English grammar exercises, Slides of English Language

Download English grammar exercises and more Slides English Language in PDF only on Docsity! The Chinese Remainder Theorem We now know how to solve a single linear congruence. In this lecture we consider how to solve systems of simultaneous linear congruences. Example. We solve the system 2x ≡ 5 (mod 7); 3x ≡ 4 (mod 8) of two linear congruences (in one variable x). Multiply the first congruence by 2−1 mod 7 = 4 to get 4 · 2x ≡ 4 · 5 (mod 7). This simplifies to x ≡ 6 (mod 7), so x = [6]7 or x = 6 + 7t, where t ∈ Z. Now substitute for x in the second congruence: 3(6+7t) ≡ 4 (mod 8). This simplifies to 5t ≡ 2 (mod 8), which we solve by multiplying both sides by 5−1 mod 8 = 5 to obtain t ≡ 2 (mod 8). So t = [2]8 or t = 2 + 8s, where s ∈ Z. Substituting t back into x gives x = 6 + 7(2 + 8s) = 20 + 56s, which gives the solution x = [20]56 or x ≡ 20 (mod 56). (Notice that 56 = 7 · 8.) Example. Solve 4x ≡ 2 (mod 6); 3x ≡ 5 (mod 8). Start by reducing the first congruence to 2x ≡ 1 (mod 3). Multiply both sides by 2 (an inverse of 2 mod 3) to solve it, which gives x ≡ 2 (mod 3), so x = 2 + 3t. Now substitute x into the second given congruence: 3(2 + 3t) ≡ 5 (mod 8). This simplifies to t ≡ −1 (mod 8) or t ≡ 7 (mod 8). So t = 7 + 8s. Substituting t into the formula for x we obtain x = 2+3(7+ 8s) = 23+24s. So x ≡ 23 (mod 24). This is the complete solution. (Notice that 24 is the least common multiple of 6, 8.) The technique of the examples can always be used to solve simultaneous congruences when there is a solution. There may be no solution, but the technique detects that as well. Example. The system x ≡ 3 (mod 4); x ≡ 0 (mod 6) has no solution. Solving the first congruence gives x = 3 + 4t, and substituting that into the second gives 3 + 4t ≡ 0 (mod 6) or 4t ≡ −3 (mod 6). This congruence has no solution, since d = gcd(4, 6) does not divide −3. Actually, in this case there is a simpler way to see there is no solution. Just notice that the first congruence implies x is odd, but the second implies that x is even. That’s a contradiction. 1 Systems that have no solution are said to be inconsistent. Theorem (Chinese Remainder Theorem). Let m1,m2, . . . ,mr be a collec- tion of pairwise relatively prime integers. Then the system of simultaneous congruences x ≡ a1 (mod m1) x ≡ a2 (mod m2) ... x ≡ ar (mod mr) has a unique solution modulo M = m1m2, · · ·mr, for any given integers a1, a2, . . . , ar. Proof of CRT. Put M = m1 · · ·mr and for each k = 1, 2, . . . , r let Mk = M mk . Then gcd(Mk,mk) = 1 for all k. Let yk be an inverse of Mk modulo mk, for each k. Then by definition of inverse we have Mkyk ≡ 1 (mod mk). Let x = a1M1y1 + a2M2y2 + · · ·+ arMryr. Then x is a simultaneous solution to all of the congruences. Since the moduli m1, . . . ,mr are pairwise relatively prime, any two simultaneous solutions to the system must be congruent modulo M . Thus the solution is a unique congruence class modulo M , and the value of x computed above is in that class. Notice that the proof is constructive! Not only does it tell us why the theorem is true, it also gives an explicit formula for the solution. Example. Find all integers x which leave a remainder of 1, 2, 3, and 4 when divided by 5, 7, 9, and 11 respectively. We are asked to solve the system of congruences: x ≡ 1 (mod 5) x ≡ 2 (mod 7) x ≡ 3 (mod 9) x ≡ 4 (mod 11). Notice that the moduli are pairwise relatively prime, as required by the theorem. We have M = 5 · 7 · 9 · 11 = 3465 and M1 = M/5 = 693, 2