Hi I've read the proof here.


On the line about division it says

$$k a = p \times \left \lfloor {\dfrac {k a} p} \right \rfloor + r$$ where $r\in S′$.

Can someone explain why $r\in S′$?


  • $\begingroup$ Because of the division theorem. $\endgroup$ – Dietrich Burde Feb 27 '18 at 19:52
  • $\begingroup$ Sorry I'm still confused. Division theorem implies that r is unique why is r necessarily in S'? $\endgroup$ – Jake Baker Feb 27 '18 at 20:03
  • $\begingroup$ The division algorithm also says that the remainder $r$ is "smaller", which means here that it is in $S'$. Because it is modulo $p$, the division algorithm gives that we can take $r$ in the set of least positive residues. $\endgroup$ – Dietrich Burde Feb 27 '18 at 20:03
  • $\begingroup$ Thanks! That cleared things up. $\endgroup$ – Jake Baker Feb 28 '18 at 10:36

This is due to the division algorithm as shown in the comments.


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