Miniconference on the Mathematics of Computation 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 an integer m is a divisor of n if n = pm for some integer p integer n > 1 is prime if its only divisors are 1 and n a real number is irrational if it is not rational that is, it isn’t a fraction of integers egs: 2 , 3 , 5 , …

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 gcd(n,0) = n If a = bq + r, then gcd(a,b) = (b,r)

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.

x = x1 + tb/d, y = y1 – ta/d, where t is an integer. How to solve ax + by = c? Calculate gcd(a,b) by using the Euclidean Algorithm. Does gcd(a,b) divide c? If NO, then there is no solution to the linear Diophantine equation. If YES then Solve ax + by = gcd(a,b) = d by working backwards through your steps in the Euclidean Algorithm. This gives a particular solution (x1,y1) The general solution is: x = x1 + tb/d, y = y1 – ta/d, where t is an integer.