Kapalci

2022-06-26

Does $\frac{x-2}{3x-6}$ really equal $\frac{1}{3}$?
In my maths lesson today we were simplifying fractions by factorising. One question was something like this: $\frac{x-2}{3x-6}$, which I simplified as $\frac{x-2}{3x-6}=\frac{x-2}{3\left(x-2\right)}=\frac{1}{3}$. It got me wondering however, whether these expressions are really equal, specifically in the case $x=2$, where the former expression is undefined but the latter takes the value $\frac{1}{3}$
Since the expressions only differ at a single point are they for all intents and purposes equal, or are they theoretically different? If I wanted to be entirely correct would I have to write $\frac{x-2}{3x-6}=\frac{1}{3}$ where $x\ne 2$?
My maths teacher explained that at $x=2$ the expression evaluates to $\frac{0}{3×0}$ and the zeros effectively cancel out. I wasn't altogether satisfied with this explanation because as far as I know $\frac{0}{0}$ is undefined.

drumette824ed

I believe putting it this way could be helpful:
One one hand, is a function that is defined on except at , and at every point it is defined it is equal to
On the other hand, can be viewed as a function that needs to accept ANY input and provide you with back. In particular, the domain restriction at  prevents to be the same thing as
Hence, simplifying wouldn't be appropriate to unless we are working in a domain where

gvaldytist

Yes... and no.
It is impractical to properly lay out all of the subtleties of what one intends with math notation, which is ambiguous. We've had ages to learn how to create notations where ambiguities frequently don't matter but do in the fine print.
There are several distinct interpretations of what ; The status of "evaluation" is where the alternatives diverge most noticeably at " ".
The most basic interpretation is that is a recipe for performing a sequence of arithmetic operations upon a given input; in this interpretation, evaluation at  is, in fact, undefined.
There are also more ways to interpret "take the continuous extension": essentially, you take the function's graph and fill in all the gaps; here you would add in . Also, if you were using the extended real numbers you would add in  and , so that evaluation at  is defined. (similarly if you were using the projective real numbers)
Your teachers description is nonsense when taken literally; however, the likely intention is that he is using "" as a stand-in for some sort of 'witness' of vanishing; e.g. we factor out  from both the numerator and denominator to get

The witnesses do "cancel" to depart if we adopt one of these "continuous extension" interpretations

Do you have a similar question?