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.
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.
PREREQUISITESMathematics students, educators, and anyone interested in deepening their understanding of modular arithmetic and its foundational principles.
Wonderful. I didn't know this form of definition but after seeing its justification (transitivity and symmetry) I am convinced.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]##.