SUPPLY AND DEMAND, Essays (high school) of English Language

Download SUPPLY AND DEMAND and more Essays (high school) English Language in PDF only on Docsity! The Chinese Remainder Theorem Chinese Remainder Theorem: If m1, m2, .., mk are pairwise relatively prime positive integers, and if a1, a2, .., ak are any integers, then the simultaneous congruences x ≡ a1 (mod m1), x ≡ a2 (mod m2), ..., x ≡ ak (mod mk) have a solution, and the solution is unique modulo m, where m = m1m2⋅⋅⋅mk . Proof that a solution exists: To keep the notation simpler, we will assume k = 4. Note the proof is constructive, i.e., it shows us how to actually construct a solution. Our simultaneous congruences are x ≡ a1 (mod m1), x ≡ a2 (mod m2), x ≡ a3 (mod m3), x ≡ a4 (mod m4). Our goal is to find integers w1, w2, w3, w4 such that: value mod m1 value mod m2 value mod m3 value mod m4 w1 1 0 0 0 w2 0 1 0 0 w3 0 0 1 0 w4 0 0 0 1 Once we have found w1, w2, w3, w4, it is easy to construct x: x = a1w1 + a2w2 + a3w3 + a4w4. Moreover, as long as the moduli (m1, m2, m3, m4) remain the same, we can use the same w1, w2, w3, w4 with any a1, a2, a3, a4. First define: z1 = m / m1 = m2m3m4 z2 = m / m2 = m1m3m4 z3 = m / m3 = m1m2m4 z4 = m / m4 = m1m2m3 Note that i) z1 ≡ 0 (mod mj) for j = 2, 3, 4. ii) gcd(z1, m1) = 1. (If a prime p dividing m1 also divides z1= m2m3m4, then p divides m2, m3, or m4.) and likewise for z2, z3, z4. Next define: y1 ≡ z1 –1 (mod m1) y2 ≡ z2 –1 (mod m2) y3 ≡ z3 –1 (mod m3) y4 ≡ z4 –1 (mod m4) The inverses exist by (ii) above, and we can find them by Euclid’s extended algorithm. Note that iii) y1z1 ≡ 0 (mod mj) for j = 2, 3, 4. (Recall z1 ≡ 0 (mod mj) ) iv) y1z1 ≡ 1 (mod m1) and likewise for y2z2, y3z3, y4z4. Lastly define: w1 ≡ y1z1 (mod m) w2 ≡ y2z2 (mod m) w3 ≡ y3z3 (mod m) w4 ≡ y4z4 (mod m) Then w1, w2, w3, and w4 have the properties in the table on the previous page.