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I woke up this morning and had this question in mind. Just curious if such function can exist.

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In case anyone has forgotten what “even” and “odd” functions are, $f$ is even if $f(x) = f(-x)$ and odd if $-f(x) = f(-x)$. See also Wikipedia on even and odd functions. –  Rory O'Kane Jun 17 '12 at 18:54
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You might find it interesting that I often used to ask this as an extra credit question on precalculus tests when even/odd function properties were covered, typically worth an extra 3 points on a 100 point scale (so a score of 103/100 was possible). I'd usually get about 2 to 5 students getting the extra points (out of a total of maybe 25-35 students) in a U.S. college precalculus class, and about half the class getting the extra points in U.S. honors level high school classes I used to teach. –  Dave L. Renfro Jun 18 '12 at 15:56

6 Answers 6

up vote 38 down vote accepted

Others have mentioned that $f(x)=0$ is an example. In fact, we can prove that it is the only example of a function from $\mathbb{R}\to \mathbb{R}$ (i.e a function which takes in real values and outputs real values) that is both odd and even. Suppose $f(x)$ is any function which is both odd and even. Then $f(-x) = -f(x)$ by odd-ness, and $f(-x)=f(x)$ by even-ness. Thus $-f(x) = f(x)$, so $f(x)=0.$

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Of course, one could argue that restrictions of the constant $0$ function to different domains symmetric about the origin are different functions, set-theoretically speaking. –  Cameron Buie Jun 17 '12 at 15:30
    
@CameronBuie That is true, I will make my answer more precise to indicate this. Thank you. –  Ragib Zaman Jun 17 '12 at 15:31
    
Funny, I never thought of f(x) = 0 as a possibility. Thanks for the answers everyone! –  bodacydo Jun 17 '12 at 21:06

Suppose $f$ odd an even. Let $x \in D$ ( D is set definition of $f$) then you have : $ f(x)=f(-x)=-f(x)$. What can you conclude about $f$ ?

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As other people have mentioned already, the real function $f(x)$ which maps every real number to zero (i.e.$f(x) = 0 \space \forall x \in \mathbb{R}$) is both even and odd because $$f(x) - f(-x) = 0 \space \space , f(x)+f(-x) = 0\space \forall x \in \mathbb{R} .$$ Also it is the only function defined over $\mathbb{R}$ to possess this property.

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Hint $\rm\ f\:$ is even and odd $\rm\iff f(x) = f(-x) = -f(x)\:\Rightarrow\: 2\,f(x) = 0.\:$ This is true if $\rm\:f = 0,\:$ but may also have other solutions, e.g. $\rm\:f = n\:$ in $\rm\:\mathbb Z/2n =\:$ integers mod $\rm 2n,$ where $\rm\: -n \equiv n.$

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+1, but note that your last $\iff$ applies (in the backwards, i.e. 'if' direction) only to $f(x) = -f(x)$, and not to the part where $f(-x)$ equals both of them. –  ShreevatsaR Jun 17 '12 at 18:04
    
Yes, I meant to write $\:\Rightarrow\: $ but it was lost in editing. Now fixed. Thanks. –  Bill Dubuque Jun 17 '12 at 18:19

Yes. The constant function $f(x) = 0$ satisfies both conditions.

Even: $$ f(-x) = 0 = f(x) $$

Odd: $$ f(-x) = 0 = -f(x) $$

Furthermore, it's the only real function that satisfies both conditions:

$$ f(-x) = f(x) = -f(x) \Rightarrow 2f(x) = 0 \Rightarrow f(x) = 0 $$

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If $K$ is a field of characteristic 2, every function $K\to K$ is both even and odd.

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i'm sorry, wouldn't that be "unequal to 2"? –  akkkk Jun 17 '12 at 15:31
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@Auke: No. I won't spoil the joke by spelling it out, sorry. –  Harald Hanche-Olsen Jun 17 '12 at 15:40
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Actually, you don't even need a field, any ring of characteristic 2 will do. –  Ilmari Karonen Jun 17 '12 at 15:42
    
@HaraldHanche-Olsen, oh, I am sorry, I misread your answer, you are completely right :) nice one –  akkkk Jun 17 '12 at 15:43
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This is a wonderful answer! –  Edward Hughes Jun 17 '12 at 23:52

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