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 — modular:residue_classes [2015/03/19 20:00] (current)aleksei imported from previous wiki 2015/03/19 20:00 aleksei imported from previous wiki 2015/03/19 20:00 aleksei imported from previous wiki 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 $n$. + + 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. ​ + ​ + As the following exercises show, all residue + classes modulo $n$ partition the set $\mathbb Z$ into $n$ parts. + + 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 $a = b + kn$. + ​ + + Show that $\overline 0, \overline 1, \dots , \overline{n − 1}$ are different residue classes. + ​ + 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 $n$. + + + Denote $\mathbb Z_n = \{\overline 0, \overline 1, \dots , \overline{n − 1}\}$. ​ + ​ + + 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. + + + Write down the addition and multiplication tables for $\mathbb Z_7$ and $\mathbb Z_8$. + ​ + + ==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 $n$**. + + + + Show that $\mathbb Z_n$, together with the addition and multiplication operations, is a commutative ring. + ​ + ==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 $\mathbb Z_n$. + + + Find noncancellative elements of $\mathbb Z_{12}$, + $\mathbb Z_7$, and $\mathbb Z_4$. + + ++Answer|For $\mathbb Z_{12}: \overline 0, \overline 2, \overline 3, \overline 4, \overline 6, \overline 8, \overline 9, \overline{10}$,​ + ++ + ​ + + 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. + + + 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.++ + ​ + 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 $\mathbb Z_7$. + + + 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 $\gcd(a,n) = 1$. + ​ + + As you may have noticed...