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1 Elementary Properties of the Integers 1. Basis Representation Theorem (Thm 1-3) 2. Euclid’s Division Lemma (Thm 2-1) 3. Greatest Common Divisor 4. Properties of Prime Numbers 5. Fundamental Theorem of Arithmetic (Thm 2-5) 2 The Integers: Z For most of this course, we restrict ourselves to the set of integers. Some helpful hints to keep in mind: • Addition, subtraction, and multiplication work normally. The sum/difference/product of two integers is always another integer. Commutative, associative, and distributive laws hold. • The reciprocal of an integer is almost never an integer. What are the exceptions? • Division/fractions must be handled with care. If you use a fraction within an equation, be ABSOLUTELY CERTAIN that the fraction represents an integer. If you are unsure, don’t use the fraction. 3 • It is valid to use constant fractions within inequalities (example: x > 12 ), but it may be confusing to do so. Consider the sets: A= 1 x∈Z:x> , 2 B = {x ∈ Z : x ≥ 1} . • The Cancellation Laws still work. If ax = ay, we can cancel the a from both sides to get x = y, as long as a 6= 0. The same is true for xa = ya. • Nonnegative integer exponents work normally. Negative or fractional exponents should generally be avoided. In particular, reciprocals and roots are not always defined, and might behave differently even when they exist. 4 The Natural Numbers: N We will occasionally restrict ourselves to non-negative integers. These are called Natural Numbers (or Whole Numbers, in older texts). • Addition and multiplication work normally. • Negation does not make sense. Subtraction is not always defined. • Reciprocals, division/fractions, exponents/roots have the same issues as in Z. • A nonempty subset of N always contains a minimal element. This is known as Well-Ordering or the Least Integer Principle. • Mathematical induction can (and often will be) used to prove that a result holds for all natural numbers. 5 Integer Division with Remainder Theorem (Euclid’s Division Lemma (Thm 2-1)). For each integer j and each positive integer k, there is a unique pair of integers q and r such that 0 ≤ r < k and j = kq + r. For j > 0, this is the familiar division problem j ÷ k = q REM r. One way to find the values of q and r: • If j < k, we can choose q = 0 and r = j. • If j ≥ k, then j − k ≥ 0 is still positive. If j − k ≥ k, subtract another copy of k (to get j − 2k ≥ 0). • Repeating this, we’ll eventually get a j − qk that is nonnegative and smaller than k. We let r = j − qk. 6 Note: This is certainly NOT the most efficient way to compute a quotient/reminder, but it makes for an easy proof (using Well-Ordering). Proof. (Existence): Given j and k, consider the subset of natural numbers S = {j − xk : x ∈ Z}. Why is this set nonempty? • By the Well-Ordering Principle, S has a smallest element. Let r = j − qk be this smallest element. • We need to show r < k. Assume this is not true, then r ≥ k and thus r − k ≥ 0. However, r − k = (j − qk) − k = j − (q + 1)k. • Thus r − k ∈ S, but this is impossible, since r − k < r and r was the smallest element of S. 7 (Uniqueness): Suppose that (q, r) and (q 0, r0) both satisfy the conclusion of the Theorem. Then kq + r = kq 0 + r0 since both sides are equal to j. Rearranging gives: r0 − r (q − q ) = . k 0 • The right-hand side must be an integer (why?) • We know that 0 ≤ r0 < k and −k < −r ≤ 0. Adding these together gives −k < r0 − r < k. • Thus −1 < r 0 −r k < 1. What does this tell us? 8 (Uniqueness, continued): • The only integer between −1 and 1 is zero. r 0 −r k • We thus have = 0, which immediately gives r0 = r. This trick is available precisely because we’ve restricted ourselves to integer values. • It also follow that q − q 0 = 0. In other words, q = q 0. This proof contains a common method for showing that a certain object is unique (in this case, the pair (q, r)): assume you have two things satisfying a given condition, and show that those two things must be equal. 9 Divisibility (Section 2-2) Definition. Given an integer n and a nonzero integer d, we say that d divides n if there is an integer c such that n = cd. We write d | n if d divides n, and d - n if d does not divide n. • Note that “d | n is equivalent to “ nd is an integer.” However, you are advised to avoid using fractions in problems about divisibility. • In terms of the Division Lemma, “d | n” is equivalent to “n leaves a remainder of zero when divided by d.” Keep this in mind: many divisibility proofs use this idea. • Divisibility is one of the key concepts of this course. It leads to the idea of congruence (modulo n), which will be another central theme. 10 Exercise: Properties of Divisibility Prove these directly from the definition (do not use fractions!): • If a is a nonzero integer, then a | 0 and 1 | a. • If a | b and b | c, then a | c. • [IMPORTANT!] If d | a and d | b, then d | (am + bn) for any m, n ∈ Z. • If a | b and b | a, then a = ±b. Find positive integers a, b, c such that a | bc, but a - b and a - c. 11 Greatest Common Divisor/Factor • You know from basic arithmetic that, given two integers, we can find their greatest common divisor (or greatest common factor ). • One way to find a g.c.d. is using prime factorization. For example: – 1512 = 23 · 33 · 7 – 4410 = 2 · 32 · 5 · 72 – gcd(1512, 4410) = 2 · 32 · 7 = 126 • This is not always so simple. Try finding – gcd(18, 42) (easy!) – gcd(629, 1517) (more work!) 12 We use the following definition of gcd(a, b): Definition (2-1). Let a and b be two integers, not both zero. The greatest common divisor of a and b is a positive integer d which satisfies the following properties: 1. d | a and d | b. In other words, d is a common divisor of a and b. 2. For any positive integer c for which c | a and c | b, we have c | d. Thus, any common divisor of a and b must also divide d. • This definition is easier to use in practice. • It can be shown that if d satisfies this definition and c is any common divisor of a and b, then d ≥ c. 13 The Euclidean Algorithm It turns out that we can compute a gcd without using prime factorization. For example, we find gcd(1517, 629) by repeated use of the Division Lemma: 1517 =(629)(2) + 259 629 =(259)(2) + 111 259 =(111)(2) + 37 111 =(37)(3) + 0 • To show that 37 is a common divisor, start with the last equation and work up. • To show any common divisor divides 37, start with the first equation and work down. 14 The Euclidean Algorthim The same process can be used to find gcd(a, b) if a and b are positive: 1. Let DIVIDEND = larger of a and b, DIVISOR = smaller of a and b. 2. Find the REMAINDER of DIVIDEND ÷ DIVISOR. 3. If REMAINDER = 0, then DIVISOR is the gcd. [Stop] 4. Otherwise, let DIVIDEND = DIVISOR, then DIVISOR = REMAINDER. 5. Go back to step 2. 15 Exercises: Euclidean Algorithm Use the Euclidean Algorithm to find the following gcd’s. Keep all of your work; we will use it for a different purpose later. • gcd(90, 25) • gcd(2499, 182) • gcd(629, 518) • gcd(4721, 1361) • gcd(89, 55) 16 Theorem (2-2). If a and b are integers, not both zero, then gcd(a, b) exists and is unique. Note that this result is trivial if either a or b is zero. Also note that changing the sign of a or b or swapping a for b does not change the gcd. Without loss of generality, we may thus assume that 0 < b ≤ a. For existence, we need to show that the Euclidean Algorithm eventually produces a remainder of zero, at which point the algorithm terminates. This follows from the Division Lemma, noting that the remainder at any stage is always smaller than the divisor at that stage. Since the remainder at one stage becomes the divisor in the next stage, the sequence of remainders decreases by at least 1 at each step. Thus, we have 0 ≤ ri ≤ b − i. So we get a remainder of zero after at most a steps (and usually, much fewer). 17 The equation ax + by = c Choose nonzero integers a, b, c. Are there integers x, y that satisfy the above equation? • Suppose such integers x and y do exist. Let d = gcd(a, b). • Since x and y are integers, the left-hand side is divisible by d. • So a necessary condition is that gcd(a, b) | c. If this is not true, then we cannot find integers x, y that satisfy the equation. • It turns out that this condition is also sufficient. • The difficult step is using the Euclidean Algorithm to solve ax + by = d; the rest is very easy. 18 Here is most of our work from gcd(1517, 629) = 37. I’ve rearranged each equation on the right to help with the upcoming process: 1517 = (629)(2) + 259 −→ 259 = 1517 − 2[629] 629 = (259)(2) + 111 −→ 111 = 629 − 2[259] 259 = (111)(2) + 37 −→ 37 = 259 − 2(111) Start with the last equation, substitute the value of 111 and simplify: 37 = 259 − 2(111) = 259 − 2(629 − 2[259]) = 5(259) − 2(629). We can now use the same process to eliminate 259: 37 = 5(259) − 2(629) = 5(1517 − 2[629]) − 2(629) = 5(1517) − 12(629). 19 The general case: ax + by = c To find a solution, assuming c = m · gcd(a, b): • Find x, y such that ax + by = gcd(a, b). You can use the method based on the Euclidean Algorithm, although you might find a solution by lucky guess. • Multiply both sides of the equation by m; rearrange to get a(xm) + b(xm) = c. Note: In each case, these methods produce only one solution. We’ll see that if there is at least one solution, then there are infinitely many.