Most of us do modulo arithmetic more often than we think, especially if we look at an analog clock to tell time. For example, if I asked you to convert 14:00 from military time into “regular” time, you’d respond “2 pm” pretty quickly. But how did you do it?
You got that answer because you went all the way around the clock and then had 2 remaining. This is modulo addition — in this case $\bmod 12$.
Modulo addition is regular addition with an extra couple steps: dividing by your “base” $n$ (on a clock $n = 12$) and then grabbing the remainder. For integers $a$,$b$, and $n$,
\[a \equiv b \bmod n \Leftrightarrow b-a= k\cdot n \text{ for an integer } k \] This is read as "$a$ is equivalent to $b$ modulo $n$ if $b-a$ is an integer multiple of $n$". Another way to write this is \[a \equiv b \bmod n \Leftrightarrow b = k\cdot n + a \text{ for an integer } k\]In other words, if $b$ can be expressed as a multiple of $n$ plus some extra amount $a$, then $a$ is equivalent to $b$ modulo $n$.
Back to our clock example, obviously $2 \neq 14$, but $2\equiv 14 \bmod 12$ because
\[14 = 1\cdot 12 + 2\]Here, our $b$, which is 14, can be expressed as an integer multiple of $12$ ($k=1$), plus some extra $a=2$, so we say that $2\equiv 14 \bmod 12$.
The base $n$ matters. If I changed the $n$, equivalence may no longer hold. For instance, $2$ is not equivalent to $14 \bmod 5$ because the remainder when you divide 14 by 5 is 4. The study of these remainders gives us mathematical objects known as finite groups.If we look at dividing any number by 12, the remainder can only be the numbers 0 – 11. Why? Well if the remainder is 12, then it’s a multiple of 12. The remainder of any integer, no matter how big it is, when dividing by $n$ has to be somewhere in the set $\{0,1,2,\ldots, n-1\}{0,1,2,…,n−1}$.
Let’s look at a few with $n=12$. As an exercise, verify each of these on your own: \[\begin{align*}1\equiv 13 \bmod 12 &\text{ because } 13 = 1\cdot 12 + 1 \\ 0\equiv 24 \bmod 12 &\text{ because } 24 = 2\cdot 12 + 0\\ 11\equiv 35 \bmod 12 &\text{ because } 35 = 2\cdot 12 + 11\\ 0\equiv 48 \bmod 12 &\text{ because } 48 = 4\cdot 12 + 0 \\ 3\equiv 123 \bmod 12 &\text{ because } 123 = 10\cdot 12 + 3 \end{align*}\] The multiple $k$ is telling us how many times we’ve gone around the clock. So we’re always stuck on the clock, no matter how many times we go around, which means that the remainder can’t be any larger than 11, or we’ve started another cycle around the clock. The set $\{0,1,2,\ldots, n-1\}$forms an algebraic structure called a group when we pair the set with the operation modulo addition.A group is a set paired with an operation that has to have three certain properties. The operation doesn’t always have to be addition, or modulo addition, and the set doesn’t have to be integers. Operations can be addition, multiplication, function composition, or some weird one you define. But to pair an operation with a set, we have to check to make sure some rules are followed:
The operation has to fit the formal definition; we cannot use any old rule we want. An operation is defined mathematically as a rule that takes any two elements in a given set $A$ and returns a unique element that is still in $A$. This means that you can’t create a rule that lets you escape the set. If my set is integers, and my rule is division, then division is not an operation on the integers because the result of $3/4$ isn’t an integer. By trying to divide 3 by 4, we left the set of integers. The uniqueness requirement ensures our rule is well-defined. If my rule is to take two integers $a$ and $b$ and return the number whose square is the product $ab$, then $2\cdot 8 = 16$ could return either 4 or -4. That’s not well-defined. The rule doesn’t guarantee that a unique element is returned.
Assuming our operation really is an operation (and modulo addition on the integers is. If you don’t believe me, try to prove it for yourself.), then we can move on. We have the set (integers modulo $n$) and the operation (modulo addition). To meet the definition of a group, the set and operation must follow these three axioms:
We can examine groups via operation tables, which are a nice visual. Think of your multiplication tables from primary school. The elements are placed on the rows and columns, and the entries are the result of applying the operation to the two elements. The row entry goes first, then the column entry. For $\mathbb{Z}_{3}$, we know the elements are $\{0,1,2\}$, so the operation table looks like this:
\[\begin{array}{c|ccc} + & 0 & 1 & 2 \\ \hline 0 & 0 & 1 & 2 \\ 1 & 1 & 2 & 0 \\ 2 & 2 & 0 & 1 \end{array} \] Since all integers modulo 3 will leave a remainder of 0, 1, or 2, these are the only three numbers we need to consider. All integers will be equivalent to one of these three. So, reading the table, the first entry is row 1 + column 1 ($\bmod 3\!$), which is $$(0 + 0) \bmod 3 \equiv 0 \bmod 3 $$ Another one is the row 2 + column 3: $$(1 + 2) \bmod 3 = 3 \bmod 3 \equiv 0\bmod 3$$ Why? Well, after we add, we have to divide by 3 and take the remainder. As an exercise, make sure you can generate this table yourself. We know that modulo addition is an operation, so let's see if combining it with $\mathbb{Z}_{3}$ fits the definition of a group. This means we have to verify the three axioms, starting with the first: Axiom 1: Associativity: We need to show the following to satisfy the first axiom: $$ [(a+b) + c] \bmod 3 \equiv [a + (b+c)]\bmod 3.$$ First, let's take any three unspecified elements of $\mathbb{Z}_{3}$. Call the $a, b,$ and $c$. We don't specify these because we have to keep things general in order to show that it doesn't matter which three elements we pick.[note]I know it's possible to verify this by trying all combinations. The set is small enough to do this, but it's a bad habit to get into. To learn to prove, you must learn to do this in general and not rely on "brute force" crutches.[/note] We are going to start with $[(a+b) + c]\bmod 3$ and manipulate it to show that it's equivalent to $[a+(b+c)]\bmod 3$. To differentiate between regular old addition and modulo addition (modulo 3 in this case), I am going to use the notation $+_{3}$ when I mean addition modulo 3. First, by the definition of modulo arithmetic, $$[(a+b)+c]\bmod 3 = (a+b)\bmod 3 +_{3} c\bmod 3$$ The right hand side of the above is the same thing as using regular addition to add $a, b,$ and $c$ together first, then take that result modulo 3. [note]Try verifying this for yourself using the table above, just so you get comfortable with this. It's true regardless of what you are modding out by, but playing with a few examples on your own is the best way to get it to sink in.[/note] So we write that $$\begin{aligned}[(a+b)+c]\bmod 3 &= (a+b)\bmod 3 +_{3} c\bmod 3\\&= (a+b+c)\bmod 3\end{aligned}$$ Look inside the parentheses. Now we're back in the nice, safe world of regular addition, and we know we can group and add any way we want to inside those parentheses. So let's do that. We know that $$(a+b+c) = (a+b) + c = a + (b+c).$$ Grab that last equality.[note] Hint: because it looks an awful lot like what we're trying to get to, just with modular arithmetic[/note] Then we can write $$\begin{aligned}[(a+b)+c]\bmod 3 &= (a+b)\bmod 3 +_{3} c\bmod 3\\&= (a+b+c)\bmod 3\\&=[a+(b+c)]\bmod 3\end{aligned}$$ We can "descend" the same ladder we walked up to start with. If $[(a+b)+c]\bmod 3 = (a+b)\bmod 3 +_{3} c\bmod 3,$ then it also holds that $[a+(b+c)]\bmod 3 = a\bmod 3 +_{3} (b+c)\bmod 3.$[note]Again, think of the stuff in the parenthesis to be a new object. Rename it is you want, because you add that first anyway. You're just reusing the definition of modulo arithmetic.[/note] $$\begin{aligned}[(a+b)+c]\bmod 3 &= (a+b)\bmod 3 +_{3} c\bmod 3\\&= (a+b+c)\bmod 3\\&=[a+(b+c)]\bmod 3\\&= a \bmod 3 +_{3} (b+c) \bmod 3\end{aligned}$$ Last step. We are now basically done. Use the definition of modulo arithmetic one more time to finish. $$\begin{aligned}[(a+b)+c]\bmod 3 &= (a+b)\bmod 3 +_{3} c\bmod 3\\&= (a+b+c)\bmod 3\\&=[a+(b+c)]\bmod 3\\&= a \bmod 3 +_{3} (b+c) \bmod 3\\&\equiv [a+(b+c)]\bmod 3\end{aligned}$$ Axiom 2: Existence of an identity This one is nice and easy here. We already did the work by calculating out that operation table. Which element when added to any other element returns the other element? Does the order matter? Use the multiplication table to verify that 0 is the identity element of $\mathbb{Z}_{3}$ Axiom 3: Every element has to have an inverse We also did the setup for this one too. Look at each element (there are only 3). Each element needs to have a corresponding pair that you can modulo add to it to retrieve the identity element $0 \mod 3$. Clearly, 0[note]I write 0, but remember that this is not really the number 0. It represents all things that are a multiple of 3. In these groups, each element is actually an object we represent lazily by a number, rather than the number as we usually think of number.[/note] is its own inverse.[note]Yes, this is totally possible. In fact, for finite groups with an odd number of elements, one element will have to be its own inverse.[/note] $(0+0) \bmod 3 = 0\mod 3$. Also, $$(1+2)\bmod 3 \equiv (2+1) \bmod 3 \equiv 0 \bmod 3.$ So 1 and 2 are inverses of each other. Therefore, every element has an inverse, and we've proven that $\mathbb{Z}_{3}$ is a group under the operation modulo addition.