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MATH 537 Class Notes

Ed Belk

Fall, 2014

1 Week One

1.1 Lecture One

Instructor: Greg Martin, Office Math 212

Text: Niven, Zuckerman & Montgomery

Conventions: N will denote the set of positive integers, and N0 the set of nonnegative integers. Unlessotherwise stated, all variables are assumed to be elements of N.

1.2 Divisibility

Definition: Let a, b Z with a 6= 0. Then a is said to divide b, denoted a|b, if there exists some c Z suchthat ac = b. If in addition a N, then a is called a divisor of b.

Properties of Divisibility: For all a, b, c Z with a 6= 0, one has:

If a|b then a| b

1|b, b|b, a|0

If a|b and b|a then a = b

If a|b and a|c, then a|(bx+ cy) for any x, y Z

If we assume that a and b are positive, we also have

If a|b then a b

The Division Algorithm: Let a, b N. Then there exist unique natural numbers q and r such that:

1. b = aq + r, and

2. 0 r < a

Proof : We prove existence first; consider the set

R = {b an : n N0} N0.

By the well-ordering axiom, R has a least element r, and we define q to be the nonnegative integer q such thatb aq = r. Then b = aq + r and r 0; moreover, if r a then one has

0 r a = (b aq) a = b a(q + 1) < b aq + r,

contradicting the minimality of r R, and we are done.

1

Now, suppose q and r are such that we have

b = aq + r = aq + r.

Without loss of generality we may assume than r r. Then

r r = (b aq) (b aq) = a(q q) a|(r r);

but 0 r r r < a, and so the above equation is a contradiction unless r r = 0, and the result isimmediate.

Greatest Common Divisor: Given any two integers a and b not both equal to zero, we define their greatestcommon divisor (commonly abbreviated gcd) to be the largest d N such that d|a and d|b; we write d = (a, b).Note that because a and b each have only finitely many divisors, the gcd is always well-defined.

Theorem 1.1.1 Let a, b Z, not both equal to zero. Then:

1. (a, b) = minS, where S = ({ax+ by : x, y Z} N), and

2. For any c Z such that c|a and c|b, we have c|(a, b).

The existence of integers x, y so that ax+ by = (a, b) as in part (1) is known as Bezouts identity.

Proof : 1. Let m = minS, with u and v such that m = au+ bv, and let g = (a, b); note that m a. Since g|aand g|b, we know from the properties of divisibility that g|m and so g m. Now, if m - a then by the divisionalgorithm we may write a = mq + r with 0 < r < m, and thus

r = amq = a q(au+ bv) = a(1 qu) + b(qv) S,

and we deduce that r m = minS, a contradiction; thus m|a. In the same fashion we show m|b, and so bydefinition m (a, b) = g, and we are done.

2. If c|a and c|b, then we know c|(ax + by) for every x, y Z, and in particular for those u, v such that(a, b) = au+ bv, whose existence is guaranteed by part 1.

2

1.2 Lecture Two

Recall: Bezouts identity states that (a, b) is the smallest positive integer that may be written ax+ by, wherex, y Z.

Proposition 1.2.1 For a, b N, one has (ma,mb) = m(a, b).

Corollary 1: If d|a, d|b, then(ad ,

bd

)= 1d(a, b); in particular,

(a

(a,b) ,b

(a,b)

)= 1.

Proof : Set g = (a, b), so that we may writeax+ by = g,

for some x, y Z. Thenmg = (ma)x+ (mb)y, thus mg (ma,mb).

Furthermore, g|a and so mg|ma; similarly mg|mb, thus mg (ma,mb), and we are done.

Definition: Two integers a and b are called relatively prime (or coprime) if (a, b) = 1.

nb. We observe that (a, b) = 1 if and only if there exist x, y such that ax+by = 1. The corresponding statementwith (a, b) = k > 1 is not, in general, true, however it is the case that

ax+ by = k (a, b)|k.

Proposition 1.2.2 If (a, n) = (b, n) = 1, then (ab, n) = 1.

Proof : Suppose we have u, v, x, y so that au+ nv = bx+ ny = 1; then we have

1 = 1 1 = (au+ nv)(bx+ ny) = ab(ux) + n(auy + bvx+ nvy),

and the result is immediate.

[Aside: Compare with the analagous result in commutative algebra. If R is a commutative, unital ring andI, J,K R are ideals such that I +K = J +K = R, then IJ +K = R.]

Proposition 1.2.3 If a|c, b|c, and (a, b) = 1, then ab|c. (Note that this is not, in general, true for (a, b) > 1,e.g. a = b = c = 2.)

Proof : Choose m,n, x, y so that c = am = bn and ax+ by = 1. Then

c = cax+ cby = (bn)ax+ (am)by = ab(nx+my),

and we deduce that ab|c.

Theorem 1.2.4 (Theorem 1.10, Niven) If d|ab and (b, d) = 1, then d|a.

Proof : Exercise.

nb. If d|a, d|b, then d|b+ax for any x Z. In fact, the condition is also necessary, as b = (b+ax)x(a).

The Euclidean Algorithm: How can we find the gcd of two integers, for example 537 and 105?

By the division algorithm, we have 537 = 5 105 + 12, and so by the above note we know (537, 105) = (105, 12).Repeating this process, we see

105 = 8 12 + 9 (105, 12) = (12, 9);

12 = 1 9 + 3 (12, 9) = (9, 3);

3

9 = 3 3 + 0 (9, 3) = (3, 0) = 3.

Thus (537, 105) = 3.

Notation: The least common multiple of a and b is denoted lcm(a, b) or, more commonly, [a, b].

Exercise: Show that (a, b)[a, b] = ab.

1.3 Primes

Definition: A natural number n is called prime if it has exactly two divisors. n is called composite if thereexists some d with 1 < d < n such that d|n. The integer n = 1 is neither prime nor composite.

Notation: Unless otherwise stated, p will denote a prime number.

Lemma 1.2.5 (Euclids lemma) If p|ab, then p|a or p|b.

Proof : Suppose p - b. Then (p, b) = 1, and so by theorem 1.2.4 we know that p|a.

Theorem 1.2.6 (The Fundamental Theorem of Arithmetic) Every n N, n > 2 may be written as the productof primes; moreover this expression is unique up to reordering of the factors.

Proof : (existence) We use strong induction. The case n = 2 is trivial from the definition of a prime, thereforesuppose n > 2. If n is prime we have the trivial factorization n = n, otherwise we may write n = ab, with1 < a < n and 1 < b < n. By the inductive hypothesis we may write a = p1p2 pk, b = q1q2 ql, with eachpi, qj prime, and the result is immediate.

(uniqueness) Let n N and suppose we have

n = p1p2 pk = q1q2 ql, each pi, qj prime.

Since p1|q1q2 ql we have by lemma 1.2.5 that p1|q1 or p1|q2 ql. Repeating this process as many times asnecessary, we find qt such that p1|qt, and by relabelling the qj if necessary we will assume t = 1. Since p1 6= 1this implies that p1 = q1, as q1 has no other factors. We then cancel p1 = q1 on both sides of the equation andwe have

p2p3 pk = q2q3 ql.

We apply the same argument to this expression to obtain p2 = q2, p3 = q3, and so on; it follows that k = l, andwe are done.

4

2 Week Two

2.1 Lecture Three

Doing a linear algebra problem backwards. Consider the augmented matrix(1 0 5370 1 105

);

this system clearly has solution

(xy

)=

(537105

). Moreover, from basic linear algebra we know that the application

of elementary row operations to this augmented system will not change the solution; therefore, with R1, R2

respectively denoting the first and second row of the matrix, we observe that

(xy

)=

(537105

)is also a solution

to the augmented matrices (1 5 120 1 105

)(R1 R1 5R2),(

1 5 128 41 9

)(R2 R2 8R1),(

9 46 38 41 9

)(R1 R1 R2),(

9 46 335 179 0

)(R2 R2 3R1).

Thus we have the matrix equation (9 4635 179

)(537105

)=

(30

).

The first entry of this equation indicates that 9(537) + (46)(105) = 3 = (537, 105), while the entries in thesecond row of the matrix are 35 = 105(537,105) and 179 =

537(537,105) . This operation is known as the extended

Euclidean algorithm.

Lemma 2.1.1 Let a, b N and use the division algorithm to write b = aq + r with 0 r < a. Then a|b if andonly if r = 0.

Proof : If r = 0 then b = aq and we are done. Conversely, if a|b then a|bax for every x, and since r = abq < a,we must have r = 0.

Theorem 2.1.2 (Euclids theorem) There are infinitely many prime numbers.

Proof : It suffices to show that every finite list of primes excludes at least one prime number. Let {p1, p2, . . . , pk}be a set of finitely many primes and let

N = p1p2 pk + 1.

Then N 2 and so by the fundamental theorem of arithmetic N is the product of primes, so there exists someprime p such that p|N . Applying the division algorithm with N and any pj yields

N = pj(p1 pj1pj+1 pk) + 1,

which (since 1 < pj) by lemma 2.1.1 implies that pj - N for any j. Thus we deduce that p 6= pj for anyj = 1, 2, . . . , k, and therefore that the set of primes {p1, p2, . . . , pk} is not exhaustive.

5

2.1 Congruences

Definition: Let m Z,m 6= 0. Given a, b Z, we say that a is congruent to b modulo m , writtena b mod m, if m|(b a). For example, we have

53 7 mod 23, but 5 6 37 mod 23.

Lemma 2.1.3 For fixed m 6= 0, congruence modulo m is an equivalence relation.

Proof : Clearly a a mod m because m|0 = a a, which proves reflexivity. Symmetry is an immediateconsequence of the fact that m|(b a) m|(a b), and to prove transitivity we observe that

a b mod m, b c mod m m|(b a),m|(c b) m|(c b) + (b a) = (c a),

and we are done.

Thus in particular, congruence modulo m (as every equivalence relation) partitions