Is modular arithmetic defined for all rational numbers (with denominators coprime to modulus)?

In the expression 1b(modm), where (b,m)=1, is 1b:

a) a rational number (and so rational numbers are defined in modulo arithmetic using multiplicative inverses)?

b) just created notation for (the inverse of b)(modm) that looks like division just to confuse us (and is used because of similarities between division and division (modm))? (same for b1)

Is it a) or b)?

Here it says it is a).

Bill Dubuque from M.SE seemingly claims it is b). So does a comment here, also this blog.


Edit: now that I thought about it, either

1) the notation ab

2) or the definition ab(modm)mab

is misleading. 1) seems a lot more likely, since I’m not sure how the mod function could be defined otherwise.

Answer

I’m not sure if there is a general consensus, but I would like to add another interpretation different from the ones already mentioned.

I tend to say it is

  • not a number, since we want it to be ‘an inverse of b modulo m‘, without specifying which inverse. I mean, why should an inverse of b modulo m be defined as \frac1b and not \frac{1+m}b or \frac1{b+m}?
  • not a congruence class modulo m. This is because in high school the relation ‘\equiv\pmod m‘ is usually introduced without even mentioning ring theory, we simply define it as some equivalence relation on integers: a\equiv b\pmod m and not (a+m\mathbb{Z})\equiv(b+m\mathbb{Z})\pmod m.

If not a number nor a congruence class, then what should it be?

I’d say it isn’t a thing, it’s just a notation used in congruences representing an unspecified inverse of b modulo m. I.e, \frac ab\equiv c\pmod m is a notation meaning as much as a\equiv bc\pmod m (provided \gcd(b,m)=1), but \frac ab has no meaning to me when pulled out the congruence.
Note that this point of view still conserves the expected properties suggested by the rational numbers, such as b\cdot\frac ab\equiv a\pmod m and \frac ab+\frac cd\equiv\frac{ad+bc}{bd}\pmod m (provided the inverses exist).

Attribution
Source : Link , Question Author : user26486 , Answer Author : Bart Michels

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