# Problem 1

We wish to prove $$\sqrt{15} \notin \mathbb {Q}$$
We will do this through a proof by contradiction. Let us assume we can represent $$\sqrt{15} = \frac{a}{b} \text{, where } a,b \in \mathbb{Z}$$. We can reduce the expression $$\frac{a}{b}$$ completely so that it is identical in value but so that a and b are coprime, sharing no common factors except 1. By the fundamental theorem of algebra, every integer can be expressed as a product of primes, so we will represent these explicitely as $$a = a_{1}^{n_{1}} \times a_{2}^{n_{2}} \times \text{. . .} \times a_{k}^{n_{k}}$$ and $$b = b_{1}^{n_{1}} \times b_{2}^{n_{2}} \times \text{. . .} \times b_{k}^{n_{k}}$$. Now if $$\sqrt{15} = \frac{a}{b}$$, then $15 = (\frac{a}{b})^2 = \frac{a^2}{b^2} = \frac{a_{1}^{2n_{1}} \times a_{2}^{2n_{2}} \times \text{. . .} \times a_{k}^{2n_{k}}}{b_{1}^{2n_{1}} \times b_{2}^{2n_{2}} \times \text{. . .} \times b_{k}^{2n_{k}}}$ $\Rightarrow 15(b_{1}^{n_{1}} \times b_{2}^{n_{2}} \times \text{. . .} \times b_{k}^{n_{k}}) = a_{1}^{n_{1}} \times a_{2}^{n_{2}} \times \text{. . .} \times a_{k}^{n_{k}}$ which implies one of the $$a_{i}$$ is 15. Let us choose $$a_{1} = 15$$ and reorder as necessary. Then $15_{1}^{2n_{1}-1} \times a_{2}^{2n_{2}} \times \text{. . .} \times a_{k}^{2n_{k}} = b_{1}^{2n_{1}} \times b_{2}^{2n_{2}} \times \text{. . .} \times b_{k}^{2n_{k}}$ implying that B is divisible by 15. But then a and b share a multiple of 15, and we already reduced them so they are coprime, which is a contradiction. Therefore $$\sqrt{15} \notin \mathbb {Q}$$ as desired.

# Problem 2

## part a)

We are given that $$\alpha = \sqrt[3]{2}$$, and $$F = \{a_{o} + a_{1}\alpha + a_{2}\alpha^2 | a_{i} \in \mathbb {C}\}$$.
Now we would like to show that if $$a, b \in F$$, then $$a+b \in F$$, $$ab \in F$$, ie. F is closed under multiplication and addition. We will start with addition.
First we will define a as $$a = a_{o} + a_{1}\alpha + a_{2}\alpha^2$$ and b similarly as $$b_{o} + b_{1}\alpha + b_{2}\alpha^2$$. Now when we add them, we can see: $a+b = a_{o} + a_{1}\alpha + a_{2}\alpha^2 + b_{o} + b_{1}\alpha + b_{2}\alpha^2=$ $=(a_{o} + b_{o}) + (a_{1}+ b_{1})\alpha + (a_{2}+ b_{2})\alpha^2$
Since $$\mathbb{Q}$$ is a subfield of $$\mathbb {C}$$, as stated in lecture, it is defined to be closed under addition, so we can redefine the coefficients as $(a_{o} + b_{0}) \equiv c_{0}, \text { where }c_{0} \in \mathbb {Q}$ $(a_{1} + b_{1}) \equiv c_{1}, \text { where } c_{1} \in \mathbb {Q}$ $(a_{2} + b_{2}) \equiv c_{2}, \text { where } c_{2} \in \mathbb {Q}$
This allows us to redefine $$a+b$$ as $c_{o} + c_{1}\alpha + c_{2}\alpha^2, c_{i} \in \mathbb {Q}$
This is of the form for F, so $$a+b \in F$$.

Now we will explore the closure of F under multiplication. Using the same definitions for $$a_{0}$$ and $$b_{0}$$, we see that $(a_{o} + a_{1}\alpha + a_{2}\alpha^2 )*(b_{o} + b_{1}\alpha + b_{2}\alpha^2 ) =$ $= a_{0}b_{0} + \alpha (a_{0}b_{1} + a_{1}b_{0}) + \alpha^2 (a_{0}b_{2} + a_{1}b_{1}) +\alpha^3(a_{1}b_{2} + a_{2}b_{1}) + a_{2}b_{2}\alpha^4$
Consider $$\alpha^3$$. $$\alpha = \sqrt[3]{2}$$, so $$\alpha^3 = 2 \in \mathbb {Q}$$. Therefore ab can be written as $ab = a_{0}b_{0} + \alpha(a_{0}b_{1}+ a_{1}b_{0}) + \alpha^2(a_{0}b_{2} + a_{1}b_{1}) + 2(a_{1}b_{2}+a_{2}b_{1}) +2a_{2}b_{2} \alpha =$ $=(a_{0}b_{0} + 2(a_{1}b_{2} + a_{2}b_{1})) + \alpha (a_{0}b_{1} + a_{1}b_{0} + 2a_{2}b_{2}) + \alpha^2 (a_{0}b_{2} + a_{1}b_{1})$
Since $$2, a_{i}, b_{i} \in \mathbb{Q}$$ and $$\mathbb {Q}$$ is closed under multiplication and addition, being a subfield of $$\mathbb {C}$$, we can redefine coeficcients as $C_{0} \equiv (a_{0}b_{0} +2(a_{1}b_{2}+a_{2}b_{1}))$ $C_{1} \equiv (a_{0}b_{1} + a_{1}b_{0} + 2a_{2}b_{2})$ $C_{2} \equiv (a_{0}b_{2} + a_{1}b_{1})$
So $$ab$$ can be rewritten as $c_{o} + c_{1}\alpha + c_{2}\alpha^2, c_{i} \in \mathbb {Q}$
which is in the form of F, so $$ab \in F$$

## Part b)

We wish to show that if $$a_{0} + a_{1} \alpha + a_{2} \alpha^2 =0$$, where $$a_{i} \in \mathbb {Q}$$, then all $$a_{i} = 0$$. This is the equivalent to saying this is a linearly independent combination of alphas. Since all $$a_{i} \in \mathbb{Q}$$, each $$a_{i}$$ can be rewritten as the quotient of two integers. Therefore, we will denote $a_{0} = \frac{b_{0}}{c_{0}}, \text { where }b_{0},c_{0} \in \mathbb {Z}$, $a_{1} = \frac{b_{1}}{c_{1}}, \text { where } b_{1},c_{1} \in \mathbb {Z}$, $a_{2} = \frac{b_{2}}{c_{2}}, \text { where } b_{2},c_{2} \in \mathbb {Z}$ This is more clear, but not as manageable, so let us try to manipulate this into terms of all integers by exploiting the zero on the right side of the equation. Thus our original equation can be rewritten by multiplying through by the product of the denominators as $c_{1}c_{2}b_{0} + c_{0}c_{2}b_{2} \alpha + c_{0}c_{1}b_{2} \alpha^2 =0$ Since all of these coefficients are integers and the set of integers are closed under multiplication, we can redefine