Foundational proofs

In section 1.1 of our text, the complex numbers are placed on a firm theoretical basis by defining $\mathbb C$ as
$$\mathbb C = \{(x,y):x\in\mathbb R\}$$
with addition defined componentwise and multiplication defined by
$$(a,b)\times(c,d) = (ac-bd,ad+bc).$$
There is a natural correspondence between this firm theoretical point of view and the way you probably think of things intuitively in terms of $i$ where $i^2=-1$. In fact, I only remember the definition of multiplication by expanding out $(a+bi)(c+di)$.

Proposition 1.1 of the text states that, given this definition, the complex numbers inherit certain properties (11 of them, in fact) from the real numbers. The text shows one of these but exercise 1.14 asks you to prove the rest. How can we prove some of these?

Note: There are plenty of these so loads of people can try it. A good answer should state the problem (don't write just 'part c') and write a clear proof using TeX snippets to typeset the formulae.

Part 1.6 asks us to show that $\mathbb C$ is commutative. That is, we need to show that
$$(x,y) + (a,b) = (a,b) + (x,y)$$
for all $a,b,x,y\in\mathbb R$. This is just a matter of working componentwise:
$$(x,y) + (a,b) = (x+a,y+b) = (a+x,b+y) = (a,b)+(x,y).$$
Note that the first and last equalities are simply the definition of complex addition while the middle equality uses the commutative property of real numbers.

Part 1.9 asks us to prove a distributive law for $\mathbb{C}$ or $$(x,y)\cdot\big((a,b)+(c,d)\big)=(x,y)\cdot(a,b)+(x,y)\cdot(c,d)\forall (x,y),(a,b),(c,d)\in\mathbb{C}.$$
Since $x,y,a,b,c,d\in\mathbb{R}$, we can use all the properties of $\mathbb{R}$ (associativity, commutativity etc.).

By the deffinition of complex addition and multiplication,
$(x,y)\cdot\big((a,b)+(c,d)\big)=(x,y)\cdot\big((a+c),(b+d)\big)$
$\ \ \ \ \ =\big(x(a+c)-y(b+d),x(b+d)+y(a+c)\big)$
$\ \ \ \ \ =\big((xa+xc)+(-yb+-yc),(xb+xd)+(ya+yc)\big)$
$\ \ \ \ \ =\big((xa-yb)+(xc-yc),(xb+ya)+(xd+yc)\big)$
$\ \ \ \ \ =(xa-yb,xb+ya)+(xc-yd,xd+yc)$
$\ \ \ \ \ =(x,y)\cdot(a,b)+(x,y)\cdot(c,d)$.

$\text{\begin{align*} and \end{align*} did not want to work for me. When I tried to use it, this is what it spit out:}$

\begin{align*}
(x,y)\cdot\big((a,b)+(c,d)\big)&=(x,y)\cdot\big((a+c),(b+d)\big)\\
&=\big(x(a+c)-y(b+d),x(b+d)+y(a+c)\big)\\
&=\big((xa+xc)+(-yb+-yc),(xb+xd)+(ya+yc)\big)\\
&=\big((xa-yb)+(xc-yc),(xb+ya)+(xd+yc)\big)\\
&=(xa-yb,xb+ya)+(xc-yd,xd+yc)\\
&=(x,y)\cdot(a,b)+(x,y)\cdot(c,d).
\end{align*}

Is there something I am doing wrong?

No, you're not doing anything wrong. There is a bug in Discourse's current MathJax processor. In fairness, it is complicated to get code working within an HTML environment, since the browser wants to interpret everything it sees. One manifestation of this is that a double slash \\ like the one you (correctly) used to terminate the lines looks like a single slash by the time MathJax sees it, because the slash is often used as an "escape character". The solution is to escape the escapes so that you have actually type four slashes to terminate a line: \\\\. I did that and it looks good now.

Another good option is to use \begin{array}{rcl} for equation arrays. The {rcl} is the alignment of each column. Just put the equals in the center column. The column delimeter is the ampersand: &, the line delimeter is three slashes : \\\.