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+===== 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.
+<WRAP def>
+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.
+</WRAP>
+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?)
+<WRAP thtask>
+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,
+$ac \mod n = (q_a n q_c n + q_a n r_c + q_c n r_a + r_a r_b) \mod n  = r_a r_b \mod n = r_b r_d \mod n = bd \mod n$.
+++++
+</WRAP>
+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:
+<WRAP thtask>
+Show that $a \equiv b \ (\mod n)$ if and only if
+$n \mid (a-b)$.
+++++Solution|
+Since $-b \equiv -b (\mod n)$, we have
+$a \equiv b \ (\mod n) \Longleftrightarrow
+a -b \equiv b - b = 0 (\mod n) \Longleftrightarrow n \mid (a-b)$.
+++++
+</WRAP>
+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).
+<WRAP thtask>
+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., $\frac{n}{\gcd(k,n)} \mid a$.
+++++
+</WRAP>
+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.
+<WRAP task>
+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),
+  *$1 \cdot 2 \cdots 10 \ (\mod 25)$.
+</WRAP>
+[[Residue classes]]

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