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Modular Arithmetic

Modular Arithmetic. Lecture 13: Oct 24. (based on slides in MIT 6.042). Last Time. Euclid’s algorithm to compute gcd(a,b). gcd(a,b) = sa + tb. 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

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Modular Arithmetic

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  1. Modular Arithmetic Lecture 13: Oct 24 (based on slides in MIT 6.042)

  2. Last Time • Euclid’s algorithm to compute gcd(a,b). • gcd(a,b) = sa + tb. • 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.

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

  4. Extended GCD Algorithm Example: a = 899, b=493 899 = 1·493 + 406 so 406 = 1·899 + -1·493 493 = 1·406 + 87 so 87 = 493 –1·406 = -1·899 + 2·493 406 = 4·87 + 58 so 58 = 406 - 4·87 = 5·899 + -9·493 87 = 1·58 + 29 so 29 = 87 – 1·58 = -6·899 + 11·493 58 = 2·29 + 0 done, gcd = 29

  5. Extended GCD Algorithm Example: a = 899, b=493 899 = 1·493 + 406 so 406 = 1·899 + -1·493 493 =1·406 + 87 so 87 = 493 – 1·406 = -1·899 + 2·493 406 = 4·87 + 58 so 58 = 406 - 4·87 = 5·899 + -9·493 87 =1·58 + 29 so 29 = 87 – 58 = -6·899 + 11·493 58 = 2·29 + 0 done, gcd = 29 s = -6, t = 11

  6. Another Example

  7. 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.

  8. Prime Divisibility Lemma:p prime and p|a·b implies p|a or p|b. 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 p| p| p| Cor : If p is prime, and p| a1·a2···amthen p|aifor some i.

  9. Fundamental Theorem of Arithmetic Every integer, n>1, has a unique factorization into primes: p0≤ p1 ≤ ··· ≤ pk p0p1 ··· pk = n Example: 61394323221 = 3·3·3·7·11·11·37·37·37·53

  10. Prime Products Claim: Every integer > 1 is a product of primes. Proof:(by contradiction) Supposenot. 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 leastnonproduct, • both are prime products, eg., • k = p1 p2p94 • m = q1 q2   q214

  11. Prime Products Claim: Every integer > 1 is a product of primes. …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

  12. Unique Factorization Claim: There is a unique factorization. pf: suppose not. choose smallestn >1: n = p1·p2···pk = q1·q2···qm p1p2···pk q1q2···qm can assume q1 < p1 so q1  pi all i

  13. Unique Factorization Claim: There is a unique factorization. pf: n = p1·p2···pk = q1·q2···qm now p1|n, soby Cor., p1|qi . so p1 = qi with i >1. so p2···pk = q1·q2···qi-1·qi+1···qm and q1  p2 contradiction! Cor : if n = p1·p2···pk , and m|n,then

  14. Modular Arithmetic Def:a  b (mod n) iff n|(a - b). For example, 15  29 (mod 7). Theorem. a  b (mod n) iff rem(a, n) = rem(b, n). • a = qn + r • b = q’n + r’ • n | (a-b) iff r = r’ Corollary. a  rem(a,n) (mod n)

  15. Modular Addition Lemma: If a  a’ (mod n), and b  b’ (mod n) then a+b  a’+b’ (mod n). pf: n| (a – a’) and n| (b – b’) implies n| ((a+b) – (a’+b’))

  16. Modular Multiplication Lemma: If a  a’ (mod n), and b  b’ (mod n) then ab  a’b’ (mod n). • 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’)

  17. 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).

  18. Multiplication Inverse The multiplicative inverse of a number a is another number a’ such that: a · a’ = 1 (mod n) Does every number has a multiplicative inverse in modular arithmetic?

  19. Multiplication Inverse Does every number has a multiplicative inverse in modular arithmetic? Nope…

  20. Multiplication Inverse What is the pattern?

  21. Multiplication Inverse Theorem. If gcd(k,n)=1, then have k’ k·k’  1 (mod n). k’ is an inversemod n of k pf: sk + tn = 1. So tn = 1 - sk This means n | 1 - sk just let k’ = s .

  22. 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), andgcd(k,n) = 1, then i  j (mod n) pf: multiply inverse of k to both sides.

  23. Fermat’s Little Theorem 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). This means that rem(k, p), rem(2k, p),…,rem((p-1)k, p) must be a permutation of 1, 2, ···, (p-1)

  24. Fermat’s Little Theorem 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)

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