# 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 $$ab \in F$$