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modular:residue_classes [2015/03/19 20:00] (current)
aleksei imported from previous wiki
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 +===== 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...
  
modular/residue_classes.txt · Last modified: 2015/03/19 20:00 by aleksei