[Congruence class] Proof of modular arithmetic theorem

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SUMMARY

The discussion centers on the proof of the modular arithmetic theorem, specifically the equivalence ##[a][x_0]=[c] \iff ax_0 \equiv c \, (mod \, m)##. Participants clarify that this relationship stems from the definition of congruence classes, where ##[x] = [y]## if ##x \equiv y \ (\mathrm{mod \ m})##. The proof utilizes the properties of congruence, including transitivity and symmetry, to establish the validity of the theorem. This understanding solidifies the foundational concepts of modular arithmetic.

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  • Understanding of congruence classes in modular arithmetic
  • Familiarity with the properties of equivalence relations
  • Basic knowledge of modular arithmetic operations
  • Ability to manipulate mathematical expressions involving modular equations
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  • Explore the concept of transitivity and symmetry in mathematical proofs
  • Learn about applications of modular arithmetic in cryptography
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Leo Liu
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Could someone explain why ##[a][x_0]=[c]\iff ax_0\equiv c\, (mod\, m)##?
My instructor said it came from the definition of congruence class. But I am not convinced.
 
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Recall that ##[x] = [y]## if ##x \equiv y \ (\mathrm{mod \ m})##. Then it's just a case of using ##[a][x_0] = [ax_0]##.
 
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ergospherical said:
Recall that ##[x] = [y]## if ##x \equiv y \ (\mathrm{mod \ m})##. Then it's just a case of using ##[a][x_0] = [ax_0]##.
Wonderful. I didn't know this form of definition but after seeing its justification (transitivity and symmetry) I am convinced.
 
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