TITLE: In stochastic calculus, why do we have $(dt)^2=0$ and other results?
QUESTION [17 upvotes]: I'm doing actuarial problems of Exam MFE and it covers some of the stochastic calculus (like Ito's Lemma). One of the frequently used results are the so-called "multiplication rules":

$(dt)^2=0$
$dZ(t)^2=dt$
$dZ(t) \, dt=0$

I tried to do some research online. There are tons of papers providing introduction to stochastic calculus, but strangely all of them seem to take these three rules as granted instead of proving them. Does anyone know the proof of these rules? Thanks.
Update: Someone mentioned that one needs to understand some analysis and measure theory to understand that. If someone could recommend relative textbooks on measure theory (I've taken analysis) and stochastic calculus, I would really appreciate it. Thanks.

REPLY [2 votes]: Other's have better answers, but in a simpler way:
$(dt)^2 ≈ 0$
Calculus looks at changes during small time steps $dt$ which when tiny, $(dt)^2 << dt$, so is approximated to be $0$. $[0.001 >> (0.001)^2 = 0.000001]$.
$dZ(t)^2 = dt$
To do with the 'scaling' chosen get a nice process. Out of three cases, $dZ(t)^2 \approx O(dt)$ provides a stable process, so we use $dZ(t)^2 = dt$.
[The two discarded cases are 

$dZ(t)^2 \rightarrow 0$ quicker than $dt$ does - process collapses to $0$, 
$dt \rightarrow 0$ quicker than $dZ(t)^2$ does - process grows indefinitely

.] 
$dZ(t)dt=0$
uses the first and second points. Pretty much anything $< dt ≈ 0$, and we know $dZ(t) = \sqrt{dt}$ (from $dZ(t)^2 = dt$), therefore $dZ(t)dt < dt$.
EDIT: I think this is actually a botched answer more related to why we pick what we do when going from the BSE to the Kolmogorov equation.