For an equality of two fractions, each has a different domain; how?

Question

Take the following equality:

$$\frac{1}{\left(\frac{1}{y_0}-t^2\right)} = \frac{y_0}{1-\left(t^2y_0\right)}$$

For the left hand side, the function is undefined for $y_0 = 0$. But for the right hand side, the function equals $0$ when $y_0 = 0$

How can this be?

Answer 1

The best way of looking at it is this: Fix $t \in \mathbb R$. Then, the expression $\dfrac 1{\frac 1{y_0} - t^2}$ is not defined in two cases : either $y_0 = 0$ or $y_0 = \frac 1{t^2}$, which does not occur if $t = 0$. So, if we define the function $f_t(y_0) = \frac{1}{\frac 1 {y_0} - t^2}$ then for $t=0$ the domain of $f_t$ is $\mathbb R - \{0\}$, and for $t \neq 0$ it is $\mathbb R \setminus \left\{0, \frac 1{t^2}\right\}$.

Now, the other expression $\frac {y_0}{1-t^2y_0}$ is well defined whenever $y_0 \neq \frac 1{t^2}$. That is, if we let $g_t(y_0) = \frac {y_0}{1 - t^2y_0}$, then $g_t$ is defined on all of $\mathbb R$ if $t=0$, and the domain misses the point $\frac 1{t^2}$ if $t \neq 0$.

Therefore, for all $t$, $g_t$ and $f_t$ have different domain! Equality of functions must imply the same domain, but this is not the case here.

Thankfully, the idea of restriction comes in: simply define $h_t(y_0)$, which has the same domain as $f_t$ but $h_t(y_0) = g_t(y_0)$. Then, $h=f$ is a correct statement, since $h$ and $f$ have the same domain.

Therefore, if you find two expressions which are equal except at a set of points where one side is undefined and the other is not, then equality between the functions is actually the equality between the restrictions of these functions to the intersection of their domains i.e. where both are well defined. It is in this sense that the above inequality should be interpreted, and not the naive way.

Answer 2

This is sloppy. The left side implies the right because it makes sure $y_0 \neq 0$. The right does not imply the left because you could have $y_0=0$ on the right and you are then multiplying by $\dfrac {\frac 10}{\frac 10}$. If you start with the right you should account for the case where $y_0=0$, then assume $y_0 \neq 0$ as a second case and proceed.