Definition 1 Let a and b be positive integers. A linear combination of a and b is any number n = ax + by, (1) where x and y are whole numbers.

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1 Greatest Common Divisors and Linear Combinations Let a and b be positive integers The greatest common divisor of a and b ( gcd(a, b) ) has a close and very useful connection to things called linear combinations of a and b (defined below) This handout will explain the connection Definition 1 Let a and b be positive integers A linear combination of a and b is any number n = ax + by, (1) where x and y are whole numbers Note that although a > 0 and b > 0, the numbers x and y are allowed to be negative or zero Examples Together in class, we will list some of the linear combinations of two pairs of integers Some linear combinations of a = 9 and b = 12: y y x 9x

2 Some linear combinations of a = 42 and b = 30: y y 0 x 42x Some linear combinations of a = 12 and b = 7: y y 0 x 12x

3 Notice that as far as we went, anyway the numbers that appear in the linear combinations table for a and b all seem to be multiples of gcd(a, b) Notice also that in each case, the multiples of gcd(a, b) all seem to be showing up Both of these statements turn out to be true, but other interesting facts will emerge first Throughout, let a and b be positive integers Theorem LC# 1 Let a and b be positive integers, and let ax + by be a linear combination of a and b For any integer d: if d a and d b, then d ax + by Proof Since d a (given) and a ax (obvious), by Property C on the Divisibility Handout, we get d ax (2) By the same token: since d b (given) and a by (obvious), by Property C on the Divisibility Handout, we get d by (3) Now, we can apply Property A of the Divisibility Handout to Equations (2) and (3): d ax + by Proof of Theorem LC#1 is complete The next step is to show that gcd(a, b) itself will always show up in the linear combinations table of a and b; we will do this by identifing a particular linear combination in the table that will turn out to be the gcd Which one will work? Well, if we are correct in our guess that all the numbers in the table are multiples of the gcd, then the gcd must be the smallest positive number 1 in the table (Note that positive means strictly positive Zero does not count here) So, we will focus on the smallest positive number in the table and try to show that this number is indeed gcd(a, b) Let us call this number d 0 Since d 0 is in the table, it is a linear combination of a and b; that is, there are integers x 0 and y 0 such that d 0 = ax 0 + by 0 To summarize: d 0 = ax 0 + by 0 is the smallest positive linear combination of a and b (4) Now, if d 0 is going to turn out to be the greatest common divisor of a and b, it must in the first place be some common divisor of a and b Proving this is the first step Theorem LC# 2 d 0 divides a Proof First, divide d 0 into a, getting a quotient q and a remainder r Recall that this means that a = qd 0 + r and (5) 0 r d There have to be positive numbers in the table, of course, so there has to be a smallest positive number This is true even though the table itself in infinite 3

4 There are two logical possibilities for r: either (a): r = 0, which would show that a = qd 0 (so that d 0 a, which is what we want to prove) or (b): 1 r d 0 1 I will show that (b) cannot happen; this will finish the proof Proof that (b) cannot happen: Solving the equation a = qd 0 + r for r gives r = a qd 0 ; (6) then, substituting (ax 0 + by 0 ) in for d 0 in (6) gives We now make some algebra moves on (7): r = a q (ax 0 + by 0 ) (7) Equation (7) repeated r = a q (ax 0 + by 0 ) multiply through r = a qax 0 qby 0 factor a out of first two terms r = a (1 qx 0 ) qby 0 factor b out of right-hand term r = a (1 qx 0 ) + b( qy 0 ) The last line shows that r is a linear combination of a and b If possibility (b) were true, then r would be a positive linear combination of a and b that is smaller than d 0, contradicting the fact that d 0 was chosen to be the smallest positive linear combination of a and b Thus, (b) cannot happen Proof of Theorem LC#2 is complete Theorem LC# 3 d 0 divides b Proof The proof of this is exactly parallel to the proof of LC#2, and will be left as an exercise Exercise 1 Prove Theorem LC#3 Up to this point, we have shown that d 0 is a common divisor of a and b, but we don t yet know that it is the greatest common divisor To establish this fact, I will show that if d 1 is any common divisor of a and b, then d 1 d 0 More is true: I will show that in fact d 1 d 0 (8) Theorem LC# 4 If d 1 a and d 1 b, then d 1 d 0 Proof Since d 1 a and d1 b, by Theorem LC#1, d1 divides any linear combination (ax + by) of a and b Since d 0 = ax 0 + by 0 is a linear combination of a and b [Equation (4)], d 1 must divide d 0 Proof of Theorem LC#4 is complete Two things are worth emphasizing at this point First, the LARGEST COMMON DIVISOR of a and b is the SMALL- EST POSITIVE LINEAR COMBINATION of a and b Second, as (8) indicates, the gcd of a and b is not only LARGER than every other common divisor; it is a MULTIPLE of every other common divisor The final theorem in this group is the one that confirms the pattern we observed, namely that all of the table entries seem to be multiples of the gcd and that all the multiples of the gcd seem to be showing up in the table 4

5 Theorem LC# 5 Let a and b be positive integers, and let d 0 = ax 0 + by 0 be the gcd 2 of a and b Then, for any integer n: n is a linear d 0 n combination of a and b Comment on how to prove this As mentioned before the proof of Theorem EA1 on the previous handout, the means two things: and If n is a linear combination of a and b, then n will also be a multiple of d 0 ( the = ); If n is a multiple of d 0, then n will also be a linear combination of a and b (the = ) These two things must be proved separately Proof (= ): By Theorems LC#2 and LC#3, we know that d 0 a and d0 b, so that by theorem LC#1, d 0 will divide every linear combination of a and b But we re given that n is a linear combination of a and b, so necessarily d 0 n Proof of = is complete Proof ( =): We re given that d 0 n, so that, for some k, n = kd 0 (9) The proof consists of replacing d 0 by (ax 0 + by 0 ) in Equation (9) and making some algebra moves: Equation (9) repeated n = kd 0 make the replacement n = k(ax 0 + by 0 ) multiply through n = kax 0 + kby 0 change order of multiplication in each term n = a(kx 0 ) + b(ky 0 ) The last equation above shows that n is a linear combination of a and b Proof of = is complete 2 d 0 was defined to be the smallest positive number in the linear combinations table, but we have now proved that d 0 = gcd(a, b) 5

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