Root 3 + root 2

1. Feb 1, 2005

lokisapocalypse

I need to show that this is an algebraic number.

In other words,

I need to show: an*x^n + an1*x^(n-1) + ... + a1 * x^1 + a0 * x^0 =
where the a terms are not ALL 0 but some of them can be.

Like for root 2 by itself,

I have 1 * (root 2) ^ 2 + 0 * (root 2)^1 + -2 * (root 2) ^ 0

2. Feb 2, 2005

mikeu

How about a5=1, a3=-10, a1=1, all others zero?

3. Feb 2, 2005

lokisapocalypse

According to Maple, that gives -.30e-7.

4. Feb 2, 2005

fourier jr

let $$x = \sqrt{3} + \sqrt{2}$$

then $$(x - \sqrt{3})^2 = x^2 - 2\sqrt{3}x + 3 = 2$$

... something like that. put one square root one one side & the rest on the other, square both sides, etc etc then you'll get a 4th-degree polynomial that has $$\sqrt{3} + \sqrt{2}$$ as a root. end of proof

5. Feb 2, 2005

Justin Lazear

Rounding error.

--J

6. Feb 2, 2005

Curious3141

The easiest way is just to find an equation with integral coefficients that has the number as a root.

What is the square of $$\sqrt{2} + \sqrt{3}$$ ? Express that in the form : integer plus a square root. Let's call this $$n_1 = a + \sqrt{b}$$

Next, find the conjugate of the number. $$n_2 = a - \sqrt{b}$$

Set up the equation $$(y - n_1)(y - n_2) = 0$$ and expand it out. You know for certain that this equation is going to have integral coefficients because that's a property of conjugate surds.

Once you have that equation in y, just put $$x^2 = y$$ and get a quartic in x. That equation will have $$\sqrt{2} + \sqrt{3}$$ as one of the roots, and you're done.

Last edited: Feb 2, 2005
7. Feb 2, 2005

robert Ihnot

You can look at this from the standpoint of Galois theory. Along with one surd we are going to get the conjugate of: $$\pm\sqrt2$$. So putting both roots in a quadratic makes the coefficients rational. $$x^2-2=0$$.

The same idea can be applied to this expression: $$\pm\sqrt{2}$$$$\pm\sqrt{3}$$. We don't want the surds to appear in the coefficients.

Maybe not as efficient as described in previous postings, but does offer some insight.

Last edited: Feb 2, 2005
8. Feb 2, 2005

Hurkyl

Staff Emeritus
It shouldn't; Maple is capable of doing symbolic arithmetic. Don't ask it to approximate. :tongue2:

By the way, please don't multiple post.

Last edited: Feb 2, 2005
9. Feb 2, 2005

HallsofIvy

Staff Emeritus
Standard method: Use the fact that (a+ b)(a- b)= a2- b2 to get rid of the square roots.
$$(x- \sqrt{3}-\sqrt{2})(x-\sqrt{3}+ \sqrt{2})= (x- \sqrt{3})^2- 2= x^2- 2\sqrt{3}x+1$$

$$(x^2+1-2\sqrt{3}x)(x^2+1+2\sqrt{3}x)= (x^2+1)^2- 12x^2= x^4+ 2x^2+ 1- 12x^2= x^4- 10x^2+ 1$$

$$\sqrt{3}+ \sqrt{2}$$
is algebraic because it satisfies
$$x^4-10x^2+ 1$$

Last edited: Feb 2, 2005
10. Feb 2, 2005

Hurkyl

Staff Emeritus
You made a typo in there somewhere.

11. Feb 2, 2005

lokisapocalypse

I apologize for the multiple post...I was in a panic. Won't happen again.

12. Feb 2, 2005

HallsofIvy

Staff Emeritus
Yep, I dropped an "x" early on. I've edited it. Thanks, Hurkyl.

13. Feb 2, 2005

Wow, I hope I never have to do these kind of problems. What course is this for?

14. Feb 2, 2005

Palindrom

Modern Algebra course of some kind. I'm doing a similar one myself this semester. Actually, I solved that very exercise a couple of hours ago.

Very nice course, by the way. Don't hope without knowing.