Constructing a Polynomial for the Sum of Square Roots - Degree 4 Homework

[happyg1]

Homework Statement

Let R be the field of real numbers and Q the field of rational numbers. In R,$$\sqrt 3$$ and $$\sqrt 2$$ are both algebriac. Exhibit a polynomial of degree 4 staisfied by$$\sqrt 2 + \sqrt 3$$.

Homework Equations

The Attempt at a Solution

I attempted to construct this by $$(\sqrt 2 + \sqrt 3)^4 = 20\sqrt 6 + 49$$
then my polynomial is $$x^4-(20\sqrt 6 + 49)=0$$
is this even close? ThenI have to show it's irreducible and I'm not quite sure where to start that.
My next problem is to prove that sin 1 is an algebriac number. I'm not sure how to construct that polynomial to prove it's irreducible, either.

Any hints or advise will be appreciated
Thanks
CC

 

[mighty2000]

Dear CC,

Great job on constructing a polynomial that satisfies \sqrt 2 + \sqrt 3! Your polynomial is indeed close, but it can be simplified further. Remember that (\sqrt 2 + \sqrt 3)^4 = (2 + 2\sqrt 6 + 3\sqrt 2 + 3\sqrt 3)^2 = (5 + 4\sqrt 6)^2 = 25 + 40\sqrt 6 + 24 = 49 + 40\sqrt 6. So your polynomial can be written as x^4 - (49 + 40\sqrt 6) = 0. This polynomial is irreducible because it has no rational roots and cannot be factored into polynomials of lower degree with rational coefficients.

For proving that sin 1 is an algebraic number, we can use the fact that sin x is a solution to the polynomial equation x - (1 - \cos x) = 0. Substituting x = 1, we get sin 1 = 1 - \cos 1. Now we can use the fact that cos x is a solution to the polynomial equation x^2 - 2 = 0. Substituting x = 1, we get \cos 1 = \sqrt 2 - 1. So sin 1 = 1 - (\sqrt 2 - 1) = 2 - \sqrt 2. Therefore, the polynomial x - (2 - \sqrt 2) = 0 has sin 1 as a root, making sin 1 an algebraic number.

I hope this helps! Let me know if you have any further questions or need clarification.