MTH 210 Test 2 Review Dr Anthony Bonato

MTH 210 Test 2 Review Dr. Anthony Bonato Ryerson University

Notes on Test 2 • Test 2 is in-class on Thursday, March 14 in VIC 205, starting at 10: 30 am • 90 minutes, 5 questions (multiple parts), 25 marks total • Material covers all material on number theory, up to and including material covered in the March 7 lecture. • Need to know: definitions, examples, exercises, assigned problems, quiz material, theorems, key facts • Two questions short answer, one question fill in the blank; two questions long answer • NOTE: scientific calculators are allowed. Phones or other internet connected devices cannot be used. You are responsible for bringing your own calculator.

Key facts •

General congruences • m > 2 an integer • a ≡ b (mod n) if n | (a - b) • congruences behave like = • [m] = {x | x ≡ m (mod n)} – called equivalence classes (mod n) or congruence classes

Greatest Common Divisors • m and n integers > 1 • among all the divisors of both m and n, the largest is called the greatest common divisor • write gcd(m, n)

Key Facts 1. gcd(n, 0) = n 2. If a = bq + r, then gcd(a, b) = (b, r) 3. gcd(n, n+1) = 1

Euclidean Algorithm • • • Goal: find gcd(a, b) Step 1: Express a = bq + r Step 2: Find gcd(b, r). Step 3: Express b = sr + t. Step 4: Find gcd(r, t). … Keep going until remainder is zero.

How to solve ax + by = c? 1. Calculate gcd(a, b) by using the Euclidean Algorithm. 2. Does gcd(a, b) divide c? a) If NO, then there is no solution to the linear Diophantine equation. b) If YES then a) Solve ax + by = gcd(a, b) = d by working backwards through your steps in the Euclidean Algorithm. This gives a particular solution (x 1, y 1) b) The general solution is: x = x 1 + tb/d, y = y 1 – ta/d, where t is an integer.