# Do holes affect the type of function (even, odd or neither)

We can know whether a function is even or odd by substituting using F(-x) but what if the function has a single hole like this

f(x) = $$\frac{x(x-2)}{x-2}$$

is such a function considered odd or neither? since substitution in the original function will confirm that it is neither while substitution after simplification will confirm that it is odd

• Why is $\dfrac {x (x - 2)} {x - 2}$ not odd? Surely $\dfrac {x - 2} {x - 2}$ is always $1$ except where $x = 2$, unless you disallow the term for odd-except-for-a-hole. Commented Nov 27, 2021 at 8:20
• Substituting with f(-x) it will be $\frac{-x(-x-2)}{-x-2}$ which is clearly not equal to -f(x) Commented Nov 27, 2021 at 8:23
• The hole at the 2 is not reflected across the origin if you got what I mean @PrimeMover Commented Nov 27, 2021 at 8:26
• Replace the word "hole" by "pole". Commented Nov 27, 2021 at 9:20
• Wouldn't it be better to replace "hole" by "singularity" as a pole is a special kind of singularity? (In our case it's a removable one.) Commented Nov 27, 2021 at 9:59

It depends upon how you defined odd function and even function. Let $$D\subset\Bbb R$$.
• If you say that a function $$f\colon D\longrightarrow\Bbb R$$ is odd if $$x\in D\implies-x\in D$$ and $$f(-x)=-f(x)$$ (this would be my definition), then $$f$$ is not odd (since $$-2$$ belongs to its domain, but $$2$$ doesn't).
• If you say that a function $$f\colon D\longrightarrow\Bbb R$$ is odd if, whenever both $$x$$ and $$-x$$ belong to $$D$$, then $$f(-x)=-f(x)$$, then your function is odd.