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When simplifying expressions, why do we add on restrictions for the simplified form if the original form was undefined at a certain point? The simplified form is defined at those points, so why should it be restricted?

An example of what I mean:

$$\frac{x^2−1}{x−1}=x+1, x≠1$$

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It's a condition on the equality (see Leibniz's law). When $A = B$ unconditionally, we can substitute $A$ for $B$ (and vice versa) unconditionally. When both $A$ and $B$ are expressions in variables $x, y,$ etc, there is an implied quantification $\forall x, y,$ etc. When either $A$ or $B$ is undefined for certain values for $x, y,$ etc, the equality does not hold unconditionally anymore. –  user2468 Jul 11 '12 at 2:38
    
+1 for thinking about these things. –  Ross Millikan Jul 11 '12 at 3:26
    
@RossMillikan Thank You! –  user26649 Jul 11 '12 at 3:31

1 Answer 1

up vote 5 down vote accepted

In short, it's because a function is all about the rule and domain of definition.

For example, let's consider the real-valued functions $f(x)=\frac{x^2}{x}$ and $g(x)=x$, with their maximal real domains of definition. In particular, then, $\mathrm{dom}(f)=\Bbb R\smallsetminus\{0\}$ and $\mathrm{dom}(g)=\Bbb R$. That means that $f$ and $g$ are not the same function, since their domains are not the same. Now, where both are defined, the rule is the same. However, this isn't enough for them to be the same function. Now, $f$ is a restriction of $g$ (specifically, to $\Bbb R\smallsetminus\{0\}$), so they are certainly related functions.

Edit: In the context of your previous question (regarding limits and restrictions), it is worth noting that $f$ can be continuously extended to $g$, since for any $\varepsilon>0$ there exists $\delta>0$ (in particular, in this case, $\delta=\varepsilon$) such that for any $0<|x|<\delta$ we have $\left|f(x)-0\right|<\varepsilon$.

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Simple and concise. +1 –  DonAntonio Jul 11 '12 at 2:42
    
@CameronBuie Thank You! –  user26649 Jul 11 '12 at 3:38

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