First thing, the set of non integers: $\mathbb{R-Z=L}$
So could we have a parity for the numbers in set $\mathbb{L}$?
I thought of it and here is my work:
Let $c$ be in $\mathbb{L}$, then $c$ is even if $\lfloor c\rfloor=2k$ where $k$ is in $\mathbb{N}$ and $c$ is odd if $\lfloor c\rfloor=2k+1$ where $k\in\mathbb{N}$.
Is this correct?
1 Answer
Parity arguments will not work in rings like $\,\Bbb R\,$ where $\,2\,$ is invertible. Indeed suppose that $\,2y=1.\,$ Then we deduce $\ x = (2y)x = 2(yx),\,$ so every element is even.
However, parity arguments will work in any ring $R$ which has a modular image that is the same as the integers modulo $2\,$ (i.e. $\,R/I \cong \Bbb Z/2).$ This is true for many familiar rings, e.g. the subring of rationals with odd denominator, various rings of algebraic integers, e.g. the Gaussian integers $\,m + n\, i\,$ for $\, m,n\in\Bbb Z.\,$ See this answer for further discussion and links (which also shows how to define parity to certain types of rings with infinite elements).
