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We know that the maximun domain of the function $\,\,\,\displaystyle{f(x)=\frac{x-1}{x-1}}\,\,\,$ is $\,\,\mathbb{R}-\{1\}$.

Which is the mathematical concept that justifies that can not be simplified the function as $f(x)=1$ and therefore say that maximun domain of $f(x)$ is $\,\,\mathbb{R}$?

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5 Answers 5

Your function is exactly the same as ($f(x)=1$ iff $x\in\Bbb R-\{1\}$).

It can be extended continuously to a (more) maximal domain, by defining also $f(1)=1$

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Berci, but why I can not simplify and say that maximun domain is $\mathbb{R}$? Why is it that I can not simplify the function? –  Roiner Segura Cubero Oct 13 '12 at 15:41
    
This is because to compute $\dfrac{x-1}{x-1}$ you need to calculate $x-1$ (numerator and denominator, in general), then take the quotient. Because the quotient $0 / 0$ is not defined, you cannot consistently evaluate $f$ at $1$; that it is equal to $g(x) := 1$ if $x \ne 1$ does not matter for the (hypothetical) value of $f$ at $1$. –  Lord_Farin Oct 13 '12 at 15:46

The axioma that builds the real numbers says that there exist reverse of numbers different of zero. Thus $ 0 /0 $ shold be $ 0 \cdot 0^{-1}.$ But $ 0^{-1}$ does not exist.

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That's simply how arithmetic works. When composing partial functions, the domain of the composite $f \circ g$ is the subset of the domain of $g$ whose image lies in the domain of $f$.

The domain of (real number) division is all pairs of real numbers whose denominator is zero. Therefore, the domain of

$$ f(x) = \frac{x-1}{x-1} $$

is the set of all numbers $a$ that satisfy:

  • $a$ is in the domain of the numerator $x-1$
  • $a-1$ is a real number
  • $a$ is in the domain of the denominator $x-1$
  • $a-1$ is a nonzero real number

And so $1$ is not in the domain of $f(x)$. If $x$ is a variable whose domain is all real numbers, then one can straightforwardly determine that the domain of $f(x)$ is the set of all real numbers other than $1$.

Thus, the restriction of $f(x)$ to any subset of $\mathbb{R} \setminus \{1\}$ is a total function.

As an aside, this all comes down to the fact division is the division of real numbers. There are other settings -- and division operations suitable to such settings -- in which such a $f(x)$ would be equal to 1.

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Just to introduce some useful terminology, your function $f(x)$ has a removable singularity at $x=1$. There is a closely related function which has the singularity removed. Often people are lazy and treat them as the same, but in some contexts the difference is important. It is as well to be precise, because it costs almost nothing and avoids careless mistakes in unfamiliar contexts.

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The rule for simplifying fractions is that you may divide the numerator and denominator by any quantity that does not equal zero. Your proposed simplification does not fit within that rule.

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