Constructing $\mathbb{R}$ from $\mathbb{Q}$ and showing $\mathbb{Q}$ is dense in $\mathbb{R}$ 

This is a very long, multi-part problem that we were told to figure out by any means possible. There are no limits on getting help or finding answers online. I haven't had much luck at all solving this for myself, can anyone help me with some of this?
Thanks!
EDIT: I have made it to number (vi), which is where I need some help. I don't really understand that I am expected to do. Can someone assist from there? Thanks!
 A: For $(i)$ if $C(\mathbb{Q})$ is the set of all cauchy sequences, we can see that the $0$ element would be the sequence $(0,0,0,...)$, and if we take two cauchy sequences $f_n,g_n\in C(\mathbb{Q})$, then $f_n+g_n=(f_1+g_1,f_2+g_2,...)$, and we can see the addition is still cauchy since: $|(f_n+g_n)-(f_m+g_m)|\le|f_n-f_m|+|g_n-g_m|\lt 2\epsilon$. 
And $|\alpha f_n -\alpha f_m|=|\alpha||f_n-f_m|\lt |\alpha|\epsilon$, thus scalar multiples are cauchy, thus $C(\mathbb{Q})$ is a vector space.
A: When you're thinking about things, understand that decimal expansions are good examples of Cauchy sequences. This justifies writing
$$
         \sqrt{2} = 1.414213562373....
$$
This is really a sequence of numbers whose square gets closer and closer to $2$. That is, in your context,
$$
        \sqrt{2} = [\{ 1, 1.4, 1.41, 1.414, 1.4142, 1.41421, \cdots\}]
$$
The square of this sequence is another Cauchy sequence that is equivalent to the constant sequence $[\{2,2,2,2,2,\cdots\}]$ because the following difference sequence tends to $0$:
$$
       \{2-1^{1},2-(1.4)^{2},2-(1.41)^{2},2-(1.414)^{2},\cdots\}
$$
Using these definitions, there are multiple ways to express the number $1$:
$$
        1 = [\{1,1,1,\cdots\}]=[\{0.9,0.99,0.999,0.9999,\cdots\}] = 0.999999\cdots
$$
To remove the ambiguity, you're lumping together all Cauchy sequences whose differences have an actual limit of $0$. That's what the equivalence relation is all about. An equivalence class $[\{ x_{n}\}]$ consists of all Cauchy sequences $\{ y_{n}\}$ such that $\{ x_{n}-y_{n}\}$ has a limit of $0$.
The first thing you have to do is show that you can add these equivalence classes of Cauchy sequences. This is done by taking one Cauchy sequence from $x=[\{ x_{n}\}]$ and another from $y=[\{ y_{n}\}]$ and showing that (a) $\{ x_{n}+y_{n}\}$ is Cauchy sequence, and (b) no matter what one you choose from $x$ and what you one you choose from $y$, you always get a Cauchy sequence which is equivalent to the first one $\{ x_{n}+y_{n}\}$. In other words, $\{ x_{n} \} \sim \{ x_{n}'\}$ and $\{ y_{n}\} \sim \{ y_{n}'\}$ implies $\{ x_{n}+y_{n} \} \sim \{ x_{n}'+y_{n}'\}$. Then you have a good definition for $x+y=[\{x_{n}+y_{n}\}]$ that doesn't depend on what members $\{ x_{n}\}\in x$ and $\{ y_{n}\}\in y$ of the equivalence classes $x$ and $y$ that you use.
A couple of things you'll need to be able to show: The sum of two Cauchy sequences is also a Cauchy sequence. And, a Cauchy sequence is bounded. Most everything else is just slogging through the details.
A: For (vi): Injectivity: Call this inclusion map $i$. Suppose $i(q)=i(p)$ for some $p,q\in \mathbb{Q}$. This means $[(p,p,\dots)]=[(q,q,\dots)]$ and therefore $(p,p,\dots)$ and $(q,q,\dots)$ are represantatives of the same equivalence class, which means by definition of the equivalence relation...
Field structure: You have shown in (iii) that the equivalence classes form a field, so check that $i(\mathbb{Q})$ has also the same field structure inherited from $\mathbb{R}$ (as defined above) and can hence be considered as copy of $\mathbb{Q}$ in $\mathbb{R}$ (as defined above).
