Proving the Chinese Remainder Theorem for Reduced Residues Modulo mn

  • Context: Graduate 
  • Thread starter Thread starter frowdow
  • Start date Start date
  • Tags Tags
    Proof
Click For Summary
SUMMARY

The discussion focuses on proving that the expression nx + my cycles through the reduced residue set modulo mn, where n and m are coprime integers. The user references George Andrews' book on number theory and outlines two key proofs: that nx1 + my1 = nx2 + my2 and that nx + my is relatively prime to mn. The user seeks guidance on demonstrating that nx + my encompasses all elements of the reduced residue set modulo mn without relying on the inequality \phi(xy) ≠ \phi(x)\phi(y).

PREREQUISITES
  • Understanding of the Chinese Remainder Theorem
  • Familiarity with reduced residue systems
  • Knowledge of the Euclidean algorithm
  • Basic concepts of number theory, particularly regarding coprime integers
NEXT STEPS
  • Study the Chinese Remainder Theorem in detail
  • Learn about reduced residue systems and their properties
  • Explore the Euclidean algorithm and its applications in number theory
  • Investigate the function \phi(n) and its significance in modular arithmetic
USEFUL FOR

This discussion is beneficial for students and enthusiasts of number theory, particularly those studying modular arithmetic, the Chinese Remainder Theorem, and related mathematical proofs.

frowdow
Messages
3
Reaction score
0
I am teaching myself number theory using George Andrews book. i am stuck in the following problem:

To prove that as x cycles thru' the reduced residue set modulo m & y cycles thru' reduced residue set modulo n, nx + my cycles thru' reduced residue set modulo mn.

I am able to prove that:
a) nx1 + my1 [STRIKE]=[/STRIKE] nx2 + my2 and
b) nx + my is relatively prime to mn

But how do i prove that nx + my cycles thru' every one of the reduced residue set modulo mn without first proving that [tex]\phi[/tex][tex]\left(xy\right)\neq[/tex][tex]\phi\left(x\right)\phi\left(y\right)[/tex]
 
Physics news on Phys.org
This description is a bit weird. Look up the chinese remainder theorem and the Euclidean algorithm. I guess that ##n,m## are coprime.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
6K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
10K
  • · Replies 3 ·
Replies
3
Views
4K