# Algebraic Numbers.

1. Feb 4, 2012

### cragar

1. The problem statement, all variables and given/known data
Show that √2 , $2^{\frac{3}{2}}$ and √3+√2 are algebraic
A real number is algebraic when it is the root of a polynomial with integer coefficients.

3. The attempt at a solution
When they say show, does that mean just give an example? Is it different than proving something.
So I took √2=x and then squared both sides same for √3
and then did the same thing for √2+√3=x . I squared both sides.
but then you have to do it again. and eventually you get a polynomial out with integer
coefficients.

2. Feb 4, 2012

### Curious3141

For each of those numbers, you have to construct a polynomial equation with integral coefficients that has that number as a root.

Start with $x = n$, where n is the number, then manipulate it until you get the required equation (with RHS equal to zero).
f
Your answers should resemble: $3x^2 - 5 = 0$ or $6x^4 + 7x^3 - 8x^2 + x + 3 = 0$ (just examples of the form).

You've got the right idea for $\sqrt{2}$, but you really need to express it as an equation.

For $\sqrt{2} + \sqrt{3}$, square both sides, simplify, then leave the surd term on one side, gather the integer and x terms on the other, then square both sides again, simplify, rearrange.

$2^{\frac{3}{2}}$ is easy, square both sides, rearrange.

Last edited: Feb 4, 2012
3. Feb 4, 2012

### cragar

so for root2 I would have $x^2-2=f(x)$ as the polynomial.

4. Feb 4, 2012

### Curious3141

Should be $x^2-2=0$. The equation is the answer.

5. Feb 4, 2012

### Deveno

no cragar is correct, to prove a number a is algebraic, you need to find a polynomial f(x) such that f(a) = 0.

for √2, the polynomial is f(x) = x2 - 2. in fact, this is the minimal polynomial for √2. note that the polynomial g(x) = x4 - 4x2 + 4 would do just as well for establishing that √2 is algebraic.

6. Feb 4, 2012

### Curious3141

Ah, fair enough.

It's understood the RHS is zero.

7. Feb 4, 2012

### Deveno

yes, that's what we mean by a root.

8. Feb 4, 2012

### Curious3141

Actually, I think the use of the word "root" is somewhat irregular. "Root of a polynomial" is acceptable, just as is "root of an equation".

For example, in this Wolfram page: http://mathworld.wolfram.com/PolynomialRoots.html, the word root is used in both contexts.

Maybe "zeroes of a polynomial function with integral coefficients" is a less contentious way of putting it.

The words "solution" is also uncontentious, since this word has to pertain to a stated equation (rather than just a function).

9. Feb 5, 2012

### Deveno

it's kind of weird...an n-th root of a is a root of xn - a, but the "root of a polynomial" may not be a (n-th) root in this sense.

but, yes, when one speaks of a "root of a function", one means an x such that f(x) = 0. for example, pi is a root of the sine function. the term "zero" might be more unambiguous.

10. Feb 5, 2012

### Curious3141

Yes, but that usage of the word "root" is understood to be distinct.

What I'm saying is "root of an equation" is also acceptable terminology, at least in common mathematical use.

In any case, for this problem, it doesn't matter a whole deal.

Cragar can answer "This number is algebraic because it is is a root of P(x)" or "...because it is a root of P(x) = 0", or "...because it is a zero of P(x)", or "...because it is a solution of P(x) = 0", or "...because it satisfies P(x) = 0".

The concept is what's important, and as long as he expresses his meaning clearly, noone can mark him wrong. The rest is just hair-splitting semantics.

11. Feb 5, 2012

### Deveno

which, no doubt, explains my plethora of split-ends.

12. Feb 5, 2012

### Curious3141

You need more trichology, less trigonometry.