Modular Arithmetic

Lecture 9: Oct 5

Def: a b (mod n) iff n|(a - b) iff a mod n = b mod n.

Modular Arithmetic

e.g. 12 2 (mod 10)

107 207 (mod 10)

7 1 (mod 2)

1 -1 (mod 2)

13 -1 (mod 7)

-15 0 (mod 5)

Lemma: If a c (mod n), and b d (mod n) then

a+b c+d (mod n).

Modular Addition

Example 1 12 2 (mod 10), 25 5 (mod 10)

=> 12 + 25 (mod 10)

2 + 5 (mod 10) 7 (mod 10)

Example 2 87 2 (mod 17), 222 1 (mod 17)

=> 87 + 222 (mod 17)

2 + 1 (mod 17)

3 (mod 17)

Example 3 101 2 (mod 11), 141 -2 (mod 11)

=> 101 + 141 (mod 11) 0 (mod 11)

Lemma: If a c (mod n), and b d (mod n) then

a+b c+d (mod n).

Modular Addition

a c (mod n) => a = c + nx for some integer x

b d (mod n) => b = d + ny for some integer y

To show a+b c+d (mod n), it is equivalent to showing that n | (a+b-c-d).

Consider a+b-c-d.

a+b-c-d = (c+nx) + (d+ny) – c –d = nx + ny.

It is clear that n | nx + ny.

Therefore, n | a+b-c-d.

We conclude that a+b c+d (mod n).

Proof

Lemma: If a c (mod n), and b d (mod n) then

ab cd (mod n).

Modular Multiplication

Example 1 9876 6 (mod 10), 17642 2 (mod 10)

=> 9876 * 17642 (mod 10)

6 * 2 (mod 10)

2 (mod 10)

Example 2 10987 1 (mod 2), 28663 1 (mod 2)

=> 10987 * 28663 (mod 2) 1 (mod 2)

Example 3 999 5 (mod 7), 674 2 (mod 7)

=> 999 * 674 (mod 7) 5 * 2 (mod 7) 3 (mod 7)

Lemma: If a c (mod n), and b d (mod n) then

ab cd (mod n).

Modular Multiplication

a c (mod n) => a = c + nx for some integer x

b d (mod n) => b = d + ny for some integer y

To show ab cd (mod n), it is equivalent to showing that n | (ab-cd).

Consider ab-cd.

ab-cd = (c+nx) (d+ny) – cd

= cd + dnx + cny + n2xy – cd = n(dx + cy + nxy).

It is clear that n | n(dx + cy + nxy). Therefore, n | ab-cd.

We conclude that ab cd (mod n).

Proof

Exercise

1444 mod 713

= 144 * 144 * 144 * 144 mod 713

= 20736 * 144 * 144 mod 713

= 59 * 144 * 144 mod 713

= 8496 * 144 mod 713

= 653 * 144 mod 713

= 94032 mod 713

= 629 mod 713

20736 * 20736 mod 713

= 59 * 59 mod 713

= 3481 mod 713

= 629 mod 713

Application

Is a number written in decimal evenly divisible by 9 if and only

if the sum of its digits is a multiple of 9?

Example 1. 9333234513171 is divisible by 9.

9+3+3+3+2+3+4+5+1+3+1+7+1 = 45 is divisible by 9.

Example 2. 128573649683 is not divisible by 9.

1+2+8+5+7+3+6+4+9+6+8+3 = 62 is not divisible by 9.

A coincidence?

This can be proved easily using modular arithmetic.

NO

Application

Claim. A number written in decimal is divisible by 9 if and only

if the sum of its digits is a multiple of 9?

Hint: 10 1 (mod 9).

Let the decimal representation of n be dkdk-1dk-2…d1d0.

This means that n = dk10k + dk-110k-1 + … + d110 + d0

Note that di10i mod 9

= (di mod 9) (10i mod 9) mod 9

= (di mod 9) (10 mod 9) (10 mod 9) … (10 mod 9) mod 9

= (di mod 9) (1 mod 9) (1 mod 9) … (1 mod 9) mod 9

= di mod 9

i terms

Application

Claim. A number written in decimal is divisible by 9 if and only

if the sum of its digits is a multiple of 9?

Let the decimal representation of n be dkdk-1dk-2…d1d0.

This means that n = dk10k + dk-110k-1 + … + d110 + d0

Note that di10i mod 9 = di mod 9.

Hence n mod 9 = (dk10k + dk-110k-1 + … + d110 + d0) mod 9

= (dk10k mod 9 + dk-110k-1 mod 9 + … + d110 mod 9 + d0 mod 9) mod 9

= (dk mod 9 + dk-1 mod 9 + … + d1 mod 9 + d0 mod 9) mod 9

= (dk + dk-1 + … + d1 + d0) mod 9

Hint: 10 1 (mod 9).

The multiplicative inverse of a number a is another number a’ such that:

a · a’ 1 (mod n)

Multiplication Inverse

Does every number has a multiplicative inverse in modular arithmetic?

For real numbers, every nonzero number has a multiplicative inverse.

For integers, only 1 has a multiplicative inverse.

An interesting property of modular arithmetic is that there are multiplicative inverse for intgers.

For example, 2 * 5 = 1 mod 3, so 5 is a multiplicative inverse for 2 under modulo 3 (and vice versa).

Multiplication Inverse

Does every number has a multiplicative inverse in modular arithmetic?

Multiplication Inverse

What is the pattern?

Multiplication Inverse

Why 2 does not have a multiplicative inverse under modulo 6?

Suppose it has a multiplicative inverse y. 2y 1 (mod 6)

=> 2y = 1 + 6x for some integer x=> y = ½ + 3x This is a contradiction since both x and y are integers.

Claim. An integer k does not have an multiplicative inverse under modulo n, if k and n have a common factor >= 2 (gcd(k,n) >= 2).

Proof. Same as above. Leave it as an exercise.

Theorem. If gcd(k,n)=1, then have k’

k·k’ 1 (mod n).

k’ is an inverse mod n of k

Proof: Since gcd(k,n)=1, there exist s and t so that sk + tn = 1.

So tn = 1 - sk

This means n | 1 – sk.

This means that 1 – sk 0 (mod n).

This means that 1 = sk (mod n).

So k’ = s is an multiplicative inverse for k.

Multiplication Inverse

What about if gcd(k,n)=1?

Would k always have an multiplicative inverse under modulo n?

gcd(k,n)=spc(k,n)

Cancellation

There is no general cancellation in modular arithmetic.

Note that (mod n) a lot like =.

If a b (mod n), then a+c b+c (mod n).

If a b (mod n), then ac bc (mod n)

However, if ac bc (mod n),

it is not necessarily true that a b (mod n).

For example, 4·2 1·2 (mod 6), but 4 1 (mod 6)

Cancellation

Why a·k b·k (mod n) when a ≠ b?

This means that ak = bk + nx.This means that (a-b)k = nx, which means a-b=(nx)/k.Since 0 < a < n and 0 < b < n, it implies that –n < a-b < n.Therefore, nx/k must be < n. For this to happen, n and k must have a common divisor >= 2!

Without loss of generality, assume 0 < a < n and 0 < b < n.Because if a·k b·k (mod n), then also (a mod n)·k (b mod n)·k (mod n).

smaller than n.

Okay, so, can we say something when gcd(n,k)=1?

Cancellation

Claim: If i·k j·k (mod n), and gcd(k,n) = 1,

then i j (mod n)

For example, multiplicative inverse always exists if n is a prime!

Proof. Since gcd(k,n) = 1, there exists k’ such that kk’ 1 (mod n).

i·k j·k (mod n).

=> i·k·k’ j·k·k’ (mod n).

=> i j (mod n)

This makes arithmetic modulo prime a field,

a structure that “behaves like” real numbers.

Arithmetic modulo prime is very useful in coding theory.

If p is prime & k not a multiple of p, then we can cancel k. So

k mod p, 2k mod p, …, (p-1)k mod p

are all different.

Fermat’s Little Theorem

This means that

k mod p, 2k mod p,…,(p-1)k mod p

must be a permutation of

1, 2, ···, (p-1)

(each number appears exactly once)

Fermat’s Little Theorem

1 kp-1 (mod p)

Theorem: If p is prime & k not a multiple of p

Proof.

1·2···(p-1) (k mod p · 2k mod p··· (p-1)k mod p) mod p

(k·2k ··· (p-1)k) mod p

(kp-1)·1·2 ··· (p-1) (mod p)

So, by cancelling 1·2 ··· (p-1) on both sides, we have

1 kp-1 (mod p)

A permutation

Wilson’s Theorem

Theorem: p is a prime if and only if

(p-1)! -1 (mod p)

First we consider the easy direction.

If p is not a prime, assume p >= 5, (for p=4, 3! 2 (mod 4) )

Then p=qr for some 2 <= q < p and 2 <= r < p.

If q ≠ r, then both q and r appear in (p-1)!, and so (p-1)! 0 (mod p).

If q = r, then p = q2 > 2q (since we assume p > 5 and thus q > 2).then both q and 2q are in (p-1)!,and so again (p-1)! 0 (mod p).

Wilson’s Theorem

Theorem: p is a prime if and only if

(p-1)! -1 (mod p)

To prove the more interesting direction, first we need a lemma.

Lemma. If p is a prime number, x2 1 (mod p) if and only if x 1 (mod p) or x -1 (mod p)

Proof. x2 1 (mod p)

iff p | x2 - 1

iff p | (x – 1)(x + 1)

iff p | (x – 1) or p | (x+1)

iff x 1 (mod p) or x -1 (mod p)

Lemma: p prime and p|a·b iff p|a or p|b.

Wilson’s Theorem

Theorem: p is a prime if and only if

(p-1)! -1 (mod p)

Let’s get the proof idea by considering a concrete example.

10! 1·2·3·4·5·6·7·8·9·10 mod 11

1·10·(2·6)·(3·4)·(5·9)·(7·8) mod 11

1·-1·(1)·(1)·(1)·(1) mod 11

-1 mod 11

Besides 1 and 10, the remaining numbers are paired up into multiplicative inverse!

Wilson’s Theorem

Theorem: p is a prime if and only if

(p-1)! -1 (mod p)

Proof.Since p is a prime, every number from 1 to p-1 has a multiplicative inverse.

By the Lemma, every number 2 <= k <= p-2 has an inverse k’ with k≠k’.

Since p is odd, the numbers from 2 to p-2 can be grouped into pairs

(a1,b1),(a2,b2),…,(a(p-3)/2,b(p-3)/2) so that aibi 1 (mod p)

Therefore, (p-1)! 1·(p-1)·2·3·····(p-3)·(p-2) (mod p)

1·(p-1)·(a1b1)·(a2b2)·····(a(p-3)/2b(p-3)/2) (mod p)

1·(-1)·(1)·(1)·····(1) (mod p)

-1 (mod p)