Isn't the inductor equation negative? The inductor "resists" change in current. So say you measure the voltage across the inductor from point A to point B - the current is flowing in from A towards B. Now say the current is increasing. The inductor will try to oppose the change by creating a current the opposite direction - from point B to A. To do this it will create a voltage, where point A has a lower voltage than point B in order to "encourage" electrons to flow the opposite way. If this is true though, the voltage measured from point A to B will be negative, so shouldn't the equation be:
$$V_{a \to b}  = -L\frac{di}{dt}$$
 A: The sign in the case of an inductor is indeed easy to be uncertain about. I would say this is a good illustration of a more general difficulty with signs in physics. The way to get signs right is sometimes not to worry over one equation or another equation of opposite sign, but rather to be clear in your mind about what happens in a simple example case.
I find it very useful to consider the simple circuit with just a resistor and an inductor in it. The voltage around the circuit is zero, so we get the equation
$IR + L dI/dt = 0$
It is easy to see that the sign is correct in this equation, because then we get
$ dI/dt = - (R/L) I$
for which the solution is exponential decay. If we had the opposite sign we would get exponential growth of the current, which is clearly wrong. But you are free to consider the first equation either in the form I wrote it, or in the form
$IR = - L dI/dt$
A: 
The inductor will try to oppose the change by creating a current the opposite direction - from point B to A.

Inductor in this case does not create some secondary current that flows opposite to the direction of the original current. There is just one current, in single direction.
What inductor does is create electromotive force that acts on the charge carriers and resists change of electric current in the wire. This electromotive force is due to curly induced electric field of the electrons in the coils. It is NOT due to voltage on the terminals of the coil.

To do this it will create a voltage, where point A has a lower voltage than point B in order to "encourage" electrons to flow the opposite way.

This is entirely wrong. For an ideal inductor with no ohmic resistance, the voltage will be such that its associated electrostatic field counteracts the induced emf over the length of the wire of the inductor.
This means that the electrostatic field has to point in direction of the current increase, thus from A to B. Thus, B is at lower electric potential than A.
That is why voltage drop across the inductor in the direction from A to B has to be given by
$$
L\frac{dI}{dt},
$$
i.e. the voltage drop is positive. (It is, after all, this positive voltage drop that is powering the current increase assumed in the beginning).
A: Consider this concrete scenario. Let's say that there's a voltage source connected to an inductor (maybe throw in a resistor in series to make it more applicable to a real-world scenario).  The voltage source is increasing in time such that it drives an increasing current from point a to point b across the inductor, as you say.  We can even do a step in voltage, from 0 to some voltage $V_s$.
The inductor will act to counter this flux and hence current, so initially, the voltage at point a must increase rapidly so that the voltage source cannot readily drive more current through it.  So this "back EMF" acts to increase the voltage at point a relative to point b.  That is consistent with NOT having a negative sign.  Your equation implies that the voltage at a is lower than the voltage at point b.
In the scenario I describe, $\frac{dI}{dt}$ is positive and $I$ is in the direction $a \rightarrow b$, without the negative sign in your equation, the voltage at $a$ is higher than the voltage at point $b$.
This is why when a step in voltage occurs across an inductor, the voltage across that inductor shoots up quickly (and the current remains relatively steady).  As time goes on, the voltage across the inductor drops and the current reaches as steady-state value.
A: The minus sign is needed when using the right-handed convention for evaluating the EMF. That is, in the same direction as the current as you've done. The inductance equation comes from Faraday's Law which also has a minus sign
$$\oint \vec E\cdot \vec {dl} = -\frac{d}{dt}\iint \vec B \cdot \vec {dA} $$
We use the right-hand convention to evaluate these integrals: pick a general direction for the $\vec{dA}s$ either inwards or outwards from the surface and align an imaginary right-handed screw such that it moves in the same direction, either inwards or outwards, when its screwed clockwise; the LHS is then evaluated in a clockwise direction from behind the screw's head. However, the minus sign is needed to get the right and left hand sides to match in sign; which wouldn't be needed if a left-hand convention had been defined by mathematicians.
For a fixed circuit geometry, $\vec B$ is proportional to the current creating it, so that the RHS can be written as 
$$-\frac{di}{dt}\iint \vec r \cdot \vec {dA} =  -L\frac{di}{dt}$$
By definition, $L$ is always chosen to be positive which requires that $\vec r$ and therefore $\vec B$ is mostly nearly aligned with $\vec {dA}$, forcing i to be clockwise with the EMF.
A: If your inductor has a 1.0 amp current and you want to increase it to 1.1 amp, you need to increase the voltage to force more current through the inductor. Since there was already 1.0 amp flowing in the inductor, then it would resist the increase in voltage forcing more current (0.1 amp) to flow. So, yes there is an inverse voltage from B to A, but its magnitude is only large enough to resist the increase, not the full voltage.
