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Old 2007-03-13, 00:02   #7
m_f_h's Avatar
Feb 2007

6608 Posts

Originally Posted by R.D. Silverman View Post
Generally, in the context in which it arises within this forum, 'mod' is NOT an operator. It is an equivalence class.
In theory this is true, but as e.g. seen in ewmayer's previous post, mostly it is in fact used as an operator (operating on both sides of the equation). I'd estimate to > 95% the percentage of occurances of "a = b (mod c)" where the r.h.s. b is a number less than c and one can resonably assume that the author thinks of the l.h.s. as "a mod c".

Also, "mod" is not an equivalence class, at best " = (mod c)" is an equivalence relation
(or "=" means "element of" and "b (mod c)" is then indeed an equivalence class, so "mod" is the operator that associates to the first operand (b) the equivalence class modulo cZ, i.e. the multiples of the second operand (c)).

[Sorry for the unqualified post, just to tease Prof. Dr. R.D.S. ]
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