Differences

This shows you the differences between two versions of the page.

@@ Line 1: / Line 1: @@
+===== Residue classes =====
+Modular arithmetics as a way of calculating modulo some number $n$ is a powerful tool. However, it can be made simpler, and thus even more effective.
+First, recall that numbers having the same remainder when divided by $n$ behave in exactly the same way in such a calculation and yield exactly the same results.
+A quick example:
+$8 * 4 + 11 \equiv 3*(-1) + 6 \equiv 3 (\mod 5)$.
+One might say that
+there are only
+$n$ //essentially different objects// for this calculation - the numbers that give distinct remainders modulo $n$. The numbers having the same remainder can be considered to be //essentially the same object//.
+The next step would be to
+//define// **addition** and **multiplication** between such	 objects in a suitable way compatible with the original addition and multiplication.
+This approach leads us to the notion of **residue classes**.
+A residue class modulo $n$ is a set of all integers that give the same remainder when divided by $n$.
+<WRAP def>
+Formally, the residue class of $a \in \mathbb Z$ modulo $n \in \mathbb N$ is
+$\overline a = \{b \in \mathbb Z \mid a \equiv b \ (\mod n)\} = \{a + kn \mid k \in \mathbb Z\}$.
+In this notation, the modulus $n$ is implicit.
+</WRAP>
+As the following exercises show, all residue
+classes modulo $n$ partition the set $\mathbb Z$ into $n$ parts.
+<WRAP task>
+Show that for all $a, b \in \mathbb Z$, the following three claims are equivalent:
+  * $\overline a = \overline b$,
+  * $\overline a \cap \overline b = \emptyset$,
+  * $a \equiv b \ (\mod n)$,
+  * exists $k \in \mathbb Z$, such that $a = b + kn$.
+</WRAP>
+<WRAP task>
+Show that $\overline 0, \overline 1, \dots , \overline{n − 1}$ are different residue classes.
+</WRAP>
+The first of those two exercises shows that there are at most $n$ different residue
+classes modulo $n$, because there are $n$ different remainders when dividing by $n$. The
+second exercise shows that the number of residue classes is at least $n$.
+<WRAP def>
+Denote $\mathbb Z_n = \{\overline 0, \overline 1, \dots , \overline{n − 1}\}$.
+</WRAP>
+We can define addition and multiplication on $\mathbb Z_n$ as
+follows:
+  *$\overline a+\overline b=\overline{a+b}$,
+  *$\overline a \cdot \overline b=\overline{a\cdot b}$.
+Thanks to the properties of $\equiv$, shown in the previous section, these operations are well-defined.
+Namely, when we write $\overline{a + b}$, the integer $a$ is only defined as an arbitrary element of the
+residue class $\overline a$. When we take an arbitrary element $a$ of $\overline a$, and an arbitrary element $b$ of
+$\overline b$, is it the case that the residue class $\overline{a + b}$ is always the same?
+Yes, because $a_1 + b_1 \equiv a_2 + b_2 \ (\mod n)$ whenever
+$a_1 \equiv a_2 \ (\mod n)$ and $b_1 \equiv b_2 \ (\mod n)$. Similarly, the multiplication of residue classes is well-defined.
+Having defined $\mathbb Z_n$, we can rewrite our quick example from the beginning of the lesson as
+$\overline{8} \cdot \overline 4 + \overline{11}= \overline 3 \cdot \overline{-1} +\overline 6 = \overline{3}$ in $\mathbb Z_5$, where the first equality holds trivially simply because the corresponding objects are equal.
+<WRAP task>
+Write down the addition and multiplication tables for $\mathbb Z_7$ and $\mathbb Z_8$.
+</WRAP>
+==Similarities to integers==
+It turns out that
+both operations of addition and multiplication on $\mathbb Z_n$ have similar properties
+as the same operations on integers:
+  *They are both **associative**, meaning $(\overline a + \overline b) + \overline c = \overline a + (\overline b + \overline c)$ and $(\overline a \cdot \overline b) \cdot \overline c = \overline a \cdot (\overline b \cdot \overline c)$.\\ This property allows the formula $\overline a + \overline b + \overline c$ to be correctly understood in a single way. The same is true for $a \cdot b \cdot c$.
+  *They are both **commutative**, meaning $\overline a + \overline b = \overline b + \overline a$ and   $\overline a \cdot \overline b = \overline b \cdot \overline a$.
+  *They both have **units**. For addition, $\overline a + \overline 0 = \overline 0 + \overline a = \overline a$. For multiplication, $\overline a \cdot \overline 1 = \overline 1 \cdot \overline a = \overline a$.
+  *There is always an **opposite element** for addition, meaning that for any $\overline a$ there exists $-\overline a = \overline{-a}$ such that $\overline a + -\overline{a} = \overline{0}$.
+  *Multiplication is **distributive** over addition:\\ \begin{align*}
+  (\overline a+\overline b)\cdot \overline c &= \overline a\cdot \overline c + \overline b\cdot \overline c\enspace,\\
+  \overline a\cdot (\overline b + \overline c) &= \overline a\cdot \overline c + \overline a\cdot \overline c\enspace.
+\end{align*}
+These properties actually mean that both $\mathbb Z$
+and $\mathbb Z_n$
+are **commutative rings** with respect to addition and multiplication. For an introduction to rings, see ?.
+Because of the latter fact $\mathbb Z_n$
+is also called the **residue class ring modulo $n$**.
+<WRAP thtask>
+Show that $\mathbb Z_n$, together with the addition and multiplication operations, is a commutative ring.
+</WRAP>
+==Differences==
+Not everything in $\mathbb Z_n$ is the same way as in $\mathbb Z$.
+Take the equation $4a = 4b$ with integers $a$ and $b$. We know that it yields $a = b$.
+More generally, for any integer $c \neq 0$
+the equality $ca = c b$
+implies $a = b$ because we can divide both sides by $c$.
+On the other hand, for example, in $\mathbb Z_{12}$
+the equality
+$\overline{4}\cdot \overline{a} = \overline{4}\cdot \overline{b}$ does not necessarily mean that
+$\overline{a} = \overline{b}$, simply because
+$\overline{4}\cdot \overline{1} = \overline{4}\cdot \overline{4}$.
+The numbers as the number $c$ above are called **cancellative**.
+Thus we made an important observation that
+every nonzero integer is cancellative while for some numbers $n$ there are nonzero nonconcellative elements of $\mathbb Z_n$.
+<WRAP task>
+Find noncancellative elements of $\mathbb Z_{12}$,
+$\mathbb Z_7$, and $\mathbb Z_4$.
+++Answer|For $\mathbb Z_{12}: \overline 0, \overline 2, \overline 3, \overline 4, \overline 6, \overline 8, \overline 9, \overline{10}$,
+++
+</WRAP>
+Similarly, consider the equation $ab = 0$ with integers $a$ and $b$. It means that at least one of the numbers $a$ and $b$ is equal to $0$. Again,
+this is not the case in $\mathbb Z_{12}$ because we can find nonzero $\overline{a}$ and $\overline b$
+such that $\overline{a} \cdot \overline{b} = \overline{0}$, e.g., $\overline{4} \cdot \overline{3} = \overline{0}$.
+Nonzero $a$ such that $ab = 0$ for some nonzero $b$ is called a **zero divisor**.
+Thus we observed that there are no zero divisors in $\mathbb Z$ but for some numbers $n$ there are zero divisors in $\mathbb Z_n$.
+It is relatively easy to see that a zero divisor is always noncancellative.
+<WRAP thtask>
+Show that a zero divisor is never cancellative.
+++Solution|Assume that a zero divisor $a$ is cancellative. Then for some $b \neq 0$, we know that $ab = 0$. On the other hand, $a0 = 0$. Then $ab = a0$ but $b \neq 0$. So our assumption was wrong and $a$ is not cancellative.++
+</WRAP>
+It turns out that in $\mathbb Z_n$ the notions of zero divisors and noncancellative nonzero elements actually coincide. In fact, a nonzero element $\overline a$ in $\mathbb Z_n$ is a zero divisor and noncancellative if and only if
+$\gcd(a,n) >  1$.
+++++Proof|
+Let us check this fact. To see that $\gcd(a,n) > 1$ implies that $\overline a$ is a zero divisor, let us recall the equality $\gcd(a,n) \cdot \operatorname{lcm}(a,n) = an$. From this we have
+$a \cdot \frac{n}{\gcd(a,n)} = \operatorname{lcm}(a,n)$. Let us denote $c :=\frac{n}{\gcd(a,n)}$ and notice that $c$ is an integer greater than $1$ but less than $n$, so that $c \not \equiv 0 (\mod n)$. In other words,
+$\overline c \neq 0$. But $\overline a \cdot \overline c = \overline{\operatorname{lcm}(a,n)} = \overline 0$. So $\overline a$ is a zero divisor.
+On the other hand, if $\gcd(a,n) = 1$, then Bezout's identity says that
+there are integers $x$ and $y$ such that $ax + ny = 1$.
+Then $ax \equiv 1 (\mod n)$ and therefore $\overline a \cdot \overline x = \overline 1$. It means that $\overline x$ is the **multiplicative inverse** of $\overline a$ in $\mathbb Z_n$.
+It follows easily that $\overline a$ is cancellative because
+$\overline a \cdot \overline b = \overline a \cdot \overline c$
+would imply $\overline x \cdot \overline a \cdot \overline b = \overline x \cdot \overline a \cdot \overline c$, then
+$\overline 1 \cdot \overline b = \overline 1 \cdot \overline c$,
+and $\overline b = \overline c$
+++++
+As we already observed in the proof, the condition $\gcd(a,n) = 1$
+actually describes those elements in $\mathbb Z_n$, which have **multiplicative inverse**. That is, those elements $\overline a$ for which there is $x \in \mathbb Z_n$ such that $\overline a \cdot \overline x = \overline x \cdot \overline a = \overline 1$. For example, in $\mathbb Z_7$
+the multiplicative inverse of $\overline 3$ is $\overline 5$, because
+$\overline 3 \cdot \overline 5 = \overline 1$ in $\mathbb Z_7$.
+Then $\overline x$ is denoted by $\overline a^{-1}$. If $\overline a$
+has a multiplicative inverse, then $\overline a$ is called a **unit** in $\mathbb Z_n$.
+In our example, $\overline 3$ is a unit and $\overline 3^{-1} = \overline 5$ in $\mathbb Z_7$.
+<WRAP task>
+Find multiplicative inverses of units in $\mathbb Z_{12}$, $\mathbb Z_7$, $\mathbb Z_{20}$. Remember that units in $\mathbb Z_n$ are exactly those elements $a$ for which $\gcd(a,n) = 1$.
+</WRAP>
+As you may have noticed...

MathWiki

Differences