Abstract Alg. Proof w/ Mod Congruence and Relative Primes

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SUMMARY

The discussion centers on proving that if \( a \equiv b \,(\text{mod } n) \), \( c \equiv d \,(\text{mod } n) \), and \( \text{gcd}(a, c) = 1 \), then \( a \equiv c \,(\text{mod } n) \). Participants reference the properties of modular arithmetic and the relationship between congruences and greatest common divisors. The proof involves manipulating congruences and leveraging the definition of relatively prime integers to establish the desired result.

PREREQUISITES
  • Understanding of modular arithmetic and congruences
  • Knowledge of the properties of the greatest common divisor (gcd)
  • Familiarity with algebraic manipulation of equations
  • Basic proof techniques in number theory
NEXT STEPS
  • Study the properties of modular arithmetic in depth
  • Learn about the Chinese Remainder Theorem and its applications
  • Explore proofs involving gcd and their implications in number theory
  • Practice solving problems related to congruences and relative primality
USEFUL FOR

Students and educators in mathematics, particularly those focusing on number theory, modular arithmetic, and proof techniques. This discussion is beneficial for anyone looking to strengthen their understanding of congruences and gcd properties.

Colleen G
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Homework Statement


If ≡(mod), ≡(mod),and gcd(,)=1,provethat ≡ (mod ).

Homework Equations


If ≡(mod)→n|ab-cd
≡(mod)→n|b-d
gcd(,)=1→ relatively prime. So bx+ny=1

Need to show n|a-c→a-c=nw

The Attempt at a Solution


If n|ab-cd, then nk=ab-cd
If n|b-d, then nl=b-d
If n|ab-cd AND n|b-d, then n|p(ab-cd)+q(b-d). So pab-pcd+qb-qd.
pad-pcd+qb-qd
=pab+qb-pcd-qd
=b(pa+q) +d(-pc-q)

I'm stuck! Don't know if this is going anywhere.
 
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Colleen G said:
If ≡(mod), ≡(mod),and gcd(,)=1,provethat ≡ (mod ).
You need to edit your post so the variables show up. ( https://www.physicsforums.com/help/latexhelp/ )
Do you mean something like ab \equiv cd \ (mod \ n) ?
 

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