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 — modular:modular_arithmetics_draft [2015/03/19 19:57] (current)aleksei imported from previous wiki 2015/03/19 19:57 aleksei imported from previous wiki 2015/03/19 19:57 aleksei imported from previous wiki Line 1: Line 1: + ===== Modular arithmetics ===== + $\renewcommand{\mod}{\operatorname{mod}}$ + Sometimes when you want to calculate something + you might only require the remainder of the answer when divided by some natural number $n$. With help of modular arithmetics finding the required remainder usually happens to be much easier than evaluating the original expression. + + Before we start off with examples, let us introduce a handy notation. + + + For integers $a, b$ and a natural number $n$ let + $a \equiv b \ (\mod n)$ + denote that $a$ and $b$ have the same remainder when divided by $n$, i.e., that $a \mod n = b \mod n$ holds. ​ + ​ + + In essence, modular arithmetics means counting in a circle. There are many real-world examples: + clock: 19:00 + 7 hours = 02:00, + + weekdays: friday (5th day) + 4 days = tuesday (2nd day), + + counting-out games, etc. + + {{url>​../​illustrations/​modular.html 300px, 250px noscroll noborder}} + + Let us consider a bit more technical example: Let's say someone asks you to check whether $5638 \times 7891 + 2731 \times 3124$ is divisible by $17$. + + One way to do this is to go ahead and calculate the expression: ​ + + $5638 \times 7891 + 2731 \times 3124 = 44489458 + 8531644 = 53021102$, ​ + + then check if $17$ divides the answer... yawn. Ok, done. + + On the other hand, a modular arithmetic approach would be to note that $\mod 17$ we have: + + $10 = 17 - 7 \equiv -7$, + + $100 = 10^2 \equiv (-7)^2 = 49 = 51 - 2 \equiv -2$, + + $5638 = (51+5) \cdot 100 + (34 + 4) \equiv 5 \cdot (-2) + 4 = -6$, + + $7891 = (85-7)\cdot 100 + (85+6) \equiv (-7)\cdot(-2) + 6 = 20 \equiv 3$, + + $2731 \equiv (-7)\cdot(-2) - 3 = 11 \equiv -6$, + + $3124 \equiv (-3)\cdot(-2) + 7 = 13 \equiv -4$, + + and + + $5638 \times 7891 + 2731 \times 3124 + \equiv (-6) \cdot 3 + (-6) \cdot (-4) = (-6) \cdot (-1) = 6$. + + So the answer is: no, it is not, as the remainder is $6$, not $0$. + + Note that in the above example in the arithmetic expressions following the equivalenvce $\equiv$, we allowed ourselves to replace any number with a number having the same remainder modulo $17$. + This is because the **“equivalence modulo + $n$”** behaves very nicely with respect to arithmetic operations. (Why not try to prove the following bit of theory yourself?) + + If $a \equiv b \ (\mod n)$ and $c \equiv d \ (\mod n)$ then + + (1) $a+c \equiv b+d \ (\mod n)$ + + and + (2) $ac \equiv bd \ (\mod n)$. + ++++Proof| + Let us denote the quotient and the remainder when dividing $a$ by $n$ by $q_a$ and $r_a$, i.e., + $a = q_a n + r_a$, and similarly for $b$, $c$, and $d$. By assumption, we have $r_a = r_b$ and $r_c = r_d$. Then + + $(a+c) \mod n = \big((q_a+q_c)n + r_a + r_c \big) \mod n = (r_a + r_c) \mod n$ + + and + + $(b+d) \mod n = \big((q_b+q_d)n + r_b + r_d \big) \mod n = (r_b + r_d) \mod n = (r_a + r_c) \mod n$. + + Similarly, + + $ac \mod n = (q_a n q_c n + q_a n r_c + q_c n r_a + r_a r_b) \mod n = r_a r_b \mod n = r_b r_d \mod n = bd \mod n$. + ++++ + ​ + + When we say that an arithmetic expression has to be computed modulo $n$, we mean + that we are interested in the remainder of the value of this expression when divided by + $n$. For example, $(2 + 5) \cdot (8 + 6)$ is equal to $10$ modulo $11$, or + $(2 + 5) \cdot (8 + 6) \equiv 10 \ (\mod 11)$. + + The proposition above allows us to perform computations modulo $n$, such that the intermediate + values do not become much larger than $n$. + + Actually, when the operations that we’re + applying are only addition, subtraction and multiplication,​ then no intermediate value + has to be larger than $n^2$ . Namely, we can compute the remainder (by $n$) after each step + of the computation. Thus, we could perform the previous computation by + *$2+5=7$ + *$8 + 6 = 14 \equiv 3 \ (\mod 11)$ + *$7 \cdot 3 = 21 \equiv 10 \ (\mod 11)$. + + In fact, we can do even more simple operations with this equivalence. + + Let us look more closely at the proposition. + Note that part (1) is reversible in the sense that + if $a+c \equiv b+d \ (\mod n)$ and + $c \equiv d \ (\mod n)$ then $a \equiv b \ (\mod n)$. ++Because| then $-c \equiv -d \ (\mod n)$ + and $a+c-c \equiv b+d-d \ (\mod n)$.++ + + You may use this observation to check the exercise: + + Show that $a \equiv b \ (\mod n)$ if and only if + $n \mid (a-b)$. + + ++++Solution| + Since $-b \equiv -b (\mod n)$, we have + + $a \equiv b \ (\mod n) \Longleftrightarrow ​ + a -b \equiv b - b = 0 (\mod n) \Longleftrightarrow n \mid (a-b)$. + ++++ + ​ + + Part (2), on the other hand, is not reversible in general: $4 \cdot 2 \equiv 2 \cdot 2 (\mod 4)$ + but $4 \not \equiv 2 (\mod 4)$. + + The following exercise explains what we should do when trying to do the reverse of part (2). + + + Let $k \in \mathbb N$. Show that + * $a \equiv b \ (\mod n)$ if and only if $ak \equiv bk \ (\mod nk)$. + * If $ka \equiv kb \ (\mod n)$, then $a \equiv b \ (\mod \frac{n}{\gcd(k,​n)} )$. + ++++Solution| + The assertions follow from the previous exercise and the facts that $nk \mid ak$ is equivalent to $a \mid n$ and $n \mid ka$ implies $\frac{n}{\gcd(k,​n)} \mid a$. The former is obvious and for the latter consider the equation $ka = qn$. Dividing by $\gcd(k,​n)$,​ we get + $\frac{k}{\gcd(k,​n)} a = q \frac{n}{\gcd(k,​n)}$. Since $\frac{k}{\gcd(k,​n)}$ and $\frac{n}{\gcd(k,​n)}$ have no common divisors, all the divisors of $\frac{n}{\gcd(k,​n)}$ must also divide $a$, i.e., $\frac{n}{\gcd(k,​n)} \mid a$. + ++++ + ​ + + In particular, if $k$ and $n$  are coprime, then + we essentially get the inverse of part (2): + $ka \equiv kb \ (\mod n)$ implies $a \equiv b \ (\mod n)$. + + In a short while, we will see a deeper algebraic meaning of this + cancelability. + + TODO: Examples on the last exercise. + + + Perform the following computations:​ + *$(5 \cdot 8 − 3 \cdot 6) \cdot (8 + 4) \ (\mod 9)$, + *$7 \cdot 7 \cdots 7 \ (\mod 11)$ (here $7$ is taken 15 times), + *$1 \cdot 2 \cdots 10 \ (\mod 25)$. + ​ + [[Residue classes]]