Math 361, Spring 2017, Assignment 11

Carefully define the following terms, then give one example and one non-example of each:

1. Compass-and-straightedge construction.
2. Constructible number.

Carefully state the following theorems (you do not need to prove them):

1. Theorem concerning sums, differences, products, and inverses of constructible numbers.
2. Theorem concerning square roots of constructible numbers.
3. Theorem concerning the degrees of constructible numbers over $\mathbb{Q}$.
4. Theorem concerning duplication of the cube.
5. Theorem concerning squaring the circle.

Describe the following compass-and-straightedge constructions:

1. Perpendicular bisector.
2. Dropping a perpendicular.
3. Constructing a parallel.
4. Multiplication of lengths.
5. Inversion of length.
6. Square root of length.

Solve the following problems:

1. Determine whether the following numbers are constructible:

(a) $\frac{\sqrt{5}-1}{4}$.

(b) $\sqrt[6]{7}$.

(c) $\alpha$ where $\alpha$ is a root of the polynomial $x^3+3x-12$. (Hint: use Eisenstein's criterion.)

--------------------End of assignment--------------------

Questions

A) $\frac{\sqrt(5) - 1}{4}$

$\alpha = \frac{\sqrt(5) - 1}{4}$

$4\alpha = \sqrt(5) - 1$

$4\alpha +1 = \sqrt(5)$

$(4\alpha +1)^2 = \sqrt(5)^2$

$16\alpha^2 + 8\alpha + 1= 5$

$16\alpha^2 + 8\alpha + 4= 0$

Therefore because the problem is of degree two it is constructible. Is my reasoning totally off?

Yes and no. We've proved that constructible numbers have degree a power of two, but not that numbers with degree a power of two are constructible. However, we know that the field of constructible numbers contains $\mathbb{Q}$ and is closed under square roots. So it contains $\sqrt{5}, 1,$ and $4$, and is closed under addition, subtraction, multiplication, and division, so it contains $\frac{\sqrt{5}-1}{4}$. -Steven.Jackson (talk) 09:40, 17 April 2017 (EDT)

The Eisenstein's criterion: Can be applied like this:

$8x^3+6x^2-9x+24$ where $p=3$ because $3^2=9$ and does not go into 24 evenly therefore the equation is irreducibility.

C)$x^3+3x-12$ where $p=3$ because $3^2=9$ and does not go into 12 evenly therefore the equation is irreducibility. How does this help us determine weather the following is constructible or not?

This shows that $x^3+3x-12$ is really the minimal polynomial of $\alpha$, and thus that $\alpha$ has degree three over the rationals. Since three is not a power of two, it follows that $\alpha$ is not constructible. -Steven.Jackson (talk) 09:40, 17 April 2017 (EDT)