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Consider the expression:

$$\dfrac{(x - c_1) \cdot x}{(x - c_1)}$$

This is often simplified as

$$x \ \text{for} \ x \neq c_1$$

This simplification step can also be done an arbitrary number of times for

$$\dfrac{(x - c_1)(x - c_2) \dots (x - c_n) \cdot x}{(x - c_1)(x - c_2) \dots (x - c_n)}$$

In which case $x \neq c_1, c_2, \dots, c_n$.

Given that it is generally valid to simplify such expressions by repeated steps of elimination of the terms, does that not imply that it is equally valid to introduce arbitrarily many such terms? And if arbitrarily many such terms are introduced, how do we know that $x$ can be defined at all?

Jackie
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  • Are you asserting that, going that way ad infinitum, we have to conclude that $x$ is different from every number ? If so, you have to take into account taht (considering for simplicity natural numbers), in order to conclude that $x \ne 1,2,\ldots$ we have to write an infinte fractional expression. – Mauro ALLEGRANZA Aug 02 '18 at 11:46
  • But consider e.g. $\dfrac {(3-1)}{(3-1)} \cdot 3$; it is a "correct" expression and its value is simply $3$. – Mauro ALLEGRANZA Aug 02 '18 at 11:49
  • @MauroALLEGRANZA Yes, that was my assertion, that since x could be rewritten as itself multiplied with a fraction of arbitrary length up to infinity, x would have to be different from all possible values and thus undefined. I have tried to explain to myself why this is nonsense, but the second fraction you gave is valid. – Jackie Aug 02 '18 at 11:58
  • But consider the simple case with only $c_1$: it has at least the value $0$ that satisfies it. – Mauro ALLEGRANZA Aug 02 '18 at 12:01
  • I am sorry, could you articulate that? Do you mean $\dfrac{x - c_1}{x - c_1} \cdot x$ with $c_1 = 0$? Then I would understand that this is a valid rewriting of x, as it hasn't changed the possible values that x can be – Jackie Aug 02 '18 at 12:06
  • What I'm saying is that the title "that any given number can be rewritten to be undefined in a domain" is simply wrong. A number is a number and not an expression. – Mauro ALLEGRANZA Aug 02 '18 at 12:09
  • Having said that, the issue is: assuming that the infinite long expression makes sense, it is true that it fails to denote a number (i.e. there is no number satisfying it) ? Yes, but only if we consider the case of the infinite long expression, because it is equiv to $x \ne 0,1,2,3 \ldots$. If we consider the finite case only,like e.g. $\dfrac {(x-1)}{(x-1)} \cdot x$, it has infinite solutions: every number except $1$. – Mauro ALLEGRANZA Aug 02 '18 at 12:12
  • In my previous comment, I'm saying that every number different from $c_1$ will do, included $0$ (if $c_1 \ne 0$). – Mauro ALLEGRANZA Aug 02 '18 at 12:14
  • @MauroALLEGRANZA I think I would modify my question to ask, under which circumstances can such an expression be constructed? I am now asserting that, as long as we do not change where $x$ is defined, $x$ can be rewritten as itself multiplied by such an expression. I am sorry if my question is/was ambiguous – Jackie Aug 02 '18 at 12:20

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I think I can actually explain to myself why this is nonsense.

Given the value x, I think it is intuitive that this could be rewritten as $\dfrac{c1}{c1} \cdot x$ for $c\neq 0$. This is permissible because we haven't arbitrarily introduced an area in a domain where x is undefined. Multiplying by $\dfrac{c}{c} =1$ does not fundamentally change anything. On the other hand, in the case of $\dfrac{x−c}{x−c} \cdot x \ , x \neq c$ , rewriting this as $x$ is only permissible in the case that the restrictions on the domain are maintained. I.e, because the area in which $x$ is not defined is the same before and after the simplification step. Introducing such a term arbitrarily however breaks this rule, and is thus invalid.

Jackie
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