# Lecture 32: Operations on modular numbers

• Addition, multiplication, negation of modular numbers is well defined

• Division is sometimes defined
• Units are numbers that you can divide by
• $$φ(m)$$ is the number of units
• Exponentiation is not well defined (we'll fix this next lecture)

• Review exercises:
• Prove that addition, multiplication, negation, subtraction of elements of $$\mathbb{Z}_m$$ are all well-defined
• Prove that if an inverse exists, it is unique (so that the "inverse" operation is a well-defined partial function)
• Prove that exponentiation of elements of $$\mathbb{Z}_m$$ is not well-defined

Last lecture, we defined modular numbers as equivalence classes of integers. In this lecture, we define basic operations on modular numbers. We will define the operations using representatives; we need to check that the operations are well defined.

Claim: Let $$+ : \mathbb{Z}_m \times \mathbb{Z}_m → \mathbb{Z}_m$$ be given by $$[a] + [b] ::= [a+b]$$. Then $$+$$ is well-defined.

Proof: We need to check that if $$[a] = [a']$$ and $$[b] = [b']$$ then $$[a + b] = [a'+b']$$. Assume $$[a] = [a']$$ and $$[b] = [b']$$. Then there exist $$c$$ and $$d$$ such that $$cm = a - a'$$ and $$dm = b - b'$$. Then $$a + b - (a' + b') = (a - a') + (b - b') = cm + dm = (c+d)m$$. Therefore $$m | (a+b)-(a'+b')$$, so $$[a+b] = [a'+b']$$.

Claim: Let the multiplication operation $$\cdot : \mathbb{Z}_m \times \mathbb{Z}_m → \mathbb{Z}_m$$ be given by $$[a][b] ::= [ab]$$. Then $$\cdot$$ is well defined.

Proof: Assume $$[a] = [a']$$ and $$[b] = [b']$$. Then as in the previous proof, there exist $$c$$ and $$d$$ with $$cm = a - a'$$ and $$dm = b - b'$$. We want to show that $$m | ab - a'b'$$. We can substitute $$b'$$ and $$a'$$ in this equation using the assumptions:

\begin{aligned} ab - a'b' &= ab - (a-cm)(b-dm) && \text{since cm = a-a' and dm = b-b'} \\ &= ab - ab +(ad+bc)m - dcm^2 = (\cdots)m && \text{algebra} \\ \end{aligned}

This shows that $$m | ab-a'b'$$ so $$\cdot$$ is well-defined.

Claim: The negation operation $$- : \mathbb{Z}_m → \mathbb{Z}_m$$ given by $$-[a] ::= [-a]$$ is well-defined.

Proof: left as exercise.

## Units and division

Definition: If $$x$$ and $$y$$ are numbers, we say that $$y$$ is an inverse of $$x$$ if $$xy = 1$$. If $$x$$ has an inverse, it is called a unit.

Note: This is a very general definition: it applies in any set that has a reasonable definition of multiplication and 1. Sets having operations called $$+$$ and $$\cdot$$ that satisfy certain conditions are called rings; one can talk about the units of any ring.

Examples: - The only units of $$\mathbb{Z}$$ are $$1$$ and $$-1$$ - All non-zero elements of $$\mathbb{Q}$$ and $$\mathbb{R}$$ are units - We will see in a later lecture that $$[a]$$ is a unit of $$\mathbb{Z}_m$$ if and only if $$gcd(a,m) = 1$$.

As another example, let's consider $$\mathbb{Z}_5$$. The elements of $$\mathbb{Z}_5$$ are $$\{[0],[1],[2],[3],[4]\}$$. $$[0]$$ is never a unit: there is nothing you can multiply 0 by to get 1. $$[1]$$ and $$[-1]$$ are always units, and are always their own inverses. In this case, $$[-1] = [4]$$. $$[2]$$ and $$[3]$$ are also units, because $$[2][3] = [6] = [1]$$.

In this case, all non-zero elements of $$\mathbb{Z}_5$$ were units, but this is not always the case. For example, in $$\mathbb{Z}_6$$, the only units are $$[1]$$ and $$[-1] = [5]$$. For example, $$[2]$$ is not a unit, because

• $$[2][0] = [0] \neq [1]$$,
• $$[2][1] = [2] \neq [1]$$,
• $$[2][2] = [4] \neq [1]$$,
• $$[2][3] = [6] = [0] \neq [1]$$,
• $$[2][4] = [8] = [2] \neq [1]$$,
• $$[2][5] = [10] = [4] \neq [1]$$.

Definition: $$φ(m)$$ is the number of units of $$\mathbb{Z}_m$$. It is called the totient of $$m$$ ($$φ$$ is also sometimes called the "Euler phi-function").

The above examples show that $$φ(5) = 4$$ and $$φ(6) = 2$$.

Claim: If $$x \in \mathbb{Z}_m$$ has an inverse, then it is unique.

Proof: Suppose $$xy = [1]$$ and $$xy' = [1]$$. We want to show $$y = y'$$. We have $y = [1]y = (xy')y = (xy)y' = [1]y' = y'$

Since the inverse is unique, we can give it a symbol: $$x^{-1}$$ is the inverse of $$x$$. We define $$x^{-n} ::= (x^{-1})^n$$ and note that $$x^nx^{-n} = [1]$$.

## Exponentiation

We have seen that addition, multiplication, and subtraction of equivalence classes, when defined in the obvious way, are well-defined. This is not true of exponentiation; we show this and fix it in the next lecture.