modular-arithmetic.pdf, Study Guides, Projects, Research of Calculus

Download modular-arithmetic.pdf and more Study Guides, Projects, Research Calculus in PDF only on Docsity! 2-9-2018 Modular Arithmetic Modular arithmetic is a way of systematically ignoring differences involving a multiple of an integer. If n is an integer, two integers are equal mod n if they differ by a multiple of n; it is as if multiples of n are “set equal to 0”. Definition. Let n, x, and y be integers. x is congruent to y mod n if n | x− y. Notation: x = y (mod n) . Remarks. n | x− y is equivalent to the following statements: (a) n | y − x. (b) x = y + jn for some j ∈ Z. (c) y = x+ kn for some k ∈ Z. I’ll often use any of these four statements as the definition of x = y (mod n). A lot of people like to write “x ∼= y (mod n)” instead of “x = y (mod n)”. I don’t think there’s any harm in using an ordinary equal sign, since the “ (mod n)” makes the meaning clear. It’s also a bit shorter to write. Example. (Examples of congruences with numbers) (a) Demonstrate that 7 = 1 (mod 6) and 57 = −13 (mod 7). (b) Express “x is even” and “x is odd” in terms of congruences. (c) What does x = 0 (mod n) means in terms of divisibility? (a) 7 = 1 (mod 6) , since 6 | 7− 1. 57 = −13 (mod 7) , since 7 | 57− (−13). (b) x is even if and only if x = 0 (mod 2) and x is odd if and only if x = 1 (mod 2). (c) x = 0 (mod n) if and only if n | x. Thus, congruences provide a convenient notation for dealing with divisibility relations. The following proposition says that you can work with modular equations in many of the ways that you work with ordinary equations. Proposition. Let n ∈ Z. (a) If a = b (mod n) and c = d (mod n), then a+ c = b+ d (mod n) . (b) If a = b (mod n) and c = d (mod n), then ac = bd (mod n) . 1 (c) If a = b (mod n), then ac = bc (mod n) . Proof. Two ideas for these kinds of proofs: 1. You can often prove statements about congruences by reducing them to statements about divisibility. 2. You can often prove statements about divisibility by reducing them to (ordinary) equations. (a) Suppose a = b (mod n) and c = d (mod n). a = b (mod n) means n | a− b and c = d (mod n) means n | c− d. By properties of divisibility, n | (a− b) + (c− d) = (a+ c)− (b+ d). Therefore, a+ c = b+ d (mod n). (b) Suppose a = b (mod n) and c = d (mod n). a = b (mod n) means n | a− b, which means a− b = jn for some j ∈ Z. c = d (mod n) means n | c− d, which means c− d = kn for some k ∈ Z. Thus, a = b+ jn, c = d+ kn, and hence ac = (b+ jn)(d+ kn) = bd+ bkn+ djn+ jkn2 = bd+ n(bk + dj + jkn). This gives ac− bd = n(bk + dj + jkn), so n | ac− bd, and hence ac = bd (mod n). (c) Suppose a = b (mod n). This means that n | a− b. By properties of divisibility, n | (a− b)c = ac− bc. Therefore, ac = bc (mod n). Example. (Solving a congruence) Solve 3x+ 4 = 2x+ 8 (mod 9). In this case, I’ll solve the modular equation by adding or subtracting the same thing from both sides. 3x + 4 = 2x + 8 (mod 9) − 4 = 4 (mod 9) 3x = 2x + 4 (mod 9) − 2x = 2x (mod 9) x = 4 (mod 9) The solution is x = 4 (mod 9). Example. Reduce 497 · 498 · 499 (mod 500) to a number in the range {0, 1, . . . 499}, doing the computation by hand. Note that 497 = −3 (mod 500) , 498 = −2 (mod 500) , 499 = −1 (mod 500) . So 497 · 498 · 499 = (−3)(−2)(−1) = −6 = 494 (mod 500) . The next result says that congruence mod n is an equivalence relation. 2 Every integer n is congruent to one of 0, 1, 2, 3, or 4 mod 5. Therefore, I have 5 cases. In each case, I want to show that 2n2 + 3n+ 2 is not divisible by 5 — or to say it in terms of congruences, I want to show that 2n2 + 3n+ 2 6= 0 (mod 5). I set n = 0, 1, 2, 3, 4 (mod 5) and “substitute” the value into 2n2 +3n+2. This substitution is justified by the properties of congruences I discussed above. For example, if n = 3 (mod 5), then n · n = 3 · 3 (mod 5) n2 = 9 = 4 (mod 5) 2 · n2 = 2 · 4 (mod 5) 2n2 = 8 = 3 (mod 5) Likewise, 3n = 3 · 3 = 9 = 4 (mod 5). So 2n2 + 3n+ 2 = 3 + 4 + 2 = 9 = 4 (mod 5) . Essentially, I can plug n = 3 into 2n2 + 3n+ 2, then reduce the result mod 5 to one of 0, 1, 2, 3, or 4. Continuing in this way, I get the following table: n (mod 5) 0 1 2 3 4 2n2 + 3n+ 2 (mod 5) 2 2 1 4 1 In all five cases, 2n2 + 3n+ 2 6= 0 (mod 5). Therefore, 2n2 + 3n+ 2 is never divisible by 5. I showed earlier how to use algebraic operations to solve simple modular equations. How would you solve something like this: 6x = 13 (mod 25)? I’d like to divide both sides by 6, but I only know how to add and multiply. I can subtract, but that’s because I can add additive inverses. Well, division is multiplication by the multiplicative inverse; what is a multiplicative inverse mod 25? Definition. Let a, b ∈ Zn. a and b are multiplicative inverses if ab = 1 (mod n) (or ab = 1 in Zn). If a is the multiplicative inverse of b, you can write a = b−1. (You don’t write “ 1 b ” unless you’re in a number system like the rational numbers where fractions are in use.) Example. (Modular multiplicative inverses) (a) Prove that 6 and 2 are multiplicative inverses mod 11. (b) Show that 8 does not have a multiplicative inverse mod 12. (a) 6 · 2 = 1 (mod 11). (b) One tedious way is to take cases: n 0 1 2 3 4 5 8n (mod 12) 0 8 4 0 8 4 n 6 7 8 9 10 11 8n (mod 12) 0 8 4 0 8 4 5 No number multiplied by 8 gives 1 mod 12. I could try all the possibilities because the numbers were small. How would you do this kind of problem if the numbers were larger? One approach is to simply appeal to the result following this example. However, I can also give a proof by contradiction. Suppose that 8 has a multiplicative inverse mod 12. Let x be the multiplicative inverse. Then 8x = 1 (mod 12). Multiplying both sides by 3, I get 24x = 3 (mod 12) , or 0 = 3 (mod 12) . This is a contradiction, since 0 and 3 do not differ by a multiple of 12. Therefore, 8 does not have a multiplicative inverse mod 12. Proposition. m ∈ Zn has a multiplicative inverse if and only if (m,n) = 1. Proof. Suppose m ∈ Zn has a multiplicative inverse, so km = 1 for some k ∈ Zn. I can regard this as a statement in Z: km = 1 (mod n) . This means that km and 1 differ by a multiple of n: km− 1 = an for some a ∈ Z. Thus, km− an = 1. This is a linear combination of m and n which gives 1. Therefore, (m,n) = 1. Conversely, suppose (m,n) = 1. I may find integers a and b such that am+ bn = 1. That is, am = 1 (mod n) . Now regarded as an equation in Zn, this says am = 1 in Zn. That is, m has multiplicative inverse a. Example. (Using the Extended Euclidean algorithm to find modular inverses) Find the multi- plicative inverse of 31 in Z52. Note that (31, 52) = 1. Apply the Extended Euclidean Algorithm: 52 - 5 31 1 3 21 1 2 10 2 1 1 10 0 6 Thus, 1 = 3 · 52 + (−5) · 31. In Z52, 52 = 0 and −5 = 47. The equation says 1 = 47 · 31. Thus, 47 is the multiplicative inverse of 31 in Z52. Theorem. If (a, n) = 1, then the following equation has a unique solution: ax = b in Zn. Proof. If (a, n) = 1, then a has a multiplicative inverse a−1 in Zn. Thus, aa −1 = 1 in Zn. First, this means that x = a−1b is a solution, since a(a−1b) = (aa−1)b = 1 · b = b. Second, if x′ is another solution, then ax′ = b. Multiplying both sides by a−1, I get a−1ax′ = a−1b, x′ = a−1b. That is, x′ = x. This means the solution is unique. Example. (Solving modular equations using modular inverses) Solve 13x = 12 (mod 15) . There is a solution, since (13, 15) = 1. I need to find a multiplicative inverse for 13 mod 15. 15 - 7 13 1 6 2 6 1 1 2 0 The Extended Euclidean Algorithm says that (−6)(15) + (7)(13) = 1. Hence, 7 · 13 = 1 (mod 15), i.e. 7 is the multiplicative inverse of 13 mod 15. Multiply the original equation by 7: 7 · 13x = 7 · 12 (mod 15) , x = 84 = 9 (mod 15) . Proposition. Suppose ac = bc (mod n) . Then a = b ( mod n (n, c) ) . 7