"Multiplying" by a dt then cancelling? Today I encountered a proof of a physics kinematic equation which contained the following steps:
$$ \frac{dx}{dt} = v $$ $$ dt \frac{dx}{dt} = v\,dt $$ $$ dx = v\,dt $$
Now I have read the previous questions which state putting a dt on both sides isn't multiply and that dy/dx isn't a ratio.
I understand dy/dx is not an actual fraction but has "convenient" fraction properties, but I don't understand why $ dx = v\,dt $ is a valid equality. Where did the left hand side $dt$ go?
 A: Here's the issue:
$\frac{dx}{dt}=v \cdot dt$ is not a valid equality. Someone made a mistake, because the equation should read $dx = v \cdot dt$
As a Mathematician, there are all kinds of things wrong with this. Instead, lets talk as Physicists.
Do dimensional analysis. The left side is in units of (Length)/(Time) while the right side is in units of [(Length)/(Time)] * (Time) = Length.
Ok, so now that we have corrected your typo, we can proceed as Mathematicians.
When you use the symbol $dt$ "freely" (without being bound as part of a "fraction") you are in fact using the Leibniz notation ("differentials"). When introduced, the notation had no formal mathematical meaning. It was only much later, during the establishment of Differential Geometry, that the symbol was "redefined" to have a precise mathematical meaning that behaved in the same way as before.
In the new formalism, $dt$ denotes a $1$-form. $\frac{d}{dt}$ denotes a vector field. The operation that "glues" together forms and fields is the inner product, to yield symbols like $\frac{dt}{dt}=1$ that have meaning as functions (which are $0$-forms).
So after all that mumbo-jumbo, we have taken the equation $\frac{dx}{dt}=v$ relating $0$-forms and turned it into an equation $dx = v dt$ relating $1$-forms. For now, think of it simply like multiplication. Later on if you choose to continue with math, what I wrote above might make more sense.
A: The intuition is that the small change in $x$ can be calculated as $v \ dt$ because there "isn't time" for the velocity to change.  Alternately, the error in $x$ is second order in $dt$, one for the change in $v$ and one for the short time that it acts.  Since $dt$ is very small, $(dt)^2$ is smaller yet and can be ignored.
More formally, expand $x$ in a Taylor series in $t$.  We have $x(t)\approx x(0)+\frac {dx}{dt}(0)t+\frac 12 \frac {d^2x}{dt^2}t^2$.  If $t$ is small, we can ignore the last term (and those that follow), so we get $x(dt)-x(0)=dx=v\ dt$
As physicists, our functions are well-behaved (everything is uniformly convergent), so we can get away with a lot that the mathematicians think is unfair to assume.  Like the Mad Hatter, we pay them better and they work harder for us.
A: The $dt$ on the LHS was moved over to the RHS, precisely where you see it. It doesn't make much physical sense where it's at, but it is arguably nicer mathematically. pre-kidney gave a nice explanation of what it all means in deeper mathematics, so I'll direct you to his post for that.
This technique of moving $dt$ to the other side (which may be foreshadowing if you've never had differential equations) is used in separation of variables, whence you are able to integrate the RHS with respect to $t$ and the LHS with respect to $x$. Doing so yields that $x=vt + c$, where $c$ is a constant, and in particular, $c=x_0$, the initial position.
