I [Congruence class] Proof of modular arithmetic theorem

AI Thread Summary
The discussion centers on proving the equivalence between the congruence classes and modular arithmetic, specifically the statement that [a][x_0] = [c] if and only if ax_0 ≡ c (mod m). Participants clarify that this stems from the definition of congruence classes, where [x] = [y] if x ≡ y (mod m). The proof involves demonstrating that [a][x_0] simplifies to [ax_0]. The justification of this relationship relies on the properties of transitivity and symmetry in congruence. Overall, the explanation successfully convinces participants of the theorem's validity.
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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