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.

Answer 1

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 $f(x)=0.$

Answer 2

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:

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

Answer 3

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 $f$ ?

Answer 4

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

Answer 5

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

Answer 6

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.