Proving the Algebraic Nature of Root 3 + Root 2 Using Polynomial Expressions"

[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

 

[mikeu]

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

 

[lokisapocalypse]

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

 

[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

 

[Justin Lazear]

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

Rounding error.

--J

 

[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.

 

[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.

 

[Hurkyl]

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

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

By the way, please don't multiple post.

 

[HallsofIvy]

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
$$x^4-10x^2+ 1$$

 

[Hurkyl]

You made a typo in there somewhere.

 

[lokisapocalypse]

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

 

[HallsofIvy]

You made a typo in there somewhere.

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

 

[adc85]

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

 

[Palindrom]

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

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.