MA 103

Introduction to Abstract Mathematics

2015/16

Slides 8a

Introduction to Modular Arithmetic

1

Clock arithmetic

You go to sleep at 11 oclock and sleep for 8 hours. At what timedo you wake?

Well, this is simple: you wake at 7 oclock.

What we are doing in this calculation is arithmetic modulo 12.

The answer is not 11 + 8 = 19,because the clock re-starts once the hour of 12 is reached.

This fairly simple idea is part of modular arithmetic.

2

Weekday arithmetic

Today, 16 November 2015, is a Monday.

What day of the week is 25 December 2015?

November has 30 days. So 25 December is 39 days after 16November.

Weekdays repeat after 7 days. Since 39 = 5 7 + 4,

25 December is 4 weekdays after Monday, which is a Friday.

To determine the day of the week, we do arithmetic modulo 7.

Many calendar problems involve repetitions after a certain cycleof days, as in a week, or the days until the new moon repeats.Some of them have been studied for thousands of years.

3

The congruence relationTheorem

Let m N be a positive integer.Define the relation Rm on Z by: a Rm b if and only if b a is amultiple of m. In other words, for a, b Z we have

a Rm b m |(b a) b a = k m,for some k Z.

Then R is an equivalence relation.

ProofWe need to prove that Rm is reflexive, symmetric and transitive:

for all a Z: a Rm a; for all a, b Z: a Rm b b Rm a; for all a, b, c Z: (a Rm b b Rm c ) a Rm c.

4

Congruent modulo m

The described equivalence relation on Z, which depends on m,has a special name and notation.

Definition

If a Rm b, that is, if m |(b a), then we say that a and b arecongruent modulo m.

We write this as: a b (mod m).

If a, b are not congruent modulo m, we write a 6 b (mod m).

We sometimes only write a b if mod m is clear from thecontext.

5

Congruence via the remainder function

Recall that if we have a function f : X Y ,then we can define an equivalence relation R on X as follows:

a R b f (a) = f (b).

We can use this to define the congruence relation as well, usingdivision with remainder.

We saw that for m N and a Z, there are unique integers qand r such that

a = q m + r and 0 r < b.

Now for a Z, let fm(a) be the remainder r upon division by m.This defines a function fm : Z {0, 1, . . . ,m 1}.

6

Congruence via the remainder function

So, the function fm : Z {0, 1, . . . ,m 1} is defined asfor a Z, fm(a) is the remainder r upon division by m.

And we can use this function to define the equivalence relationcongruent modulo m on Z:

a b (mod m) fm(a) = fm(b).

In order to make sure this is all correct, we need to prove:

for all a, b Z: fm(a) = fm(b) a b (mod m).

7

Same last digit in base m

So weve seen that a b (mod m) if and only if a and b differ by a multiple of m, or, equivalently, if they have the same remainder upon division by m. (Note

that this remainder is a number in {0, 1, . . . ,m 1}.)

If a and b are positive, then this remainder can also be seen asthe last digit when a and b are written in base m.

The most familiar case is m = 10: a b (mod 10) means thata and b have the same last digit (when written in decimals).

Note that when a < 0 this is no longer true.

For example: 6 4 (mod 10).

8

Add and multiply modulo m

An important property of remainders is that we can add, subtractand multiply them just as we can do this with the original numbers.

Theorem

Let m N and a, a, b, b Z such that a a (mod m) andb b (mod m). Then we have(i) (a + b) (a + b) (mod m);(ii) (a b) (a b) (mod m);(iii) a b a b (mod m);(iv) c Z : c a c a (mod m);(v) n N : an (a)n (mod m).

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Proof of the theorem

The proof is fairly straightforward, provided we use precisely whatx y (mod m) means.

We are given that a a (mod m) and b b (mod m). This means that m |(a a) and m |(b b). And this means in turn that there exist k , ` Z such thata a = k m and b b = ` m.

And now we can start looking at the statements we have to prove. . .

Part (v) is done by induction on n, using that for n = 2 if followsfrom (iii).

10

One corollary testing divisibility by 9

Since 10 = 9 + 1, we have that 10 1 (mod 9).By (v), this means that 10n 1n 1 (mod 9) for all n N.

So we can calculate for a decimal number amam1 . . . a1a0:

amam1 . . . a1a0 am 10m + am1 10m1 + + a1 10 + a0 am 1 + am1 1 + + a1 1 + a0 am + am1 + + a1 + a0 (mod 9)

11

One corollary testing divisibility by 9

In particular we get:

amam1 . . . a1a0 is divisible by 9

amam1 . . . a1a0 0 (mod 9) am + am1 + + a1 + a0 0 (mod 9)

Example

Which of the following numbers (written in normal decimalnotation) are divisible by 9:

1232 ? 12672 ? 120672 ?

12

Second corollary testing divisibility by 11

We can do something similar for divisibility by 11.

For that we observe that 10 1 (mod 11).

Which leads to 10n (1)n {

1, if n even1, if n odd (mod 11).

So we can calculate for a decimal number amam1 . . . a1a0:

amam1 . . . a1a0 a0 + a1 10 + a2 102 + + am 10m a0 + a1 (1) + a2 1 + + am (1)m a0 a1 + a2 a3 + am (mod 11)

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Second corollary testing divisibility by 11

In particular we get:

amam1 . . . a1a0 is divisible by 11

amam1 . . . a1a0 0 (mod 11) a0 a1 + a2 a3 + am 0 (mod 11)

Example

Which of the following numbers are divisible by 11:

1232 ? 12672 ? 120672 ?

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