Does $f(x) = 1 = \frac{x-1}{x-1}$ have a hole at $x=1$? While learning about asymptotes and holes in rational functions in Precalculus, I came across a problem that shouldn't happen, but I don't understand. Something just doesn't add up in my head... Here we go:
Let f(x)=1.
Since $\frac aa=1$ and $a*1=a$, obviously $a={a*b \over b}$.
Let $b=(x-1)$.
We can multiply f(x) by 1, technically, by saying that: $$f(x)=1={x-1 \over x-1}$$
However, a bit of precalc knowledge shows us that ${x-1 \over x-1}$ has a vertical asymptote at x=1, and an x-intercept at x=1 as well. This means that the graph has a hole, or is undefined at, x=1. (Because plugging in 1 for x yields a denominator of 0, which is undefined.)
However, this is absurd because the graph of $f(x)=1$ obviously has no holes, anywhere.
Somehow, multiplying this extremely simple function by something equivalent to 1 makes it undefined at a particular point...
Whaaaaa...?
EDIT: I realize that the above function does not have a vertical asymptote. However, my statement was referring to the fact that it appears to have both a vertical asymptote and an x-intercept at the same x-value, which means that it has neither of those things, just a hole.
 A: There is no vertical asymptote. A function $f(x) = \frac{P(x)}{Q(x)}$, where $P$ and $Q$ are continous, has a vertical asymptote at $x_0$ if $Q(x_0)=0$ and $P(x)\neq 0$, which is not true in your case.
In your case, the function $f(x)=\frac{x-1}{x-1}$ simply has a hole.

As for why that hole appears:
Simple: It's because $1=\frac{x-1}{x-1}$ is not true for $x=1$. It is only true for $x\in\mathbb R\setminus \{1\}$.
You say in the beginning that "obviously", it is true that $a=\frac{a\cdot b}{b}$, which is an equality that is only true for $b\neq 0$.
In other words, your sentence:

Somehow, multiplying this extremely simple function by something equivalent to 1 makes it undefined at a particular point...

is not true, because you didn't multiply it by something equivalent to $1$. You multiplied it with something that equivalent to $1$ in most points and is undefined at a particular point.
A: When you make divisions like $\frac{f(x)}{p(x)}$, always make the claim that p(x) should be different from 0.
