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I have a book which says:

If a function $f$ satisfies $f(-x)=f(x)$ for all $x$ in its domain, then $f$ is called an even function. However, if $f(-x)=-f(x)$ for every $x$ in the domain of $f$, then $f$ is called an odd function.

But this is not clear. Because it does not mention what happens when, for a particular function, $a$ is in its domain but $-a$ isn't. The ambiguity arises in a function like $f:\{-5,-1,0,1,6\} \to \{-6,0,1,5\}$ given by $f(-5)=5, f(-1)=f(1)=1, f(0)=0$ and $f(6)=-6.$ We cannot decide in which one of the following categories the function $f$ falls:
1. Both odd and even,
2. Odd,
3. Even, and
4. Neither odd nor even.
[This is because I assumed that no one wants a fifth category like "can't decide".]
For that, can anybody please state the formal non-ambiguous definitions of odd and even fuctions? And also, regarding a problem such as this, can anybody tell me if there is any international body of Mathematics (like IUPAC is there for Chemistry), which decides which definitions and conventions to adopt and if it exists, how can I find those rules?

I know such a question might sound silly, but I have a set of questions for interviews which has the following question:
"Give an example of a function which is both odd and even. Is your choice unique?" - The answer to the second part of the question depends on the precise definition of odd and even functions.

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2 Answers 2

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The definition of odd and even function requires the domain $D_f$ of $f$ be such that $\forall x\in D_f\left(-x\in D_f\right)$. When $D_f$ satisfies this property, then $f$ is said to be odd if $\forall x\in D_f(f(-x)=-f(x))$ and it is said to be even if $\forall x\in D_f(f(-x)=f(x))$.

And also, regarding a problem such as this, can anybody tell me if there is any international body of Mathematics (like IUPAC is there for Chemistry), which decides which definitions and conventions to adopt and if it exists, how can I find those rules?

I don't think there exists such thing, but there are some universal unwritten conventions.

Related: 1, 2.

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  • $\begingroup$ You can weaken the definition of course; e.g. $f$ is even if, whenever $x$ and $-x$ are in its domain, we have $f(x)=f(-x)$. Also, to add to your last comment: definitions and conventions vary greatly in mathematics, so in general the best practice is to understand your audience, and define what you mean whenever your definitions or conventions may be different from theirs. $\endgroup$
    – Sean Clark
    Apr 10, 2014 at 15:10
  • $\begingroup$ @user1306 Yes, they do vary a lot. And I realize you're not contradicting what I said, but I still want to point out there are some universal conventions. $\endgroup$
    – Git Gud
    Apr 10, 2014 at 15:11
  • $\begingroup$ Except that is not the same definition. Requiring that for any x in the domain, -x is in the domain, and that f(x) = f(-x) is not the same as saying "whenever x and -x are in the domain, f(x) = f(-x). $\endgroup$
    – Jonathan Hebert
    Apr 10, 2014 at 15:15
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    $\begingroup$ @Doop Yes, it is not the same. The way I define it, however, is the most common. $\endgroup$
    – Git Gud
    Apr 10, 2014 at 15:16
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    $\begingroup$ @IndrayudhRoy The codomain doesn't matter. If you want to take $\sqrt \colon \mathbb R\to \mathbb C, x\mapsto \sqrt x$, then for all $x\in \mathbb ]-\infty, 0[$, one has $\sqrt x=i\sqrt{-x}$ and $\sqrt{-x}=\sqrt{-x}$, so it isn't even. It obviously isn't odd either. The problems you're trying to allude to are not with the codomain, but with the definition of $\sqrt \cdot$, which, when well determined, should pose no problem. $\endgroup$
    – Git Gud
    Apr 10, 2014 at 15:58
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The definition is perfectly clear. If -x is not in the domain then clearly f(x) = f(-x) is not true.

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  • $\begingroup$ This is wrong, if $x$ isn't in the domain, $f(x)=f(-x)$ isn't false and it isn't true. It is senseless. Saying this is false is like saying "car the 23 red" is false. $\endgroup$
    – Git Gud
    Apr 10, 2014 at 15:04
  • $\begingroup$ How do you prove that? $\endgroup$
    – Indrayudh Roy
    Apr 10, 2014 at 15:05
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    $\begingroup$ Perhaps I should have said that it isn't true, not that it is false. $\endgroup$
    – Jonathan Hebert
    Apr 10, 2014 at 15:07
  • $\begingroup$ @Indrayudh Roy This requires little proof. If x is in the domain then f(x) is in the range. If f(-x) = f(x) then f(-x) is in the range, therefore -x is in the domain. "For all x in D, f(x) = f(-x)" is a complete and unambiguous definition. $\endgroup$
    – Jonathan Hebert
    Apr 10, 2014 at 16:16

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