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All functions $f(x):\mathbb{R}\to\mathbb{R}$ can be decomposed into an even and an odd part $f(x)=E(x)+O(x)$.

The proof I see here, and on Wikipedia requires $2$ to have an inverse, however I want to know if this property is true of functions in a ring.

For a ring $R$ for every function $f(x):R\to R$, can I find even and odd functions $E(x)$ and $O(x)$ such that $f(x)=E(x)+O(x)$?

Using a similar method to the ordinary proof I can find that $2E(x)$ and $2O(x)$ are defined uniquely by $f$, but this only shows that $2f$ has a decomposition into an even and an odd part. Because $2$ does not necessarily have an inverse, I'm not certain how to solve this.

I think that a decomposition seems possible, but it would not necessarily be unique?

Numeral
  • 1,344
  • The characteristic $2$ problem is real. Saying that a function is Even over a field of characteristic $2$ says nothing. We always have $F(x)=F(x)$, after all. So, sure. You have a decomposition of the form you want, but it has no content. – lulu Mar 12 '23 at 20:39
  • @lulu I hadn't realised all functions are even over a field of characteristic 2. But if the ring is not of characteristic 2 does an even+odd decomposition exist? – Numeral Mar 12 '23 at 20:45
  • Well, if the field is not characteristic $2$ then you can divide by $2$, so the usual argument works. – lulu Mar 12 '23 at 20:46
  • I'm only thinking of fields, I should stress. Not sure what happens in a more general ring. – lulu Mar 12 '23 at 20:47

1 Answers1

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It is not always possible to find such a decomposition.

If $f(x)=E(x)+O(x)$ where $E$ is even and $O$ is odd, then $f(x)+f(-x)=2E(x)$ is always divisible by $2$. So for instance if $R=\mathbb{Z}$ then this is not possible for any function with $f(1)=0$ and $f(-1)=1$.

On the other hand, if $2=0$ in $R$, then every function is both even and odd, so such decompositions are not unique (assuming $R$ has more than one element).

Numeral
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Eric Wofsey
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