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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 $1$.

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    $\begingroup$ This is an interesting question, but posted on the wrong website. You may want to think about what it means precisely for two functions to be "the same". $\endgroup$ Jan 11, 2017 at 15:16
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    $\begingroup$ The crux of the matter is that you have been deceived: if they told you that a function is simply a formula, then they lied... Rather, a function f is: 1. a source set S, 2. a target set T, 3. a way to associate to each element x in S a unique element of T, denoted by f(x). In your example, 2. and 3. coincide but one generally chooses for S the real line for the function f(x)=1. Then you see that one cannot choose this same set S for the function f(x)=(x-1)/(x-1) since (1-1)/(1-1) does not exist, hence the two functions are indeed different. $\endgroup$
    – Did
    Jan 11, 2017 at 15:26
  • $\begingroup$ Related : math.stackexchange.com/questions/462199/… $\endgroup$
    – Arnaud D.
    Jan 11, 2017 at 16:40
  • $\begingroup$ "so they're the same function but with different domain" That answers your question. The domain is a required part of the determination of a function. The "same function but with different domain" is a bit like saying "24 and 48 are the same number but with a different power of 2". If the domains are different... then the functions are different. It's that simple. $\endgroup$
    – fleablood
    Jan 11, 2017 at 17:11
  • $\begingroup$ Almost identical: math.stackexchange.com/questions/1670139/…. $\endgroup$
    – Alex M.
    Jan 11, 2017 at 20:20

3 Answers 3

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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.

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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.

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    $\begingroup$ Indeed, except that the target set can be any set containing the range, not necessarily the range. $\endgroup$
    – Did
    Jan 11, 2017 at 15:30
  • $\begingroup$ @Did Thank for making it more precise :-) $\endgroup$
    – Math-fun
    Jan 11, 2017 at 15:30
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$f(x)=(x-1)/(x-1)$ does not have a value when $x=1$, different thing happens to $f(x)=1$

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