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On the Wikipedia page for Modular Arithmetic, the Basic Properties section contains descriptions for the properties listed, except the following one:

k a ≡ k b (mod k m) for any integer k

What would be a good description for that entry?

Dess
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  • The property is stated incompletely. It says, if we have $a\equiv b\bmod m$, then $ka \equiv kb\bmod km$ for every integer. So it is just multiplication by $k$. – Dietrich Burde Jun 07 '25 at 15:45
  • That's the only one where they change from one modulus to another. The others are about about compatibility within one modulus. I'd describe it scales to other moduli. – fleablood Jun 07 '25 at 17:08

1 Answers1

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The other names ("compatibility with xyz operation") don't apply here since this scaling (which scales the modulus too) is not an operation on $\Bbb Z_{\large m}$ (since the codomain is $\,\Bbb Z_{\large \color{#c00}k\:\!m})$. There is no standard name that distinguishes this general form of the congruence scaling law from the operational special case. Both are called "scaling a congruence". Note that in the operation case scaling compatibility is a special case of multiplication compatibility. $\ \ $

Generally for any algebraic structure defined by purely equational (vs. relational) axioms, the notion of ring-congruence generalizes in a straightforward way to that of an equivalence relation that is compatible with all operations of the structure, i.e. $\, A\equiv a,\ B\equiv b\,\Rightarrow\, A\oplus B = a\oplus b\,$ for all binary operations $\,\oplus\,$ of the algebra, and similarly for all other $\,n$-ary operations of the structure. Such compatibility means all the structure's operations are well-defined on the quotient set, which yields a quotient algebra - which reifies the (modular) congruence arithmetic within an algebraic structure of the same type, just like even/odd parity arithmetic of integers mod $\,2\,$ is ring-theoretically reified as arithmetic in the quotient/residue ring $\,\Bbb Z/2.$

Bill Dubuque
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  • When you say "operational special case", you mean the following? $k a \equiv k b (mod; m)$ – Dess Jun 07 '25 at 16:52
  • @Dess $\ $ Yes, when the scaling codomain is also $\Bbb Z_m$ then it is a scaling operation on $\Bbb Z_m$ (by definition an ($n$-ary) operation on a set $S$ is a map $,S^n\to S).,$ Yes, that equivalence is true, being equivalent to $,km\mid k(a!-!b)!\iff! m\mid a!-!b,,$ for $,k\neq 0.,$ Generally see here for congruence scaling and cancellation. $\ \ $ – Bill Dubuque Jun 07 '25 at 17:04