Why aren't the functions $f(x) = \frac{x-1}{x-1}$ and $f(x) = 1$ the same? I understand that division by zero isn't allowed, but we merely just multiplied $f(x) = 1$ by $\frac{x-1}{x-1}$ to get $f(x) = \frac{x-1}{x-1}$ and $a\cdot 1 = 1\cdot a = a$ so they're the same function but with different domain how is this possible?
Or in other words why don't we simplify $f(x) = \frac{x-1}{x-1}$ to $f(x) = 1$ before plotting the points.
Is it just defined this way or is there a particular reason ?
Note: my book says the domain of $f(x) = 1$ is $\mathbb{R}$ and the domain of 
$f(x) = \frac{x-1}{x-1}$ is $\mathbb{R}$ except $1$.
 A: Question: What is a function? 
Answer: Maybe simply said it is a map (receipe), $f(x)$, that projects some elements, $x$, contained in a specifically defined set, Domain $D$, into another set, Range $R$.
Discussion: Hence when defining a function one must define the Domain as well as the functional form. Otherwise the function is not defined.
Conclusion: If two funtions have the same domain and the same receipe then they are the same "maps" otherwise they are not.
A: They are the same almost everywhere. But clearly one of them does not exist for $x=1$ (since "$\tfrac{0}{0}$" is undefined), while the other one is simply $1$ at $x=1$.

I understand that division by zero isn't allowed, but we merely just multiplied f(x) = 1 by (x-1)/(x-1)

You can multiply by any fraction $\tfrac{a}{a}$; but not when $a=0$ because the fraction you want to multiply with, isn't even defined then. So multiplying by $\tfrac{x-1}{x-1}$ is fine, but only valid for $x \ne 1$.

why don't we simplify f(x) = (x-1)/(x-1) to f(x) = 1 before plotting the points.
  Is it just defined this way or is there a particular reason ?

You can simplify, but recall that simplifying is actually dividing numerator and denominator by the same number: you can simplify $\tfrac{ka}{kb}$ to $\tfrac{a}{b}$ by dividing by $k$. But also then: this only works for $k \ne 0$ since you can't divide by $0$. So "simplifying" $\tfrac{x-1}{x-1}$ to $1$ is fine, for $x-1 \ne 0$ so for $x \ne 1$.


Note: my book says the domain of $f(x) = 1$ is $\mathbb{R}$ and the domain of 
  $f(x) = \frac{x-1}{x-1}$ is $\mathbb{R}$ except $1$.

Technically, the domain is a part of the function: it should be given (as well as the codomain). It is very common though that when unspecified, in the context of real-valued functions of a real variable, we assume the 'maximal domain' is intended (and $\mathbb{R}$ is taken as codomain). Then look at:
$$f : \mathbb{R} \to \mathbb{R} : x \mapsto f(x) = 1$$
and
$$g : \mathbb{R} \setminus \left\{ 1 \right\} \to \mathbb{R} : x \mapsto g(x) = \frac{x-1}{x-1}$$
The functions $f$ and $g$ are different, but $f(x) = g(x)=1$ for all $x$ except when $x=1$, where $g$ is undefined.
A: $f(x)=(x-1)/(x-1)$ does not have a value when $x=1$, different thing happens to $f(x)=1$
