In this whole worksheet, suppose n₁ = 3 and n₂ = 5. 1. Compute N = ₁+₂= Write out the list of possible remainders modulo ₁: Write out the list of possible remainders modulo 1₂: Choose one of the entries from the first list and call it a₁. You have chosen a Choose one of the entries from the second list and call it a2. You have chosen a2 - 2. The Chinese Remainder Theorem says there is exactly one solution modulo N to the system of congruences: mod z = 0₂ mod n₂. Let's see this in action. In the left hand column, write the numbers from 0 to N-1. This will take the place of z. For every value of z, calculate 2 mod ny and a mod n2 and put the value in the correct column. You will see patterns appearing. I r mod n mod n₂ Now compare the entries in the rows to your choice of an and a2 in Question 1. a mod ny andra2 mod n₂? In which row is Do any other rows have that same pair of values? You have found your *unique* solution to the system of congruences. 3. Some questions for thought: (a) For what value z is z = 0 mod n and = 0 mod n₂? (b) For what value z is z = 1 mod ₁ and 2 = 1 mod n₂? (c) For what value az is r= -1 mod n and x = -1 mod n₂?
In this whole worksheet, suppose n₁ = 3 and n₂ = 5. 1. Compute N = ₁+₂= Write out the list of possible remainders modulo ₁: Write out the list of possible remainders modulo 1₂: Choose one of the entries from the first list and call it a₁. You have chosen a Choose one of the entries from the second list and call it a2. You have chosen a2 - 2. The Chinese Remainder Theorem says there is exactly one solution modulo N to the system of congruences: mod z = 0₂ mod n₂. Let's see this in action. In the left hand column, write the numbers from 0 to N-1. This will take the place of z. For every value of z, calculate 2 mod ny and a mod n2 and put the value in the correct column. You will see patterns appearing. I r mod n mod n₂ Now compare the entries in the rows to your choice of an and a2 in Question 1. a mod ny andra2 mod n₂? In which row is Do any other rows have that same pair of values? You have found your *unique* solution to the system of congruences. 3. Some questions for thought: (a) For what value z is z = 0 mod n and = 0 mod n₂? (b) For what value z is z = 1 mod ₁ and 2 = 1 mod n₂? (c) For what value az is r= -1 mod n and x = -1 mod n₂?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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