Confusion regarding the definition of addition modulo $n$ on the group $Z_n$.

abstract-algebrabinary operationsgroup-theorymodular arithmetic

Consider the group $Z_n$$= \{[0],[1],[2],…,[n-1]\}$. Where the elements of $Z_n$ are equivalence classes, and $Z_n$ is a partition of the integers. So, to me I understand it as: If $[r]$ belongs to $Z_n$, then $[r]$ contains all the integers that have a reminder -when divided by n- of $r\pmod n$. Notice the reminder is not just r, it's $r\pmod n$, to guarantee that the notation $ [r] $ still make sense when $r>n$. Then the definition of addition modulo n is: $[a]+[b] = [a+b]$. Which means that if you add the set with the elements of reminder $a\pmod n$ with the set of the elements with the reminder $b\pmod n$ you get the set of the elements with reminder of $(a+b)\pmod n$, which make sense.

But the problem is that I see books define $[r]$ to mean: the set of numbers with reminder $r$. And then they define $[a]+[b]=[a+b]$. But the problem is $a+b$ could be bigger than $n$, so under this definition of $[r]$, $[a+b]$ means the set of numbers with reminder $a+b$, which doesn't make sense when $a+b>n$. So which is the correct definition of the operation addition modulo n in the context of abstract algebra ?

Best Answer

Well, both are correct. If $a,b$ are integers, the addition mod $n$ is defined as $$[a]+[b] = [a+b]$$ where $[a+b]$ is the congruence or residue class mod $n$. So if $a+b$ is not a remainder mod $n$, a number from $0$ to $n-1$, then divide $a+b$ by $n$ with remainder: $a+b =qn+r$, where $0\leq r<n$, and so $$[a+b]=[r].$$

There are two basic facts to use:

(1) Each integer $a$ lies in the same residue class as its remainder $r$ mod $n$, i.e., if $a=qn+r$ with $0\leq r<n$, then $[a]=[r]$.

(2) For distinct remainders $r,r'$ mod $n$, the residue classes are distinct, i.e., $[r]\ne[r']$.