1
$\begingroup$

In various different places, I've seen the notation $$a \equiv b \equiv c \pmod n,$$ with the intended meaning obviously being $$a \equiv b \pmod n \quad\land\quad b \equiv c \pmod n \quad\Longrightarrow\quad a \equiv c \pmod n,$$ as in

$$x \equiv x - 1 + 1 \equiv 1 \pmod{x - 1}.$$

However, others do not share my views on this matter, saying it's ambiguous and notation abuse, among other things, with their primary problem being that multiple equivalence signs are corresponding to a single mod. As such, I am looking for a notable source, preferably a scientific paper, that uses (or discusses) this kind of notation to educate myself and/or the others in question on the matter.

$\endgroup$
8
  • 2
    $\begingroup$ I have never seen anyone object to such "chaining". $\endgroup$ Nov 28, 2018 at 22:47
  • 2
    $\begingroup$ Do they give any examples where it is ambiguous? I can't think of any, and certainly similar chaining of transitive relations is very common, as for example in $a=b<c<d≤e=f$, therefore $a<f$. $\endgroup$
    – MJD
    Nov 28, 2018 at 22:51
  • $\begingroup$ @MJD I don't exactly follow their logic, but it has something to with how there is just one $\pmod n$, and it is apparently not clear that it applies to all congruence relations. $\endgroup$
    – Maya
    Nov 28, 2018 at 22:54
  • 3
    $\begingroup$ If it's good enough for Hardy and Wright, surely, it should be good enough for anybody - and I find an instance of this kind of "chaining" on page 57 of their An Introduction to the Theory of Numbers (fourth edition 1960): $$(vn'^2+v'n^2)\bar{H} \equiv vn'\bar{h} + v'n\bar{h'} \equiv K \pmod{nn'}.$$ I also found it on page 36 of H. E. Rose, A Course in Number Theory (second edition 1994): $$x \equiv c_im_i'm_i'' \equiv x c_i \pmod{m_i}.$$ Those were the first (indeed only) books I looked at. Instances of this convention in respectable mathematical literature don't seem to be hard to find. $\endgroup$ Nov 28, 2018 at 23:58
  • 2
    $\begingroup$ From page 50 of the third book I looked at, H. Davenport's The Higher Arithmetic (eighth edition 2008): $$(ab)^{lk} \equiv (a^l)^k(b^k)^l \equiv 1 \pmod{p}.$$ Page 108 of the fourth book, Tom M. Apostol's Introduction to Analytic Number Theory (1976): $$23{,}716 \equiv 640 \equiv -1 \pmod{641}.$$ Unscientific conclusion: this convention is found everywhere. :) $\endgroup$ Nov 29, 2018 at 2:41

0

You must log in to answer this question.

Browse other questions tagged .