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If $f(x) = \frac{x^2}{x}$, what is the domain of $x$?

Should we simplify it to $f(x) = x$ so that $x$ is any real number?

Or we do not simplify it and that the $x$ is any real number and $x \ne 0$.

But depend on the web service such as this example from Wolfram Alpha, the domain of $x$ does not include 0.

Allen
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    Do you mean "what is the domain of $f$"? Personally, if I was being very strict, I would say that the definition of $f$ was "take $x$, square it, and then divide the result by $x$", which always gives you $x$ back except when $x = 0$, when the calculation doesn't make sense. So the domain is the set of non-zero real numbers. (In practice, I am never this strict, except when I am teaching students how to be this strict.) – Billy Jul 29 '13 at 06:58
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    Think of it this way: will your calculator get upset when you try to divide $0^2$ by $0$. Yes it will. So $0$ is not in the domain. – André Nicolas Jul 29 '13 at 07:00
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    Technically, you have not defined a function until you have also specified the domain and codomain. And clearly once you have specified a domain, the question of what the domain is becomes moot. – Tobias Kildetoft Jul 29 '13 at 07:01

1 Answers1

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The usual convention in such cases is that the domain of the function is the largest subset of $\Bbb R$ on which the function is defined. Thus, the function $f(x)=\frac{x^2}x$ is not identical to the function $g(x)=x$: while they agree on $\Bbb R\setminus\{0\}$, the domain of $f$, the domain of $g$ is all of $\Bbb R$. Two functions with different domains are by definition not the same function. Thus, in general you may not simplify $f$ to $g$. Of course if you’re only interested in the behavior of $f$ on some subset of $\Bbb R\setminus\{0\}$, then it’s perfectly fine to replace $f$ by $g$, since $f=g\upharpoonright(\Bbb R\setminus\{0\})$.

Brian M. Scott
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