 

[Dick]

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

Start with x=3+sqrt(2). That's x-3=sqrt(2). Square both sides.

 

[Dick]

x^2-6x+7

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

 

[kathrynag]

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

 

[Dick]

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

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.