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When I was wondering about if Constant Functions were even or odd, I thought about the function:

f(x) = 0 , It's simultaneously odd and even,

f(3) = 0 , f(-3) = 0, f(1) = 0 , -f(1) = 0,

(It's Indeed a counter-example) I'm pretty sure I haven't solved the profound paradoxes of math. Is what I have concluded right?, Has anyone heard of it?

P.S ( I was taught it can't be both, so YEAH)

MohaGain
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2 Answers2

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As Clive Newstead says here:

Unlike integers, "not even" does not mean the same thing as "odd". You'll find situations like this all over the place: always work from the definitions, don't let your knowledge of the English language deceive you.

His advice goes for all mathematical concepts.

From the very definitions, there's no reason why a function can't be both odd and even. Recall that a function $f\colon\Bbb R\to \Bbb R$ is:

  1. Odd if, and only if, $\forall x\in \Bbb R(f(-x)=-f(x))$;
  2. Even if, and only if, $\forall x\in \Bbb R(f(-x)=f(x))$.

'Odd' and 'even' are just words to represent the above definitions. The meaning it has in the English language (or even in other mathematical concepts) has no bearing here.

You should also note that the concept of odd and even functions is only defined for symmetric domains, that is, domains such that if $x$ is the domain, so must $-x$ be too. See this link for further clarification.

Community
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Git Gud
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If a function $f\colon\mathbb{R}\to\mathbb{R}$ (the domain could also be a symmetric interval) is odd and even, you have, for all $x$ in the domain,

$$ f(x)=f(-x)=-f(-x) $$

that is,

$$f(x)=-f(x).$$

Can you draw a conclusion from this?

egreg
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