$$ R(x) = \frac{x^{2} - 1}{x - 1} = \frac{(x +1)(x - 1)}{x - 1} = x + 1$$

The reduced function has a domain of $\mathbb{R}$ whereas the original function excludes $1$ from the domain. Therefore the functions are not equal. But if you reduce a rational number $\frac42$ to $2$ they are equivalent. Why is this so?

  • 3
    $\begingroup$ What you really have is that $R(x) = x+1, x \neq 1$. You can't add points to the domain of a function by simplifying its algebraic representation. $\endgroup$
    – Randall
    Feb 19, 2023 at 0:45
  • 6
    $\begingroup$ When you reduce $4/2$ to $2$, your divisor is $2$, but when you simplify your function, your divisor is $x-1$, and $x-1$ may be $0$. $\endgroup$ Feb 19, 2023 at 0:50
  • 1
    $\begingroup$ "But if you reduce a rational number ... Why is this so?" Because functions aren't individual numbers? You shouldn't expect the same principles to apply to both. A function is (ontologically) a set of pairs of arguments and their respective values, e.g. $\{(0,1.24), (2, 3.05), \ldots\}$. Your original function had no point $(1,y)$; it was missing from the domain as it is a division by zero. And doing algebra on the function doesn't incorporate it; it should be missing from the new domain too. Any function with that argument in its domain is a different function. $\endgroup$
    – Jam
    Feb 19, 2023 at 0:51

1 Answer 1


The difference is that $R(x)$ describes an entire range of outputs for a domain or inputs. A function is defined by its domain, so even if other functions are algebraically identical for any given input that doesn't make their domains and thus the functions themselves equivalent.

Another example is $(\sqrt{x})^2$, which is algebraically identical to $x$ but has a restricted domain to positive real numbers.


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