Why is congruence modulo n defined for n > 1?

Question

Many literature gives the definition of congruence modulo n for n > 1.

Is there a serious reason to avoid n = 1? Otherwise, do we get rid of the case just because the congruence modulo 1 is too trivial?

EDIT: My apologies, "Many literature" would be wrong. I've investigated several textbooks and found none of those restricts modulus as n > 1. Only Wikipedia and articles that may refer to Wikipedia give the restriction.

In Gauss's Disquisitiones Arithmeticae, the origin of the congruence relation, modulus are numbers with no restriction. A few textbooks remark modulo 1 is trivial, and so we usually assume n > 1 in practice, but it is just an assumption and not a definition.

mod 1 for integers is trivial, but for reals is quite useful: it's the fractional part. — lhf, May 26 '22 at 01:35
On integers, mod $0$ is equality and mod $1$ everything is congruent. Mod $-n$ is the same as mod $n$. — Arturo Magidin, May 26 '22 at 01:53
Though if you ask many programming languages, $\bmod 0$ is undefined and not identity: Why is $n\bmod 0$ undefined? — peterwhy, May 26 '22 at 02:08
@peterwhy In Arturo's claim that "$!!\bmod 0:!$ is equality" he means they are the same equivalence relation, i.e. he refers to $!\bmod!$ as a relation, not an operator. But you refer to the binary operation $!\bmod!$. This answer explains how the two are related. — Bill Dubuque, May 26 '22 at 10:04
Note: the proposed dupe is wrong. This question concerns $!\bmod 1$ but the dupe concerns $!\bmod 0,,$ which are very different congruences. Please read more carefully when proposing dupes. — Bill Dubuque, Jun 08 '22 at 19:14

score 1 · Answer 1 · edited May 26 '22 at 02:24

1

Let $m, n \in \mathbb{Z}$. The remainder of $m/1$ is 0, and the remainder of $n/1$ is also 0. Since $0=0$, then $m ≡ n \pmod 1$ . It's a perfectly cromulent operation, just not used much because it's trivial.

Though, as lhf pointed out in their comment, $\bmod 1$ is more interesting on $\mathbb{R}$ than on $\mathbb{Z}$.

edited May 26 '22 at 02:24

Arturo Magidin

398,050

answered May 26 '22 at 02:05

Dan

14,978

1

"mod" is not an operation here, it is a ternary relation. – Arturo Magidin May 26 '22 at 02:23

Bill Dubuque · Accepted Answer · 2022-05-26T09:49:20.847

Not only is there no serious reason to avoid modulus $\,n = 1\,$ but it may prove cumbersome to do so. In many modular arguments it is natural to work modulo an unknown modulus $\,n,\,$ including the possibility $\,n = 1$.

For example, a nice technique to prove that $\,d = \gcd(a,b)= 1\,$ is to use equational reasoning $\!\bmod d\,$ to prove $\,a\equiv 0\equiv b\,\Rightarrow\, 1\equiv 0\Rightarrow |d| = 1.\,$ This allows us to replace less intuitive manipulation of divisibility statements by more intuitive manipulation of equations (congruences) - allowing us to employ equational logic familiar since grade school. For a simple example see this answer posted a few days ago, which essentially exploits uniqueness of inverses (to prove $r\equiv -b),\,$ an argument that would be a bit harder to discover while thinking in divisibilty language. Compare also the conciseness of that two-line proof to the other answers using alternative arguments (one of many examples one could give illustrating the "cumbersome" claim above).

Please use due diligence in searching for dupes, before answering. — amWhy, Jun 08 '22 at 18:23
@amWhy The proposed dupe is incorrect. This question is about modulus $n = 1$ not $n = 0.,$ If you are aware of a correct dupe then post it. — Bill Dubuque, Jun 08 '22 at 18:29

Why is congruence modulo n defined for n > 1?

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