Difference between revisions of "Modular arithmetic/Introduction"

Revision as of 19:27, 27 April 2016

Modular arithmetic is a special type of arithmetic that involves only integers. This goal of this article is to explain the basics of modular arithmetic while presenting a progression of more difficult and more interesting problems that are easily solved using modular arithmetic.

Motivation

Let's use a clock as an example, except let's replace the $12$ at the top of the clock with a $0$ .

$[asy]picture pic; path a; a = circle((0,0), 100); draw (a); draw((0,0), linewidth(4)); pair b; b = (0,100); for (int i = 0; i < 12; ++i) { label (pic, (string) i, b); b = rotate(-30,(0,0)) * b; } pic = scale(0.8) * pic; add(pic); draw ((0,0) -- 50*dir(90)); draw ((0,0) -- 70*dir(90));[/asy]$

Starting at noon, the hour hand points in order to the following:

$1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, \ldots$

This is the way in which we count in modulo 12. When we add $1$ to $11$ , we arrive back at $0$ . The same is true in any other modulus (modular arithmetic system). In modulo $5$ , we count

$0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, \ldots$

We can also count backwards in modulo 5. Any time we subtract 1 from 0, we get 4. So, the integers from $-12$ to $0$ , when written in modulo 5, are

$3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0,$

where $-12$ is the same as $3$ in modulo 5. Because all integers can be expressed as $0$ , $1$ , $2$ , $3$ , or $4$ in modulo 5, we give these integers their own name: the residue classes modulo 5. In general, for a natural number $n$ that is greater than 1, the modulo $n$ residues are the integers that are whole numbers less than $n$ :

$0, 1, 2, \ldots, n-1.$

This just relates each integer to its remainder from the Division Theorem. While this may not seem all that useful at first, counting in this way can help us solve an enormous array of number theory problems much more easily!

Residue

We say that $a$ is the modulo- $m$ residue of $n$ when $n\equiv a\pmod m$ , and $0\le a<m$ .

Congruence

There is a mathematical way of saying that all of the integers are the same as one of the modulo 5 residues. For instance, we say that 7 and 2 are congruent modulo 5. We write this using the symbol $\equiv$ : In other words, this means in base 5, these integers have the same residue modulo 5:

2(base 5) $\equiv$ 12(base 5) $\equiv$ 22(base 5) $\equiv$ 32(base 5) $\equiv$ 42(base 5)

$7 \equiv 2 \pmod{5}.$

The (mod 5) part just tells us that we are working with the integers modulo 5. In modulo 5, two integers are congruent when their difference is a multiple of 5. Thus each of the following integers is congruent modulo 5:

$-12 \equiv -7 \equiv -2 \equiv 3 \equiv 8 \equiv 13 \equiv 18 \equiv 23 \pmod{5}$

In general, two integers $a$ and $b$ are congruent modulo $n$ when $a - b$ is a multiple of $n$ . In other words, $a \equiv b \pmod{n}$ when $\frac{a-b}{n}$ is an integer. Otherwise, $a \not\equiv b \pmod{n}$ , which means that $a$ and $b$ are not congruent modulo $n$ .

Examples

$31 \equiv 1 \pmod{10}$ because $31 - 1 = 30$ is a multiple of $10$ .

$43 \equiv 22 \pmod{7}$ because $\frac{43 - 22}{7} = \frac{21}{7} = 3$ , which is an integer.

$8 \not\equiv -8 \pmod{3}$ because $8 - (-8) = 16$ , which is not a multiple of $3$ .

$91 \not\equiv 18 \pmod{6}$ because $\frac{91 - 18}{6} = \frac{73}{6}$ , which is not an integer.

Sample Problem

Find the modulo $4$ residue of $311$ .

Solution:

Since $311 \div 4 = 77$ R $3$ , we know that

$311 \equiv 3 \pmod{4}$

and $3$ is the modulo $4$ residue of $311$ .

Another Solution:

Since $311 = 300 + 11$ , we know that

$311 \equiv 0+11 \pmod{4}$

We can now solve it easily

$11 \equiv 3 \pmod{4}$

and $3$ is the modulo $4$ residue of $311$

Making Computation Easier

We don't always need to perform tedious computations to discover solutions to interesting problems. If all we need to know about are remainders when integers are divided by $n$ , then we can work directly with those remainders in modulo $n$ . This can be more easily understood with a few examples.

Addition

Problem

Suppose we want to find the units digit of the following sum:

$2403 + 791 + 688 + 4339.$

We could find their sum, which is $8221$ , and note that the units digit is $1$ . However, we could find the units digit with far less calculation.

Solution

We can simply add the units digits of the addends:

$3 + 1 + 8 + 9 = 21.$

The units digit of this sum is $1$ , which must be the same as the units digit of the four-digit sum we computed earlier.

Why we only need to use remainders

We can rewrite each of the integers in terms of multiples of $10$ and remainders:
$2403 = 240 \cdot 10 + 3$
$791 = 79 \cdot 10 + 1$
$688 = 68 \cdot 10 + 8$
$4339 = 433 \cdot 10 + 9$ .
When we add all four integers, we get

$(240 \cdot 10 + 3) + (79 \cdot 10 + 1) + (68 \cdot 10 + 8) + (433 \cdot 10 + 9)$ $= (240 + 79 + 68 + 433) \cdot 10 + (3 + 1 + 8 + 9)$

At this point, we already see the units digits grouped apart and added to a multiple of $10$ (which will not affect the units digit of the sum):

$= 820 \cdot 10 + 21 = 8200 + 21 = 8221$ .

Solution using modular arithmetic

Now let's look back at this solution, using modular arithmetic from the start. Note that
$2403 \equiv 3 \pmod{10}$
$791 \equiv 1 \pmod{10}$
$688 \equiv 8 \pmod{10}$
$4339 \equiv 9 \pmod{10}$
Because we only need the modulo $10$ residue of the sum, we add just the residues of the summands:

$2403 + 791 + 688 + 4339 \equiv 3 + 1 + 8 + 9 \equiv 21 \equiv 1 \pmod{10},$

so the units digit of the sum is just $1$ .

Addition rule

In general, when $a, b, c$ , and $d$ are integers and $m$ is a positive integer such that

$a \equiv c \pmod{m}$ $b \equiv d \pmod{m}$

the following is always true:

$a + b \equiv c + d \pmod{m}$ .

And as we did in the problem above, we can apply more pairs of equivalent integers to both sides, just repeating this simple principle.

Proof of the addition rule

Let $a-c=m\cdot k$ , and $b-d=m\cdot l$ where $l$ and $k$ are integers. Adding the two equations we get: $\begin{eqnarray*} mk+ml&=&(a-c)+(b-d)\\ m(k+l)&=&(a+b)-(c+d) \end{eqnarray*}$

Which is equivalent to saying $a+b\equiv c+d\pmod{m}$