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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.

MJD
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Indrayudh Roy
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2 Answers2

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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.

Jonathan Hebert
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  • 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. – Git Gud Apr 10 '14 at 15:04
  • How do you prove that? – Indrayudh Roy Apr 10 '14 at 15:05
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    Perhaps I should have said that it isn't true, not that it is false. – Jonathan Hebert Apr 10 '14 at 15:07
  • @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. – Jonathan Hebert Apr 10 '14 at 16:16
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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.

Community
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Git Gud
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  • 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.

     – Sean Clark Apr 10 '14 at 15:10
  • @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. – Git Gud Apr 10 '14 at 15:11
  • 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). – Jonathan Hebert Apr 10 '14 at 15:15
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    @Doop Yes, it is not the same. The way I define it, however, is the most common. – Git Gud Apr 10 '14 at 15:16
  • @GitGud [citation needed]. – Sean Clark Apr 10 '14 at 15:17
  • @Doop I'm not claiming they are the same, I'm just saying it is a weaker way to define a notion of even or odd. – Sean Clark Apr 10 '14 at 15:19
  • @Git Gud I agree with your definition, my comment was directed at the other commenter. – Jonathan Hebert Apr 10 '14 at 15:19
  • @user1306 Alright, I follow you. But we have to pick one or the other to answer the OPs question. – Jonathan Hebert Apr 10 '14 at 15:20
  • @user1306 For what it's worth, and I'm not claiming this implies that the definition I propose is better, thinking about real functions, the symmetry over the origin would require to use the definition I used. In other contexts it might be indifferent or maybe yours works out better. – Git Gud Apr 10 '14 at 15:22
  • So that means according to GitGud's answer, the function that I mentioned is neither even nor odd and according to @user1306's "weaker" definition, the function is even. I hope I am getting it right. – Indrayudh Roy Apr 10 '14 at 15:28
  • @IndrayudhRoy That is correct. – Git Gud Apr 10 '14 at 15:32
  • Consider √x. Is this not both even and odd, vacuously, according to user1306 definition? This function is not symmetrical over either the origin or y-axis, and therefore I think our definition should imply that this function is neither even or odd. – Jonathan Hebert Apr 10 '14 at 15:34
  • @Doop You're right. Different definitions might yield different properties. – Git Gud Apr 10 '14 at 15:36
  • @Git Gud yes, and I assert that the better definition is the one that agrees with our notion of symmetry, because that is what we are trying to describe. – Jonathan Hebert Apr 10 '14 at 15:37
  • @Doop Yes, your example did convince me about that. – Git Gud Apr 10 '14 at 15:38
  • @Doop, in reply to your comment on the example $f(x)=\sqrt {x}$, if we consider the domain of $f$ to be $\Bbb R^{+}$, then things become clear. But if we allow the domain to be $\Bbb R$ and codomain $\Bbb C$, then things become different. So, we must also mention the range of $f$ while defining odd and even functions. Neither of the two definitions mention that. – Indrayudh Roy Apr 10 '14 at 15:47
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    @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. – Git Gud Apr 10 '14 at 15:58