Modular Arithmetic

Lecture 13: Oct 24 (based on slides in MIT 6.042)

Last Time

1. Euclid’s algorithm to compute gcd(a,b).

2. gcd(a,b) = sa + tb.

3. Use this to solve Die Hard’s water jug problem.

How to find s and t so that gcd(a,b) = sa + tb??

Method: apply Euclidean algorithm, finding

coefficients as you go.

Example: a = 899, b=493899 = 1·493 + 406 493 = 1·406 + 87 406 = 4·87 + 58 87 = 1·58 + 29 58 = 2·29 + 0 done, gcd = 29

GCD Algorithm

Example: a = 899, b=493899 = 1·493 + 406 so 406 = 1·899 + -1·493493 = 1·406 + 87 so 87 = 493 –

1·406 = -1·899 + 2·493406 = 4·87 + 58 so 58 = 406 - 4·87 = 5·899 + -9·49387 = 1·58 + 29 so 29 = 87 – 1·58 = -6·899 + 11·49358 = 2·29 + 0 done, gcd = 29

Extended GCD Algorithm

Example: a = 899, b=493899 = 1·493 + 406 so 406 = 1·899 + -1·493493 = 1·406 + 87 so 87 = 493 – 1·406 = -1·899 + 2·493406 = 4·87 + 58 so 58 = 406 - 4·87 = 5·899 + -9·49387 = 1·58 + 29 so 29 = 87 – 58 = -6·899 + 11·49358 = 2·29 + 0 done, gcd = 29 s = -6, t = 11

Extended GCD Algorithm

Another Example

Die Hard

Given jug of 21 and jug of 26, is it possible to have exactly 11 gallons in one jug?

gcd(21,26) = 1

5x21 – 4x26 = 1

55x21 – 44x26 = 11

Can we save some water??

26x21 – 21x26 = 0

52x21 – 42x26 = 0

Therefore, 3x21 – 2x26 = 11.

Prime Divisibility

pf: say p does not divide a. so gcd(p,a)=1.

So by the Theorem, there exist s and t such that

sa + tp = 1

(sa)b + (tp)b = b

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

p| p| p|

Cor : If p is prime, and p| a1·a2···am then p|ai for some i.

Every integer, n>1, has a unique factorization into primes:

p0 ≤ p1 ≤ ··· ≤ pk

p0 p1 ··· pk = n

Fundamental Theorem of Arithmetic

Example:

61394323221 = 3·3·3·7·11·11·37·37·37·53

Claim: Every integer > 1 is a product of primes.

Prime Products

Proof: (by contradiction)

Suppose not. Then set of non-products is nonempty.

By WOP, there is a least n > 1 that is not a product of

primes.

In particular, n is not prime.

So n = k·m for integers k, m where n > k,m >1.

Since k,m smaller than the least nonproduct,

both are prime products, eg.,

k = p1 p2 p94

m = q1 q2 q214

Prime Products

…So

n = k m = p1 p2 p94 q1 q2 q214

is a prime product, a contradiction.

The set of nonproducts > 1 must be empty.

QED

Claim: Every integer > 1 is a product of primes.

Claim: There is a unique factorization.

Unique Factorization

pf: suppose not. choose smallest n >1:

n = p1·p2···pk = q1·q2···qm

p1p2···pk

q1q2···qm

can assume q1 < p1

so q1 pi all i

pf: n = p1·p2···pk = q1·q2···qm

now p1|n, so by Cor., p1|qi .

so p1 = qi with i >1.

so p2···pk = q1·q2···qi-1·qi+1···qm

and q1 p2

contradiction!

Unique Factorization

Cor : if n = p1·p2···pk , and m|n, then

Claim: There is a unique factorization.

1 2 ji i im = p ·p ···p

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

Modular Arithmetic

Theorem. a b (mod n) iff rem(a, n) = rem(b, n).

For example, 15 29 (mod 7).

Corollary. a rem(a,n) (mod n)

1. a = qn + r

2. b = q’n + r’

3. n | (a-b) iff r = r’

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

a+b a’+b’ (mod n).

Modular Addition

pf: n| (a – a’) and n| (b – b’) implies

n| ((a+b) – (a’+b’))

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

ab a’b’ (mod n).

Modular Multiplication

pf: n| (a – a’) and n| (b – b’) implies

a – a’ = pn and b – b’ = qn

implies

ab = (pn + a’)(qn + b’)

= pqn + a’qn + b’pn + a’b’

= (pq+aq’+b’p)n + a’b’

n | (ab – a’b’)

Application

Why is a number written in decimal evenly divisible by 9 if and

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

Multiplication Inverse

Nope…

Does every number has a multiplicative inverse in modular arithmetic?

Multiplication Inverse

What is the pattern?

Theorem. If gcd(k,n)=1, then have k’

k·k’ 1 (mod n).

k’ is an inverse mod n of k

pf: sk + tn = 1.

So tn = 1 - sk

This means n | 1 -

sk

just let k’ = s .

Multiplication Inverse

Cancellation

So (mod n) a lot like =.

main diff: can’t cancel

4·2 1·2 (mod 6)

4 1 (mod 6)

No general cancellation

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

then i j (mod n)

pf: multiply inverse of k to both sides.

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

k, 2k, …, (p-1)k

are all different (mod p).

So their remainders on division by p are all different (mod p).

Fermat’s Little Theorem

This means that

rem(k, p), rem(2k, p),…,rem((p-1)k, p)

must be a permutation of

1, 2, ···, (p-1)

so 1·2···(p-1) =rem(k,p)·rem(2k,p)···rem((p-

1)k,p) (k)·(2k) ··· ((p-1)k) (mod p) (kp-1)·1·2 ··· (p-1) (mod p)so

1 kp-1 (mod p)

Fermat’s Little Theorem