Can there be a function that's even and odd at the same time?

Question

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

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
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

Ragib Zaman · Accepted Answer · 2012-06-17T15:32:24.967

70

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 $f(x)=0.$

edited Jun 17 '12 at 15:32

answered Jun 17 '12 at 15:25

Ragib Zaman

36,065

5

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

score 36 · Answer 2 · answered Jun 17 '12 at 15:29

36

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

answered Jun 17 '12 at 15:29

Harald Hanche-Olsen

32,608
2
61
87

i'm sorry, wouldn't that be "unequal to 2"? – akkkk Jun 17 '12 at 15:31
2

@Auke: No. I won't spoil the joke by spelling it out, sorry. – Harald Hanche-Olsen Jun 17 '12 at 15:40
11

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
1

This is a wonderful answer! – Edward Hughes Jun 17 '12 at 23:52
+1 That's a wonderful answer with which to start my day. Thanks you! – Shalom Craimer Jun 18 '12 at 05:39
I don't get it :( – Agos Jun 18 '12 at 07:19
1

@Agos: Have a look at this wikipedia page, and try to work out the consequences of the characteristic being 2. – Harald Hanche-Olsen Jun 18 '12 at 11:33
@HaraldHanche-Olsen thanks. I lol'd. – Agos Jun 18 '12 at 13:37
If $K={0,1}$ then how is the function $f(x) = x + 1$ both even and odd? – Mateen Ulhaq Oct 28 '16 at 10:58
1

@MateenUlhaq $f(-1)=f(1)=-f(1)=0$ and $f(-0)=f(0)=-f(0)=1$. Perhaps what surprises you is that in characteristic 2, and odd function need not satisfy $f(0)=0$. – Harald Hanche-Olsen Oct 28 '16 at 23:17

Ayman Hourieh · Answer 3 · 2012-06-17T15:34:51.207

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:

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

edited Jun 17 '12 at 15:34

answered Jun 17 '12 at 15:24

Ayman Hourieh

40,828

score 10 · Answer 4 · edited Jun 17 '12 at 18:19

10

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 $\rm\: -n \equiv n.$

edited Jun 17 '12 at 18:19

answered Jun 17 '12 at 15:50

Bill Dubuque

282,220

+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

Mohamed · Answer 5 · 2013-11-07T01:26:38.110

4

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 $f$ ?

edited Nov 07 '13 at 01:26

answered Jun 17 '12 at 15:25

Mohamed

3,771

wamiq reyaz · Answer 6 · 2013-01-28T17:09:23.587

4

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.

edited Jan 28 '13 at 17:09

answered Jan 28 '13 at 17:04

wamiq reyaz

195

score 2 · Answer 7 · answered Sep 09 '15 at 06:23

2

Let $R$ be a Boolean ring and $X$ be an arbitrary set. Then every function $f:R\rightarrow X$ is both even and odd.

answered Sep 09 '15 at 06:23

Say what........? – bodacydo Sep 09 '15 at 07:06
@bodacydo for example: any polynomial $P(x)\in R[x]$. – Sep 09 '15 at 07:31
Thanks that is useful to know! – bodacydo Sep 09 '15 at 20:12

score 2 · Answer 8 · edited Jun 12 '20 at 10:38

I post this as an addendum to the awesome answers already present for this Q.

Most trivial example:

For $f(x)=0$ we have:

$f(x)= 0 =-f(-x)$ Hence, Odd

$f(x)= 0 =f(-x)$ Hence, Even

I was searching for an answer to my question.

$f(x) = \sin(x) + \sin(\pi + x)$

Here, the function is also even and odd at the same time (another example or representation of the same). This is due to the value resulting in zero over the entire domain. So, it can be concluded that all the functions that have their $Range=\{0\}$ should be both even and odd at the same time despite their notation is of a constant function($f(x)=0$) or not.

Another example,

$f:\{1,-1\}\to \mathbb{R}$

$f(x) = x^{2}-1$

$f(x) = 0$ for all values in the domain. So, it is both even and odd at the same time because while deciding even odd functions the domain to which the function is restricted by definition must be considered.

score 1 · Answer 9 · answered Sep 02 '18 at 19:21

We could also think even and odd functions as the following to get $f(x) = 0$ is a function that is both odd and even and go for some different functions:

Odd functions have graphs that are symmetric with respect to origin.

Even functions have graphs that are symmetric with respect to $y$-axis.

So, graph of $f(x) = 0$ satisfies both conditions, therefore it is both odd and even. Furthermore, we can define some piecewise functions that satisfies this condition using their graphs. For example, let $f(x)$ not defined on intervals $(-1,-2)$ and $(1,2)$; and whenever it is defined, $f(x) = 0$. Then $f(x)$ is, again, both even and odd. And as long as we define a piecewise function by removing intervals that are symmetric with respect to $y$-axis, and rest of the function $= 0$ even though they are not continuous.

Can there be a function that's even and odd at the same time?

9 Answers9

Most trivial example:

I was searching for an answer to my question.

Linked

Related