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ADVANCED CALCULUS

An Introduction to Classical Analysis

Dover Publications, Inc.

The Number System

1. Groups. An equivalence relation (=) between mathematical elements a, b, c, ··· (such as numbers, matrices, permutations) must be

E1(reflexive): a = a;

E2(symmetric): a = b implies b = a;

E3 (transitive): a = b, b = c imply a = c.

An operation ([??]) relates every ordered pair of elements a, b to a third element a [??] b = c. If a' = a, b' = b, and a' [??] b' = c', the operation is said to be well defined if c' = c.

A group is composed of a finite or infinite set S of elements, an equivalence relation, and a well-defined operation ("multiplication") which obeys the four postulates M1 to M4:

M1 (Closure). If a and b are in S,

(1) a [??] b is in S.

M2. The operation is associative:

(2) (a [??] b) [??] c = a [??] (b [??] c).

M3. The set S includes an identity element e such that

(3)' e [??] a = a.

M4. Every element a has an inverse a-1 in S such that

(4)' a-1 [??] a = e

The group may satisfy the additional postulate

M5. The operation is commutative:

(5) a [??] b = b [??] a.

In this event the group is called commutative or abelian.

From the postulates M1 to M4 we now make a series of important deductions. We shall call the operation "multiplication," and write ab for a [??] b in the proofs.

I. The cancelation law for left factors:

(6) c [??] a = c [??] b implies a = b.

Proof. Multiply (6) by c-1 on the left. From (2), (4)', and (3)':

(c-1c)a = (c-1c)b,ea = eb,a = b.

II. The identity element may be a right factor:

(3)" a [??] e = a.

Proof. a-1(ae) = (a-1a)e = ee = e = a-1a; hence ae = a from (6).

III. Element a is an inverse of a-1:

(4)" a [??] a-1 = e.

Proof. a-1(aa-1) = (a-1a)a-1 = ea-1 = a-1e; hence aa-1 = e from (6).

IV. The cancelation law for right factors:

(7) a [??] c = b [??] c implies a = b.

Proof. Multiply (7) by c-1 on the right. From (2), (4)", and (3)":

a(cc-1) = b(cc-1), ae = be,a = b.

V. A group has only one identity element.

Proof. If xa = a = ea, x = e from (7).

VI. An element of a group has only one inverse.

Proof. If xa = e = a-1a, x = a-1 from (7).

VII. The inverse of a [??] b is b-1 [??] a-1.

Proof. (b-1a-1)(ab) = b-1(a-1a)b = b-1eb = b-1b = e.

VIII. The equations a [??] x = b and y [??] b have the unique solutions x = a-1 [??] b, y = b [??] a-1.

Proof. The correctness of the given solutions follows by direct substitution; their unicity follows from the cancelation laws.

A group is the simplest mathematical system of any consequence. Its essential properties may be summarized as follows:

A group is closed under an associative, well-defined operation ([??]) and has a unique identity element e such that

(3) a [??] e = e [??] a = a;

moreover, every element a has a unique inverse a-1 such that

(4) a-1 [??] a = a [??] a-1 = e

Note that the operation is commutative in the special cases (3) and (4). But in general the operation need not be commutative (cf. Ex. 3).

Example 1. The matrices

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form a finite group under matrix (row-by-column) multiplication. The matrix e is the identity, and each matrix is its own inverse. Note that ab = c, bc = a, ca = b, and that the group is commutative.

Example 2. The integers (positive, negative, and zero) form a group under addition (+); for the integers are closed under addition, addition is associative, the identity element is zero, and the integers ±n are inverse to each other. Since a + b = b + a, the group is commutative.

The integers do not form a group under multiplication; although we have closure, the associative law, and the identity element 1, an element other than ±1 has no inverse (reciprocal).

Example 3. Permutation Groups. Consider the six permutations of 1, 2, 3, namely:

p1 = 123, p2 = 231, p3 = 312, p4 = 132, p5 = 321, p6 = 213.

We regard each pi as replacing 123 by the digits in its symbol. Thus we may write more explicitly

p2 = 123/231 = 231/312 = 213/321 etc.,

where the upper digits are carried into the corresponding digits below. Such "fractions" are not altered when the upper and lower digits are subjected to the same permutation.

The product of two permutations pipj is defined as the permutation of 123 obtained by first applying pi, then pj. For example,

p2p4 = 123/231 123/132 = 123/231 231/321 = 123/321 = p5.

This "fraction" notation shows that such products are associative; that is, (pipj)pk = pi(pjpk). But the commutative law does not hold in general; thus,

p4p2 = 123/132 123/231 = 123/132 132/213 = 123/213 = p6.

The six elements p1 ··· p6, form a group; for under the product operation just defined, postulates M1 and M2 are fulfilled, p1 is the identity element, and each pi has a unique inverse defined by the "reciprocal" of its fraction symbol; thus,

p2-1 = 231/123 = 123/312 = p3.

The multiplication table for this group, where pi is replaced by i, is as follows

The table shows that p1, p2, p3 alone form a group—a commutative subgroup of the given group.

Similar considerations show that the n! permutations of 1, 2, 3, ···, n form a non-commutative group under the product operation just defined. Moreover the n cyclic permutations 1 2 ··· n, 2 3 ··· 1, ···, form a commutative subgroup.

2. Fields. A field is composed of a set S of elements, an equivalence relation, andtwo well-defined operations, addition (+) and multiplication (·), which satisfy the following postulates.

A. The set S forms a commutative group under addition, whose identity is called zero.

M. The set S with zero omitted forms a commutative group under multiplication.

D. Multiplication is distributive with respect to addition:

(1) a · (b + c) = a · b + a · c.

The identity under addition is written 0 ("zero"), and the additive inverse of a is written -a and called the negative of a:

(2)(3) 0 + a = a, -a + a = 0.

The identity under multiplication is written 1 ("unity"), and the multiplicative inverse of a is written a-1 and called the reciprocal of a:

(4)(5) 1 · a = a, a-1 · a = 1.

The field postulates may now be displayed as follows:

A1. a + b belongs to S;

A2. (a + b) + c = a + (b + c);

A3. 0 + a = a;

A4. -a + a = 0;

A5. a + b = b + a;

M1. a · b belongs to S;

M2. (a · b) · c = a · (b · c);

M3. 1 · a = a;

M4. a-1 · a = 1, (a ≠ 0);

M5. a · b = b · a;

D. a · (b + c) = a · b + a · c.

In a field both cancelation laws,

(6) c + a = c + b implies a = b,

(7) c · a = c · b implies a = b if c ≠ 0,

are valid (§ 1). Note that zero is not an element of the multiplicative group.

Zero has the important property

(8) a · 0 = 0

for all elements of the field. For, from 0 + b = b and D, we have

a · 0 + a · b = a · b = 0 + a · b,

and on canceling a · b we get (8).

Equation (8) shows that zero has no reciprocal. This explains the exclusion of zero from the multiplicative group. We also have the important

Theorem. If a, b are elements of a field, a · b = 0 when and only when at least one factor is zero.

Proof: If b = 0, the equation is satisfied for any a.

If b ≠ 0, multiply the equation by its reciprocal b-1; then

(a · b) · b-1 = 0, a · (b · b-1) = 0,a = 0.

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Excerpted from ADVANCED CALCULUS by LOUIS BRAND. Copyright © 2014 Dover Publications, Inc.. Excerpted by permission of Dover Publications, Inc..
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