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Say $f(x)$ is defined as $f(x) \equiv x+1\pmod m$. My notes say that $f(x)$ has domain $\{0,\ldots, m-1\}$ based off of this definition, and that all of its outputs are mapped here too. However, I am not seeing how $\{0,\ldots, m-1\}$ is the domain based off of just the above definition of $f$, and am altogether unsure of how to inteprret its definition. I would assume that its domain is all real numbers, (or some other universal quantification over a large set like this) and given a number $x$, it outputs a number such that it is congruent mod $m$ to $x+1$.

Also, I am not seeing how its output must be in $\{0,\ldots, m-1\}$; it seems its output is merely defined to be congruent to numbers in that set in that it differs by a multiple of $m$ to any numbers in that set, but could be any real number. How should I interpret this definition, and how is it the case that $f(x)$ has both domain and codomain $\{0,\ldots, m-1\}$?

EDIT: my course notes go on to say that "the unique pre-image of $y$ is $y-1$", with reference to $f$. How could this be the case when $y$ could be $0$ and $x$ could be $m-1$?

Princess Mia
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  • The domain and codomain are both that set, and $(m-1) + 1 \equiv 0 \pmod{m}$ – Brevan Ellefsen Aug 21 '23 at 05:42
  • @BrevanEllefsen sorry, I am new to modular arithmetic and would appreciate an explanation as to the relevance of $(m-1) +1 \equiv 0 \pmod m$ towards determining the domain and codomain for $f(x)$ – Princess Mia Aug 21 '23 at 05:45
  • you dont determine the domain and codomain of a function, they are part of its definition. In this case you are told the domain of $f$ is ${0, \ldots, m-1}$ and the image (and thus an implicit choice of codomain) is obviously seen to be ${0, \ldots, m-1}$ by calculation so doesn't need to be specified. – Brevan Ellefsen Aug 21 '23 at 05:48
  • @BrevanEllefsen I was never told the domain explicitly of this function; all I was given is $f(x) \equiv x+1\pmod m$. From this and this alone, how do we deduce that its domain is ${0, \ldots, m-1}$? – Princess Mia Aug 21 '23 at 06:00
  • if you were never given the domain (and target), then you can’t define the function, and whoever did the lecturing/notes writing or whatever is making a very huge pedagogical mistake. Maybe they thought context was sufficient to fill in the blanks, but they shouldn’t expect the readers to read their mind (especially if this is an introductory set of lectures/notes); it really is quite disappointing. – peek-a-boo Aug 21 '23 at 06:52
  • The correct definition for the increment operation on $,\Bbb Z_m,$ is $,f(x) := (x+1)\bmod m.\ $ Your critiques no longer apply to this correct definition. – Bill Dubuque Aug 21 '23 at 06:57
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    The author might intend this: since $f$ is an operation on $\Bbb Z_n = {0,1,\ldots,m-1}$ it takes values in $\Bbb Z_m$ so the definition means that the value of $f(x)$ is the (unique) integer $\in \Bbb Z_m,$ that is $\equiv x+1\pmod{!m}.,$ Alternatively, the author might intend $,f(x)\equiv\ ... $ to denote a definition, not a congruence. If so, they have confused notation for $!\bmod!$ the congruence relation vs. remainder operation (cf. here). Fixing both yields the correct definition above. – Bill Dubuque Aug 21 '23 at 07:43
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    Re: your edit: the author likely means the subtraction operation on $,\Bbb Z_m,,$ i.e. $,y-1\bmod m.,$ Again, the exposition is very sloppy and likely to cause much confusion for those new to the subject. – Bill Dubuque Aug 21 '23 at 18:36

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