I want to give a (simple) map $f:\mathbb{Z}_k \to \{d,d+1,\ldots,d+k-1\}$ for $d,k \in \mathbb{N}$ and such that $\forall i \in \mathbb{Z}_k. f(i) \bmod k = i$.

Is there a particularly simple way of doing this?


The goal is to formally prove that if $f$ is periodic with period $k$ then $\sum_{m}^{m+k-1} f \; l = \sum_{m+d}^{m+d+k-1} f \; l$. The above is the particular case $m = 0$.

Current solution (suggested by @kingW3)

$f(i) = (i-d) \; mod \; k + d$ and if $g(j) = j \; mod \; k$ then $g \circ f = Id$.

I'm interested in showing $f \circ g = Id$. How can I do it?

  • $\begingroup$ Let $f(a+k\mathbb{Z}) = a \text{ mod } k$? $\endgroup$ – Yanko Apr 21 at 12:20
  • $\begingroup$ @Yanko there $f$ is not mapping onto $\{d,d+1,\ldots,d+k-1\}$ $\endgroup$ – Javier Apr 21 at 12:31
  • 2
    $\begingroup$ $f(i)=((i-d)\mod k)+d$ $\endgroup$ – kingW3 Apr 21 at 12:44
  • $\begingroup$ @kingW3 seems like working, the same should work for $f:\{m,\ldots,m+k-1\} \to \{m+d,\ldots,m+d+k-1\}$ with $m \in \mathbb{N}$ right? $\endgroup$ – Javier Apr 21 at 12:57

There's only one way to map $\Bbb{Z}_k$ onto a set of consecutive $k$ integer in such a way $$ f(\bar x)\equiv x\bmod k. $$ That's because in any given set $K$ of $k$ consecutive integers there's only one $y\in D$ satisfying the displayed congruence for any given $\bar x$. Thus, let $f(\bar x)=y$.


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