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Inverses and GCDs

Inverses and GCDs. Supplementary Notes. Prepared by Raymond Wong. Presented by Raymond Wong. 7 | 30. e.g.1 (Page 4). E.g., 30 can be expressed as 1 x 2 x 3 x 5. composite. 1 | 30. 1 divides 30. 6 | 30. 7 does not divide 30. 6 divides 30. 10 | 30. 10 divides 30.

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Inverses and GCDs

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  1. Inverses and GCDs Supplementary Notes Prepared by Raymond Wong Presented by Raymond Wong

  2. 7 | 30 e.g.1 (Page 4) • E.g., 30 can be expressed as 1 x 2 x 3 x 5 composite 1 | 30 1 divides 30 6 | 30 7 does not divide 30 6 divides 30 10 | 30 10 divides 30 2 | 30 2 divides 30 15 | 30 15 divides 30 3 | 30 3 divides 30 30 | 30 30 divides 30 5 | 30 5 divides 30

  3. 7 | 24 e.g.2 (Page 4) • E.g., 24 can be expressed as 1 x 2 x 2 x 2 x 3 composite 1 | 24 1 divides 24 4 | 24 7 does not divide 24 4 divides 24 6 | 24 6 divides 24 2 | 24 2 divides 24 3 | 24 3 divides 24 8 | 24 8 divides 24 12 | 24 12 divides 24 24 | 24 24 divides 24

  4. 7 | 11 e.g.3 (Page 4) • E.g., 11 can be expressed as 1 x 11 prime 1 | 11 1 divides 11 7 does not divide 11 11 | 11 11 divides 11

  5. e.g.4 (Page 4) • E.g., Is the following correct? 7 | 0 0 can be expressed as 0 x 7

  6. e.g.5 (Page 5) • E.g., What is gcd(7, 0)? 7 | 7 and 7 | 0 • E.g., Let n be a non-negative integer. What is gcd(n, 0)? n | n and n | 0

  7. e.g.6 (Page 7) • Illustration of Theorem 2.15 • E.g., j = 27 k = 58 if 27 and 58 are relatively prime (i.e., gcd(27, 58) = 1) then there exists two integers x and y such that 27x + 58y = 1 x = -15 y = 7 there exists two integers x and y such that 27x + 58y = 1 if then 27 and 58 are relatively prime (i.e., gcd(27, 58) = 1)

  8. e.g.6 (Page 7) • Illustration of Corollary 2.16 • E.g. a = 27 n = 58 if 27 has a multiplicative inverse (with respect to 58) then gcd(27, 58) = 1 if gcd(27, 58) = 1 then 27 has a multiplicative inverse (with respect to 58)

  9. e.g.7 (Page 10) • E.g., m = 21 n = 9 21 can be expressed as 9 x 2 + 3 (i.e., nq + r) r is defined to be 21 mod 9 q = 2 r = 3 21 mod 9 is equal to 3 0  r < n

  10. e.g.8 (Page 11) • Illustration of “Proof by Contradiction” We are going to prove that a claim C is correct Proof by Contradiction: Suppose “NOT C” …. Derive some results, which may contradict to 1. “NOT C”, OR 2. some facts e.g., we derived that C is true finally e.g., we derived that “1 = 4”

  11. e.g.9 (Page 11) Suppose that I want to prove that the above claim is correct by “Proof by Contradiction”. • Illustration of “Proof by smallest counter example” We are going to prove the following claim C: statement P(m) is true for each non-negative integer m, namely 0, 1, 2, … P(0) true If we can prove that statement P(m) is true for each non-negative integer separately, then we can prove the above claim C is correct. P(1) true P(2) true P(3) true P(4) true … true

  12. e.g.9 Suppose that I want to prove that the above claim is correct by “Proof by Contradiction”. • Illustration of “Proof by smallest counter example” We are going to prove the following claim C: statement P(m) is true for each non-negative integer m, namely 0, 1, 2, … P(0) true Suppose “NOT C”. We can assume that there exists a non-negative integer k’ such that P(k’) is false P(1) true P(2) true false P(3) true There may exist another non-negative integer k such that P(k) is false P(4) true false … true

  13. e.g.9 Suppose that I want to prove that the above claim is correct by “Proof by Contradiction”. • Illustration of “Proof by smallest counter example” We are going to prove the following claim C: statement P(m) is true for each non-negative integer m, namely 0, 1, 2, … P(0) true Suppose “NOT C”. P(1) true We can assume that there exists a smallest non-negative integer k such that P(k) is false P(2) true false P(3) true Why? P(4) true false This is called by “Proof by smallest counter example”. … true

  14. e.g.10 (Page 11) • We want to prove the following theorem. Theorem 2.12 (Euclid’s Division Theorem): Let n be a positive integer. For every nonnegative integer m, there exist unique integers q, r such that m = nq + r and 0  r < n

  15. e.g.10 • We want to prove the following theorem. Theorem 2.12 (Euclid’s Division Theorem): Let n be a positive integer. For every nonnegative integer m, there exist unique integers q, r such that m = nq + r and 0  r < n Claim 1: For every nonnegative integer m, there exist integers q, r such that m = nq + r and 0  r < n Claim 2: This pair q, r is unique.

  16. e.g.10 Claim 1: For every nonnegative integer m, there exist integers q, r such that m = nq + r and 0  r < n

  17. Claim 1: For every nonnegative integer m, there exist integers q, r such that m = nq + r and 0  r < n e.g.10 P(m) Proof by contradiction. Claim C Suppose that there exists an integer m that P(m) is false Proof by smallest counter example. Suppose that there exists a “smallest” integer m that P(m) is false There do not exist integers q, r such that m = nq + r and 0  r < n Consider two cases. Case 1: m < n Case 2: m  n We can write m = 0 + m = n.0 + m = nq + r where q = 0 and r = m We conclude that there exist integers q, r such that m = nq + r and 0  r < n Contradiction

  18. Claim 1: For every nonnegative integer m, there exist integers q, r such that m = nq + r and 0  r < n e.g.10 P(m) Proof by contradiction. Claim C Suppose that there exists an integer m that P(m) is false Proof by smallest counter example. Suppose that there exists a “smallest” integer m that P(m) is false There do not exist integers q, r such that m = nq + r and 0  r < n Consider two cases. Case 2: m  n

  19. Claim 1: For every nonnegative integer m, there exist integers q, r such that m = nq + r and 0  r < n e.g.10 P(m) Proof by contradiction. Claim C Suppose that there exists an integer m that P(m) is false Proof by smallest counter example. Suppose that there exists a “smallest” integer m that P(m) is false There do not exist integers q, r such that m = nq + r and 0  r < n Consider m-n = nq’ + r’ Consider two cases. Case 2: m  n m = nq’ + n + r’ We know that m-n  0 = n(q’ + 1) + r’ = nq + r Thus, m-n is a non-negative integer. where q = q’+1 and r = r’ Since m-n is smaller than m, We conclude that there exist integers q, r such that m = nq + r and 0  r < n there exist integers q’, r’ such that m-n = nq’ + r’ and 0  r’ < n Contradiction

  20. Claim 1: For every nonnegative integer m, there exist integers q, r such that m = nq + r and 0  r < n e.g.10 P(m) Proof by contradiction. Claim C Suppose that there exists an integer m that P(m) is false Proof by smallest counter example. Suppose that there exists a “smallest” integer m that P(m) is false There do not exist integers q, r such that m = nq + r and 0  r < n Consider two cases. In both cases, there are contradictions. This implies that Claim 1 is correct.

  21. e.g.10 • We want to prove the following theorem. Theorem 2.12 (Euclid’s Division Theorem): Let n be a positive integer. For every nonnegative integer m, there exist unique integers q, r such that m = nq + r and 0  r < n Claim 1: For every nonnegative integer m, there exist integers q, r such that m = nq + r and 0  r < n Claim 2: This pair q, r is unique.

  22. e.g.10 Claim 1: For every nonnegative integer m, there exist integers q, r such that m = nq + r and 0  r < n Claim 2: This pair q, r is unique.

  23. Claim 1: For every nonnegative integer m, there exist integers q, r such that m = nq + r and 0  r < n e.g.10 Claim 2: This pair q, r is unique. Proof by contradiction. Suppose that this pair q, r is not unique. There exists a pair (q, r) and another pair (q’, r’) (where (q, r)  (q’, r’)) such that m = nq + r …(*) and 0  r < n and m = nq’ + r’ …(**) and 0  r’ < n What is the greatest possible value? Consider (*) – (**) Consider r’ - r < n - r m - m = (nq+r) – (nq’ + r’) r’ – r < n n - 0 0 = nq+r – nq’ - r’ = n 0 = n(q-q’)+(r - r’) What is the smallest possible value? r’ - r = n(q-q’) r’ – r > -n Consider r’ - r > r’ - n n(q-q’)= r’ - r 0 - n -(r’ – r) < n We conclude that |r’ – r| < n = -n

  24. Claim 1: For every nonnegative integer m, there exist integers q, r such that m = nq + r and 0  r < n e.g.10 Claim 2: This pair q, r is unique. Proof by contradiction. Suppose that this pair q, r is not unique. There exists a pair (q, r) and another pair (q’, r’) (where (q, r)  (q’, r’)) such that m = nq + r …(*) and 0  r < n and m = nq’ + r’ …(**) and 0  r’ < n We conclude that |r’ – r| < n Consider (*) – (**) integer |n(q-q’)| < n m - m = (nq+r) – (nq’ + r’) We conclude that q – q’ = 0 0 = nq+r – nq’ - r’ q = q’ 0 = n(q-q’)+(r - r’) Note that n(q-q’)= r’ - r r’ - r = n(q-q’) 0 = r’ – r n(q-q’)= r’ - r r = r’ Contradiction We conclude that q = q’ and r = r’ (i.e., (q, r) = (q’, r’)) We conclude that |r’ – r| < n

  25. e.g.11 (Page 17) • Illustration of Lemma 2.13 k = 102 j = 70 Consider two integers 102 and 70. Suppose that we can write 102 as 102 = 70.1 + 32 q = 1 r = 32 According to the lemma, we have gcd(102, 70) = gcd(70, 32)

  26. e.g.12 (Page 17) • Prove the following lemma is correct. If j, k, q and r are non-negative integers such that k = jq + r then gcd(j, k) = gcd(r, j)

  27. If j, k, q and r are non-negative integers such that k = jq + r then gcd(j, k) = gcd(r, j) e.g.12 Consider two cases. Case 2: r > 0 Case 1: r = 0 e.g., if 10 = 2q then gcd(2, 10) = 2 Since k = jq + r, we have k = jq Consider gcd(j, k) = j e.g., gcd(0, 7) = 7 Consider gcd(r, j) = gcd(0, j) = j Thus, gcd(j, k) = gcd(r, j)

  28. If j, k, q and r are non-negative integers such that k = jq + r then gcd(j, k) = gcd(r, j) e.g.12 Consider two cases. Case 2: r > 0

  29. If j, k, q and r are non-negative integers such that k = jq + r then gcd(j, k) = gcd(r, j) e.g.12 Consider two cases. Case 2: r > 0 We want to prove the following. Claim 1: If d is a common divisor of j and k, then d is a common divisor of r and j. Claim 2: If d is a common divisor of r and j, then d is a common divisor of j and k.

  30. If j, k, q and r are non-negative integers such that k = jq + r then gcd(j, k) = gcd(r, j) e.g.12 Claim 1: If d is a common divisor of j and k, then d is a common divisor of r and j. Consider two cases. Case 2: r > 0 Let d be a common divisor of j and k d is a divisor of j j can be written as j = i1d where i1 is a non-negative integer d is a divisor of k k can be written as k = i2d where i2 is a non-negative integer Consider k = jq + r r = k – jq =i2d – i1d.q =(i2 – i1q)d We conclude that d is a divisor of r Since d is a divisor of j d is a common divisor of r and j

  31. If j, k, q and r are non-negative integers such that k = jq + r then gcd(j, k) = gcd(r, j) e.g.12 Claim 1: If d is a common divisor of j and k, then d is a common divisor of r and j. Consider two cases. Claim 2: If d is a common divisor of r and j, then d is a common divisor of j and k. Case 2: r > 0 Let d be a common divisor of r and j d is a divisor of r r can be written as r = i3d where i3 is a non-negative integer d is a divisor of j j can be written as j = i1d where i1 is a non-negative integer Consider k = jq + r = i1d.q + i3d = (i1q + i3)d We conclude that d is a divisor of k Since d is a divisor of j d is a common divisor of j and k

  32. 2 2 5 5 3 11 11 3 7 … … 7 If j, k, q and r are non-negative integers such that k = jq + r then gcd(j, k) = gcd(r, j) e.g.12 Claim 1: If d is a common divisor of j and k, then d is a common divisor of r and j. Consider two cases. Claim 2: If d is a common divisor of r and j, then d is a common divisor of j and k. Case 2: r > 0 From Claim 1 and Claim 2, we conclude that d is a common divisor of j and k if and only if d is a common divisor of r and j. d is not a common divisor of j and k if and only if d is not a common divisor of r and j. A set of common divisors of j and k A set of common divisors of r and j A set of non-common divisors of r and j A set of non-common divisors of j and k We conclude that gcd(j, k) = gcd(r, j)

  33. e.g.13 (Page 17) • How to use Lemma 2.13 for Euclid’s GCD algorithm Suppose that we want to find gcd(102, 70) k = 102 J = 70 We can use Lemma 2.13 to compute gcd(102, 70) Consider two integers 102 and 70. Suppose that we can write 102 as 102 = 70.1 + 32 q = 1 r = 32 This corresponds to r. r decreases and finally its value becomes 0. According to the lemma, we have gcd(102, 70) = gcd(70, 32) Note that 70 = 32.2 + 6 gcd(70, 32) = gcd(32, 6) Note that 32 = 6.5 + 2 gcd(32, 6) = gcd(6, 2) Note that 6 = 2.3 + 0 gcd(6, 2) = gcd(2, 0) Thus, gcd(102, 70) = gcd(2, 0) = 2

  34. e.g.13 Suppose that we want to find gcd(102, 70) k = j.q + r k j q r 102 70 1 32 102 = 70.1 + 32 70 32 2 6 70 = 32.2 + 6 32 = 6.5 + 2 32 6 5 2 6 2 3 0 6 = 2.3 + 0 gcd(102, 70) = gcd(2, 0) = 2

  35. e.g.14 (Page 24) • Definition of Multiplicative Inverse • Given a positive integer n, • we define Zn = {0, 1, 2, …, n-1} • Given a value a  Zn, • a is said to have a multiplicative inverse a’ in Zn if a’ .n a = 1

  36. e.g.14 • E.g., n = 9 Z9 = {0, 1, 2, …, 8} Does 2 have a multiplicative inverse in Z9? We may try all possible values in Z9 0 .9 2 = 0 0 is not a multiplicative inverse of 2 in Z9 1 .9 2 = 2 1 is not a multiplicative inverse of 2 in Z9 2 .9 2 = 4 2 is not a multiplicative inverse of 2 in Z9 3 .9 2 = 6 3 is not a multiplicative inverse of 2 in Z9 4 .9 2 = 8 4 is not a multiplicative inverse of 2 in Z9 5 .9 2 = 1 5 is a multiplicative inverse of 2 in Z9 6 .9 2 = 3 6 is not a multiplicative inverse of 2 in Z9 7 .9 2 = 5 7 is not a multiplicative inverse of 2 in Z9 8 .9 2 = 7 8 is not a multiplicative inverse of 2 in Z9 Yes 2 has a multiplicative inverse 5 in Z9.

  37. e.g.14 • E.g., n = 9 Z9 = {0, 1, 2, …, 8} Does 3 have a multiplicative inverse in Z9? We may try all possible values in Z9 0 .9 3 = 0 0 is not a multiplicative inverse of 3 in Z9 1 .9 3 = 3 1 is not a multiplicative inverse of 3 in Z9 2 .9 3 = 6 2 is not a multiplicative inverse of 3 in Z9 3 .9 3 = 0 3 is not a multiplicative inverse of 3 in Z9 4 .9 3 = 3 4 is not a multiplicative inverse of 3 in Z9 5 .9 3 = 6 5 is not a multiplicative inverse of 3 in Z9 6 .9 3 = 0 6 is not a multiplicative inverse of 3 in Z9 7 .9 3 = 3 7 is not a multiplicative inverse of 3 in Z9 8 .9 3 = 6 8 is not a multiplicative inverse of 3 in Z9 No 3 does not have a multiplicative inverse in Z9.

  38. e.g.15 (Page 25) • Illustration of Lemma 2.5 • Suppose that we want to find a value x in Z9 such that 2 .9x = 3 ……………(*) If 2 has a multiplicative inverse 5 in Z9 Why is it correct? then x = 5 .9 3 and this solution is unique. 2 .9x = 3 Why is this solution unique? 5 .9 (2 .9x) = 5 .9 3 The computation/derivation in the right-hand-side box is valid for anyx that satisfies equation (*). (5 .9 2) .9x = 5 .9 3 1 .9x = 5 .9 3 Thus, we conclude that onlyx that satisfies the equation (*) is 5 .9 3 x = 5 .9 3

  39. e.g.16 (Page 26) • Illustration of Theorem 2.7 If 2 has a multiplicative inverse 5 in Z9 then the inverse 5 is unique. Why is it correct? According to Lemma 2.5 Consider 2 .9x = b ……(*) If 2 has a multiplicative inverse 5 in Z9 then x = 5 .9 b and this solution is unique. If we set b = 1, the equation (*) becomes 2 .9x = 1 According to the inverse definition,x is an inverse of 2 According to Lemma 2.5, we have x = 5 .9 1 and this solution is unique.

  40. e.g.17 (Page 27) • Please find each non-zero value a  Z5 such that a has a multiplicative inverse a’ in Z5. (i.e., a .5 a’ = 1) For each non-zero a  Z5 and each non-zero b  Z5, we compute a .5 b if the above answer = 1 then we know that a has a multiplicative inverse b in Z5 or b has a multiplicative inverse a in Z5

  41. a = 3 and b = 1 a = 3 and b = 2 a = 3 and b = 3 a = 3 and b = 4 3 .5 1 = 3 3 .5 2 = 1 3 .5 3 = 4 3 .5 4 = 2 a = 4 and b = 1 a = 4 and b = 2 a = 4 and b = 3 a = 4 and b = 4 4 .5 1 = 4 4 .5 2 = 3 4 .5 3 = 2 4 .5 4 = 1 For each non-zero a  Z5 and each non-zero b  Z5, we compute a .5 b e.g.17 if the above answer = 1 then we know that a has a multiplicative inverse b in Z5 or b has a multiplicative inverse a in Z5 Z5 = {0, 1, 2, 3, 4} a = 1 and b = 1 a = 1 and b = 2 a = 1 and b = 3 a = 1 and b = 4 1 .5 1 = 1 1 .5 2 = 2 1 .5 3 = 3 1 .5 4 = 4 a = 2 and b = 1 a = 2 and b = 2 a = 2 and b = 3 a = 2 and b = 4 2 .5 1 = 2 2 .5 2 = 4 2 .5 3 = 1 2 .5 4 = 3

  42. a = 3 and b = 1 a = 3 and b = 2 a = 3 and b = 3 a = 3 and b = 4 3 .5 1 = 3 3 .5 2 = 1 3 .5 3 = 4 3 .5 4 = 2 a = 4 and b = 1 a = 4 and b = 2 a = 4 and b = 3 a = 4 and b = 4 4 .5 1 = 4 4 .5 2 = 3 4 .5 3 = 2 4 .5 4 = 1 For each non-zero a  Z5 and each non-zero b  Z5, we compute a .5 b e.g.17 if the above answer = 1 then we know that a has a multiplicative inverse b in Z5 or b has a multiplicative inverse a in Z5 3 2 4 1 Z5 = {0, 1, 2, 3, 4} 1 has a multiplicative inverse 1 in Z5 a = 1 and b = 1 a = 1 and b = 2 a = 1 and b = 3 a = 1 and b = 4 2 has a multiplicative inverse 3 in Z5 3 has a multiplicative inverse 2 in Z5 1 .5 1 = 1 1 .5 2 = 2 1 .5 3 = 3 1 .5 4 = 4 a = 2 and b = 1 a = 2 and b = 2 a = 2 and b = 3 a = 2 and b = 4 2 .5 1 = 2 2 .5 2 = 4 2 .5 3 = 1 2 .5 4 = 3 3 has a multiplicative inverse 2 in Z5 2 has a multiplicative inverse 3 in Z5 4 has a multiplicative inverse 4 in Z5

  43. e.g.18 (Page 27) • Please find each non-zero value a  Z6 such that a has a multiplicative inverse a’ in Z6. (i.e., a .6 a’ = 1) For each non-zero a  Z6 and each non-zero b  Z6, we compute a .6 b if the above answer = 1 then we know that a has a multiplicative inverse b in Z6 or b has a multiplicative inverse a in Z6

  44. a = 1 and b = 1 a = 1 and b = 2 a = 1 and b = 3 a = 1 and b = 4 a = 1 and b = 5 1 .6 1 = 1 1 .6 2 = 2 1 .6 3 = 3 1 .6 4 = 4 1 .6 5 = 5 a = 2 and b = 1 a = 2 and b = 2 a = 2 and b = 3 a = 2 and b = 4 a = 2 and b = 5 2 .6 1 = 2 2 .6 2 = 4 2 .6 3 = 0 2 .6 4 = 2 2 .6 5 = 4 a = 3 and b = 1 a = 3 and b = 2 a = 3 and b = 3 a = 3 and b = 4 a = 3 and b = 5 3 .6 1 = 3 3 .6 2 = 0 3 .6 3 = 3 3 .6 4 = 0 3 .6 5 = 3 a = 4 and b = 1 a = 4 and b = 2 a = 4 and b = 3 a = 4 and b = 4 a = 4 and b = 5 4 .6 1 = 4 4 .6 2 = 2 4 .6 3 = 0 4 .6 4 = 4 4 .6 5 = 2 a = 5 and b = 1 a = 5 and b = 2 a = 5 and b = 3 a = 5 and b = 4 a = 5 and b = 5 5 .6 1 = 5 5 .6 2 = 4 5 .6 3 = 3 5 .6 4 = 2 5 .6 5 = 1 For each non-zero a  Z6 and each non-zero b  Z6, we compute a .6 b e.g.18 if the above answer = 1 then we know that a has a multiplicative inverse b in Z6 or b has a multiplicative inverse a in Z6 Z6 = {0, 1, 2, 3, 4, 5}

  45. a = 1 and b = 1 a = 1 and b = 2 a = 1 and b = 3 a = 1 and b = 4 a = 1 and b = 5 1 .6 1 = 1 1 .6 2 = 2 1 .6 3 = 3 1 .6 4 = 4 1 .6 5 = 5 a = 2 and b = 1 a = 2 and b = 2 a = 2 and b = 3 a = 2 and b = 4 a = 2 and b = 5 2 .6 1 = 2 2 .6 2 = 4 2 .6 3 = 0 2 .6 4 = 2 2 .6 5 = 4 a = 3 and b = 1 a = 3 and b = 2 a = 3 and b = 3 a = 3 and b = 4 a = 3 and b = 5 3 .6 1 = 3 3 .6 2 = 0 3 .6 3 = 3 3 .6 4 = 0 3 .6 5 = 3 a = 4 and b = 1 a = 4 and b = 2 a = 4 and b = 3 a = 4 and b = 4 a = 4 and b = 5 4 .6 1 = 4 4 .6 2 = 2 4 .6 3 = 0 4 .6 4 = 4 4 .6 5 = 2 a = 5 and b = 1 a = 5 and b = 2 a = 5 and b = 3 a = 5 and b = 4 a = 5 and b = 5 5 .6 1 = 5 5 .6 2 = 4 5 .6 3 = 3 5 .6 4 = 2 5 .6 5 = 1 For each non-zero a  Z6 and each non-zero b  Z6, we compute a .6 b e.g.18 if the above answer = 1 then we know that a has a multiplicative inverse b in Z6 or b has a multiplicative inverse a in Z6 X 5 X X 1 Z6 = {0, 1, 2, 3, 4, 5} 1 has a multiplicative inverse 1 in Z6 5 has a multiplicative inverse 5 in Z6

  46. e.g.18 Z5: Z6: Z7: Z8: Z9:

  47. Lemma 2.5Consider equation 2 .6x = b If 2 has a multiplicative inverse x’ in Z6 equation “2 .6x = b” has a solution x = x’ .6 b e.g.19 (Page 30) The equation “2x mod 6 = 3” does not have a solution • Illustration of Corollary 2.6 • If there is a b  Z6 (e.g., 3) such that 2 .6x = b ………… (*)does not have a solution, then 2 does not have a multiplicative inverse in Z6 2x is equal to an even number. 2x mod 6 is also equal to an even number. Why is it correct? Proof by contradiction Suppose that 2 has a multiplicative inverse x’ in Z6 By Lemma 2.5, we know that equation “2 .6x = b” has a solution x = x’ .6 b This leads to a contradiction that equation “2 .6x = b” does not have a solution.

  48. X 5 X X 1 In some of our previous slides, we derive that 2 does not have a multiplicative inverse in Z6by checking the table. Z6 e.g.19 The equation “2x mod 6 = 3” does not have a solution • Illustration of Corollary 2.6 • If there is a b  Z6 (e.g., 3) such that 2 .6x = b ………… (*)does not have a solution, then 2 does not have a multiplicative inverse in Z6 2x is equal to an even number. 2x mod 6 is also equal to an even number. How will we use this corollary? Consider that the exam question asks you whether 2 has a multiplicative inverse in Z6. Suppose that we find that the equation “2x mod 6 = 3” does not have a solution (i.e., 2 .6 x = 3 does not have a solution) According to this corollary, we conclude that 2 does not have a multiplicativeinverse in Z6.

  49. e.g.20 (Page 36) • Illustration of Lemma 2.8 The modular equation 2 .7 x = 1 has a solution in Z7 if and only if there exist integers x, y such that 2x + 7y = 1 Only if Suppose that we know the modular equation 2 .7 x = 1 has a solution x = 4 We know that there exist integers x, y such that 2x + 7y = 1(In this case, x = -3 and y = 1) If Suppose that we know that there exist integers x, y such that 2x + 7y = 1(In this case, x = -3 and y = 1) We know the modular equation 2 .7 x = 1 has a solution x = 4

  50. e.g.20 • Illustration of Lemma 2.8 The modular equation 2 .7 x = 1 has a solution in Z7 if and only if there exist integers x, y such that 2x + 7y = 1 Why is it correct? Only if The modular equation 2 .7 x = 1 has a solution x in Z7 We can write as 2x mod 7 = 1 We can re-write as 2x = 7q + 1 where q is an integer 2x – 7q = 1 2x + 7(-q) = 1 Thus, there exist integers x, y such that 2x + 7y = 1 where y = -q

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