Do all odd functions have symmetric zeros?

Question

Is there odd functions where a zero of positive x-axis doesn't have necessarily a symmetric?

What I mean is, consider an odd function where its positive zeros are x=1, x=2, x=3. Since it is odd, it has a symmetry related to origin of referencial, so there are also the zeros x=-1, x=-2, x=-3. But is it possible to have an odd function with x=1,x=2,x=3, x=-1 and x=-2? Where x=3 doesn't have a symmetric.

Answer 1

It's correct that for odd functions, $f(a) = 0 \iff f(-a) = 0$. We call $f$ odd if $f(-x) = -f(x)$, so if $f(a) = 0$, then $f(-a) = -f(a) = 0$.

Answer 2

An odd function $f $ satisfies $f (-x)=-f (x) $, so if $3$ is a zero of $f$, $f (-3)=-f (3)=-0=0$. This works for any zero, so yes, the zeroes are symmetric.

Answer 3

If an odd function exists in $x=a$ and in $x=-a$ and if it has a zero in one of them, then necessarily it has one in its symmetric counterpart since $f(-a)=-f(a)$ by definition of odd.

The only way to make this go wrong, is if $f$ would be defined in only one of those points. This is something you may or may not (want to) allow for a function to be called odd.

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What I mean is, consider an odd function where its positive zeros are x=1, x=2, x=3. Since it is odd, it has a symmetry related to origin of referencial, so there are also the zeros x=-1, x=-2, x=-3. But is it possible to have an odd function with x=1,x=2,x=3, x=-1 and x=-2? Where x=3 doesn't have a symmetric.

With the usual understanding (definition) of odd, you cannot have the situation you describe. We call a function odd if $f(-x)=-f(x)$ "for all $x$", but this is only meaningful if for every $x$ in the domain of $f$, you also have $-x$ in that domain.

This means we're only considering functions defined on either $\mathbb{R}$ or with a domain that is symmetric with respect to the origin such as intervals of the form $[-a,a]$, with $a \in \mathbb{R}$.

Take for example the odd function which has (at least) the zeroes you describe: $$f : \mathbb{R} \to \mathbb{R} : x \mapsto (x+3)(x+2)(x+1)x(x-1)(x-2)(x-3)$$ You could limit the domain to an asymmetric interval, such as $\left[ -2.5, 3.5 \right]$: $$g : \left[ -2.5, 3.5 \right] \to \mathbb{R} : x \mapsto (x+3)(x+2)(x+1)x(x-1)(x-2)(x-3)$$ Now this function $g$ has a zero in $x=3$ without a symmetric counterpart, since it is undefined in $x=-3$. Whether or not you would still call this function odd depends on the definition: do you require the domain to be of the specific form as described above or not?