Do we have $\frac{1}{a} - \frac{1}{b} = b - a$?

Question

I am attempting to prove that $$\frac{1}{E'} - \frac{1}{E} = \frac{1}{m_e c^2} \cdot (1-\cos\theta)$$ can be derived from

$$E + m_ec^2 - E' = c^2(p^2 - 2pp'\cos\theta + p'^2) + m_e^2c^4 $$

where $p\cdot c = E$ and $p'\cdot c = E'$.

And have become found that $$E - E' = \frac{1}{m_ec^2} \cdot (1-\cos\theta)$$

This leads to the question, Is $\frac{1}{A} - \frac{1}{B} = B - A$?

Additionally, upon manipulating $\frac{1}{a} - \frac{1}{b} = b - a$, I've found that $1 = ab$. Does this mean that $\frac{1}{a} - \frac{1}{b} = b - a$ as long as $1 = ab$?

Answer 1

If $1=ab,$ then $b=\frac1a,$ so $\frac1a-\frac1b=b-\frac1b.$ On the other hand, we also have $a=\frac1b,$ so $b-a=b-\frac1b.$ Hence, if $ab=1,$ then $\frac1a-\frac1b=b-a.$ However, the reverse implication does not hold.

Note that $$\frac1a-\frac1b=\frac{b}{ab}-\frac{a}{ab}=\frac1{ab}(b-a),$$ so that $$\frac1{ab}(b-a)=b-a$$ if and only if $ab\ne0$ and $$0=\left(1-\frac1{ab}\right)(b-a)=\frac1{ab}(ab-1)(b-a).$$ Equivalently, $a$ and $b$ are both non-zero, and $ab=1$ or $a=b.$