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# Elementary Mathematics from an Advanced Standpoint: Geometry

## Overview

"Nothing comparable to it." — *Mathematics Teacher. *This comprehensive three-part treatment begins with a consideration of the simplest geometric manifolds: line-segment, area, and volume as relative magnitudes; the Grassmann determinant principle for the plane and the Grassmann principle for space; classification of the elementary configurations of space according to their behavior under transformation of rectangular coordinates; and derivative manifolds. The second section, on geometric transformations, examines affine and projective transformations; higher point transformations; transformations with change of space element; and the theory of the imaginary. The text concludes with a systematic discussion of geometry and its foundations. 1939 edition. 141 figures.

## Product Details

ISBN-13: | 9780486434810 |
---|---|

Publisher: | Courier Corporation |

Publication date: | 06/18/2004 |

Series: | Dover Books on Mathematics Series |

Pages: | 226 |

Sales rank: | 1,034,720 |

Product dimensions: | 5.50(w) x 8.50(h) x 0.48(d) |

## Read an Excerpt

#### Elementary Mathematics From An Advanced Standpoint Geometry

**By Felix Klein, E. R. Hedrick, C. A. Noble**

**Dover Publications, Inc.**

**Copyright © 2004 Dover Publications, Inc.**

All rights reserved.

ISBN: 978-0-486-16741-1

All rights reserved.

ISBN: 978-0-486-16741-1

CHAPTER 1

**PART ONE**

**THE SIMPLEST GEOMETRIC MANIFOLDS**

**I. LINE-SEGMENT, AREA, VOLUME AS RELATIVE MAGNITUDES**

You will notice by this chapter heading that I am following the intention announced above, of examining simultaneously the corresponding magnitudes on the straight line, in the plane, and in space. At the same time, however, we shall take into account the principle of fusion by making use at once of the *rectangular system of coordinates* for the purpose of analytic formulation.

If we have a *line-segment,* let us think of it as laid upon the *x* axis. If the abscissas of its endpoints are *x*1 and *x*2, its *length* is *x*1 - *x*2, and we may write this difference in the form of the determinant

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Similarly, the *area of a triangle* in the *xy* plane which is formed by the three points 1, 2, 3, with coordinates (*x*1, *y*1), (*x*2, *y*2), (*x*3, *y*3), will be

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Finally, we have, for the *volume of the tetrahedron* made by the four points 1, 2, 3, 4, with coordinates (*x*1, *y*1, *z*1), ..., (*x*4, *y*4, *z*4), the formula

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

We say ordinarily that the length, or, as the case may be, the area or the volume, is equal to the *absolute value* of these several magnitudes, whereas, actually, our formulas furnish, over and above that, a *definite sign,* which depends upon the order in which the points are taken. We shall make it a fundamental rule always to take into account in geometry the signs which the analytic formulas supply. We must accordingly inquire as to the *geometric significance of the sign in these determinations of content.*

It is important, therefore, how we choose the *system of rectangular coordinates.* Let us, then, at the outset, adopt a convention, which is, of course, arbitrary, but which must be binding in all cases. In the case of *one dimension,* we shall think of the positive *x* axis as always pointing to the right. In the plane, the positive *x* axis will be directed toward the right, the positive *y* axis upward (see **Fig. 1**). If we were to let the *y* axis point downward, we should have an essentially different coordinate system, one which would be a reflection of the first and not superimposable upon it by mere motion in the plane, i.e., without going out into space. Finally, the *coordinate system in space* will be obtained from the one in the plane by adding to the latter a *z* axis directed positively to the *front* (see **Fig. 2**). A choice of the *z* axis pointing positively to the rear would give, again, an essentially different coordinate system, one which could not be made to coincide with ours by any movement in space.

If we always adhere to these conventions, we shall find the *interpretation of our signs in simple geometric properties of the succession of points as these are determined by their numbering.*

For the *segment* (1, 2) this property is obvious: *The expression x*1 - *x*2*for its length is positive or negative according as point* 1 *lies to the right or to the left of point* 2.

In the case of the *triangle,* we obtain: *The formula for area has the positive or the negative sign according as a circuit about the triangle from the vertex* 1 *to* 3 *via* 2 *turns out to be counterclockwise or the reverse.* We shall prove this by taking, first, a conveniently placed special triangle, evaluating the determinant which expresses its area, and then, through consideration of continuity, passing to the general case. We consider that triangle which has, as its first vertex, the unit point on the *x* axis (*x*1 = 1,*y*1 = 0), as its second, the unit point on the *y* axis (*x*2 = 0, *y*2 = 1), and as its third the origin (*x*3 = 0, *y*3 = 0). According to our agreement about the system of coordinates, we must traverse the boundary of this triangle in the counterclockwise sense (see **Fig. 3**), and our formula for its area yields the positive value:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Now we can bring the vertices of this triangle, by continuous deformation, into coincidence with those of any other triangle traversed in the same sense, and we can do this in such a way that the three vertices of the triangle shall at no time be collinear. In this process, our determinant changes value continuously, and since it vanishes only when the points 1, 2, 3 are collinear, it must always remain positive. This establishes the fact that the area of any triangle whose boundary is traversed in counterclockwise sense is positive. If we interchange two vertices of the original triangle, we see at once that every triangle which is traversed *in clockwise sense* has negative area.

We can now treat the *tetrahedron* in analogous fashion. We start, again, with a conveniently placed tetrahedron. As first, second, and third vertices, we choose, in order, the unit points on the *x, y,* and *z* axes, and as fourth vertex, the origin. Its volume is therefore

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

It follows, as before, that every tetrahedron which can be obtained from this one by continuous deformation during which the four vertices are never complanar (i.e., during which the determinant never vanishes), has positive volume. But one can characterize all these tetrahedrons by the sense in which the one face (2, 3, 4) is traversed when it is looked at from the vertex 1. In this way we obtain the result: *The volume of the tetrahedron* (1, 2, 3, 4) *which our formula yields is positive if the vertices* 2, 3, 4, *looked at from vertex* 1, *follow one another in counterclockwise sense; otherwise it is negative.*

We have thus, from our analytic formulas, actually deduced geometric rules which permit us to assign a definite sign to any segment, any triangle, any tetrahedron, if the vertices are given in a definite order. Great advantages are thus gained over the ordinary elementary geometry which considers length and contents as absolute magnitudes. Indeed, we can mention general simple theorems where elementary geometry must distinguish numerous cases according to the appearance of the figure.

Let me begin with a very primitive example, the *ratio of the segments* made by three points on a line, say the *x* axis. Denoting the three points by 1, 2, and 4 (see **Fig. 5**), as is convenient in view of what is to follow, we see that the ratio in question will be given by the formula *S* = (*x*1 - *x*2)/(*x*1 - *x*4), and it is clear that this quotient is positive or negative according as the point 1 lies outside or inside the segment (2, 4). If, as is customary in elementary presentations, we give only the absolute value | *S* | = | *x*1 - *x*2 |/| *x*1 - *x*4|, we must always either refer to the figure, or state in words whether we have in mind an inside or an outside point, which is, of course, more complicated. *The introduction of the sign thus takes account of the different possible orders of the points on the line,* a fact to which we shall often have to refer in the course of these lectures.

If we now add a fourth point 3, we can set up the *cross ratio* of the four points, that is,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

This expression has again a definite sign, and we see at once that *D*< 0 when the pair of points 1 and 3, on the one hand, and the pair 2 and 4, on the other hand, mutually separate one another; and that *D* > 0 in the opposite case, i.e., when 1 and 3 lie both outside or both inside the segment 2, 4. (See **Figs. 6** and **7**.) Thus there are always two essentially different arrangements which yield the same absolute value *D.* If this absolute value alone is given, we must give the arrangement also. For example, if we define harmonic points by the equation *D* = 1, as is still the custom, unfortunately, in the schools, one must include in the definition the demand that the two pairs of points separate each other, whereas in our plan the *one* statement *D* = -1 is sufficient. This practice of taking account of the sign is especially useful in *projective* geometry, in which, as you know, the cross ratio plays a leading role. There we have the familiar theorem that four points on a line have the same cross ratio as the four points which arise when we project the given points from a center upon another line (perspective). If we now consider the cross ratio as a relative magnitude, affected by a sign, the converse of this theorem holds without exception: If each of two sets of four points lies on one of two lines, and if they have the same cross ratio, they can be derived one from the other by projection, either single or repeated. For example, in **Fig. 8**, the sets 1, 2, 3, 4, and 1", 2", 3", 4" are in perspective to 1', 2', 3', 4' if we use the centers *P* and *P'*. If, however, we know only the absolute value of *D,* the corresponding theorem does not hold in this simple form; we should have to make a special assumption about the arrangement of the points.

We have a more fruitful field if we consider *applications of our triangle formula.* Let us first select any point 0 in the interior of a triangle (1, 2, 3) and let us join 0 to each of the vertices (see **Fig. 9**). Then the sum of the areas of the three partial triangles, thought of in the elementary sense as absolute magnitudes, is equal to the area of the original triangle. Thus we may write | (1, 2, 3) | = | (0, 2, 3) | + | (0, 3, 1) | + | (0, 1, 2) |. The figure shows that, in all the triangles, the order of the vertices, as they appear in the above equation, is counterclockwise. Hence the areas (1, 2, 3), (0, 2, 3), (0, 3, 1), (0, 1, 2), are all positive in the sense of our general definition, so that we may write our formula in the form

(1, 2, 3) = (0, 2, 3) + (0, 3, 1) + (0, 1, 2).

Now I assert *that the same formula also holds when* 0 *lies outside the triangle, and, further, when* 0, 1, 2, 3 *are any four points whatever in the plane.* If we take **Fig. 10**, for example, we see that the boundaries of (0, 2, 3) and (0, 3, 1) are traversed in counterclockwise sense, but that of (0, 1, 2) is traversed in the clockwise sense, so that our formula for the absolute areas would give | (1, 2, 3) | = | (0, 2, 3) | + | (0, 3, 1) | - | (0, 1, 2) |. The figure verifies the correctness of this equation.

We shall give a general proof of our theorem by means of the *analytic definition,* whereby we shall recognize in our formula a well known theorem of the theory of determinants. For convenience, let us take the point *O* as our origin *x* = 0, *y* = 0, which is obviously no essential specialization, and let us substitute for each of the four triangle areas the appropriate determinant. Then, omitting everywhere the factor , we are to prove that, for arbitrary *x*1, ..., *y*3, the following relation holds:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

The value of each of the determinants on the right will remain unchanged if we replace the second and third 1 of the last column by zeros, since these elements enter only those minors which are multiplied by zero when we develop according to the top row. If we now make a cyclic interchange of rows in the last two determinants, which is permissible in determinants of the third, or, in fact, of any odd order, we can write our equation in the following form:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

But this is an identity, for on the right there are only the minors of the last column of the first determinant, so that we have merely the well known development of this determinant according to the elements of a column. Thus, at one stroke, we have proved our theorem for all possible positions of the four points.

We can generalize this formula so that it will give the *area of any polygon.* Imagine that you had, say, the following problem in surveying: To determine the area of a rectilinear field after having measured the coordinates of the corners 1, 2, ..., *n* - 1, *n* (see **Fig. 11**). One who is not accustomed to operate with signs would then sketch the shape of the polygon, divide it up into triangles by drawing diagonals, perhaps, and then according to the particular shape of the field, paying especial regard to whether some of the angles are re-entrant, find the area as the sum or difference of the areas of the partial triangles. However, we can give at once a general formula which will give the correct result quite mechanically without any necessity of looking at the figure: If *O* is any point in the plane, say the origin, then the area of our polygon, the boundary being traversed in the sense 1, 2, ..., *n,* will be (1, 2, 3, ..., *n*) = (0, 1, 2) + (0, 2, 3) + ... + (0, *n* 1, *n*) + (0, *n,* 1), whereby each triangle is to be taken with the sign determined by the sense in which the circuit about it is made. *The formula yields the area of the polygon positively or negatively according as the circuit of the polygon in the sense* 1, 2, ..., *n is counterclockwise or not.* It will suffice to write this formula. You yourselves can easily supply the proof.

Instead of pursuing this example further, I prefer to take up some especially interesting cases, which, to be sure, could not arise in surveying, namely, cases of *polygons which overlap themselves* as in the adjoining quadrilateral (see **Fig. 12**). If we wish here to talk at all about definite area, it can only be the value which our formula yields. Let us consider what this value means geometrically. At the outset we notice that this must be independent of the particular location of the point *O.* Let us place *O,* as conveniently as possible, at the point where the overlappings cross. Then the triangles (0, 1, 2) and (0, 3, 4) will be zero and there remains:

(1, 2, 3, 4) = (0, 2, 3) + (0, 4, 1).

The first triangle has negative area, the second positive area; hence the area of our overlapping quadrilateral, if we prescribe a circuit in the sense (1, 2, 3, 4), is equal to the absolute value of the area of the part (0, 4, 1) that was traversed in counterclockwise sense, *diminished* by that of the part (0, 2, 3) that was traversed in clockwise sense.

As a second example, let us examine the adjoined *star pentagon* (see **Fig. 13**). If we take *O* in the middle part, all the partial triangles in the sum

(0, 1, 2) + (0, 2, 3) + ... + (0, 5, 1)

are traversed in the positive sense; their sum covers the five-cornered central part of the figure twice, and each of the five tips once. If we again consider a positive circuit around our polygon (1, 2, 3, 4, 5, 1), we see that every part of the boundary is traversed counterclockwise and that, in particular, we have passed twice around the portion of the polygon which is doubly counted in the area, but only once around the remaining portions.

From these two examples we can infer the following *general rule: For any rectilinear polygon with arbitrary overlappings, our formula yields, as total area, the algebraic sum of the separate partial areas bounded by the polygonal line, whereby each of these partial areas is counted as often as we pass around its boundary when the circuit* (1, 2, 3, ..., *n,* 1) *is made once, this counting to be made positively or negatively according as we pass around the partial area in counterclockwise or clockwise sense.* You will have no difficulty in establishing the truth of this general theorem.

*(Continues...)*

Excerpted fromElementary Mathematics From An Advanced Standpoint GeometrybyFelix Klein, E. R. Hedrick, C. A. Noble. Copyright © 2004 Dover Publications, Inc.. Excerpted by permission of Dover Publications, Inc..

All rights reserved. No part of this excerpt may be reproduced or reprinted without permission in writing from the publisher.

Excerpts are provided by Dial-A-Book Inc. solely for the personal use of visitors to this web site.

## First Chapter

#### Elementary Mathematics From An Advanced Standpoint Geometry

**By Felix Klein, E. R. Hedrick, C. A. Noble**

**Dover Publications, Inc.**

**Copyright © 2004 Dover Publications, Inc.**

All rights reserved.

ISBN: 978-0-486-16741-1

All rights reserved.

ISBN: 978-0-486-16741-1

CHAPTER 1

**PART ONE**

**THE SIMPLEST GEOMETRIC MANIFOLDS**

**I. LINE-SEGMENT, AREA, VOLUME AS RELATIVE MAGNITUDES**

You will notice by this chapter heading that I am following the intention announced above, of examining simultaneously the corresponding magnitudes on the straight line, in the plane, and in space. At the same time, however, we shall take into account the principle of fusion by making use at once of the *rectangular system of coordinates* for the purpose of analytic formulation.

If we have a *line-segment,* let us think of it as laid upon the *x* axis. If the abscissas of its endpoints are *x*1 and *x*2, its *length* is *x*1 - *x*2, and we may write this difference in the form of the determinant

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Similarly, the *area of a triangle* in the *xy* plane which is formed by the three points 1, 2, 3, with coordinates (*x*1, *y*1), (*x*2, *y*2), (*x*3, *y*3), will be

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Finally, we have, for the *volume of the tetrahedron* made by the four points 1, 2, 3, 4, with coordinates (*x*1, *y*1, *z*1), ..., (*x*4, *y*4, *z*4), the formula

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

We say ordinarily that the length, or, as the case may be, the area or the volume, is equal to the *absolute value* of these several magnitudes, whereas, actually, our formulas furnish, over and above that, a *definite sign,* which depends upon the order in which the points are taken. We shall make it a fundamental rule always to take into account in geometry the signs which the analytic formulas supply. We must accordingly inquire as to the *geometric significance of the sign in these determinations of content.*

It is important, therefore, how we choose the *system of rectangular coordinates.* Let us, then, at the outset, adopt a convention, which is, of course, arbitrary, but which must be binding in all cases. In the case of *one dimension,* we shall think of the positive *x* axis as always pointing to the right. In the plane, the positive *x* axis will be directed toward the right, the positive *y* axis upward (see **Fig. 1**). If we were to let the *y* axis point downward, we should have an essentially different coordinate system, one which would be a reflection of the first and not superimposable upon it by mere motion in the plane, i.e., without going out into space. Finally, the *coordinate system in space* will be obtained from the one in the plane by adding to the latter a *z* axis directed positively to the *front* (see **Fig. 2**). A choice of the *z* axis pointing positively to the rear would give, again, an essentially different coordinate system, one which could not be made to coincide with ours by any movement in space.

If we always adhere to these conventions, we shall find the *interpretation of our signs in simple geometric properties of the succession of points as these are determined by their numbering.*

For the *segment* (1, 2) this property is obvious: *The expression x*1 - *x*2*for its length is positive or negative according as point* 1 *lies to the right or to the left of point* 2.

In the case of the *triangle,* we obtain: *The formula for area has the positive or the negative sign according as a circuit about the triangle from the vertex* 1 *to* 3 *via* 2 *turns out to be counterclockwise or the reverse.* We shall prove this by taking, first, a conveniently placed special triangle, evaluating the determinant which expresses its area, and then, through consideration of continuity, passing to the general case. We consider that triangle which has, as its first vertex, the unit point on the *x* axis (*x*1 = 1,*y*1 = 0), as its second, the unit point on the *y* axis (*x*2 = 0, *y*2 = 1), and as its third the origin (*x*3 = 0, *y*3 = 0). According to our agreement about the system of coordinates, we must traverse the boundary of this triangle in the counterclockwise sense (see **Fig. 3**), and our formula for its area yields the positive value:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Now we can bring the vertices of this triangle, by continuous deformation, into coincidence with those of any other triangle traversed in the same sense, and we can do this in such a way that the three vertices of the triangle shall at no time be collinear. In this process, our determinant changes value continuously, and since it vanishes only when the points 1, 2, 3 are collinear, it must always remain positive. This establishes the fact that the area of any triangle whose boundary is traversed in counterclockwise sense is positive. If we interchange two vertices of the original triangle, we see at once that every triangle which is traversed *in clockwise sense* has negative area.

We can now treat the *tetrahedron* in analogous fashion. We start, again, with a conveniently placed tetrahedron. As first, second, and third vertices, we choose, in order, the unit points on the *x, y,* and *z* axes, and as fourth vertex, the origin. Its volume is therefore

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

It follows, as before, that every tetrahedron which can be obtained from this one by continuous deformation during which the four vertices are never complanar (i.e., during which the determinant never vanishes), has positive volume. But one can characterize all these tetrahedrons by the sense in which the one face (2, 3, 4) is traversed when it is looked at from the vertex 1. In this way we obtain the result: *The volume of the tetrahedron* (1, 2, 3, 4) *which our formula yields is positive if the vertices* 2, 3, 4, *looked at from vertex* 1, *follow one another in counterclockwise sense; otherwise it is negative.*

We have thus, from our analytic formulas, actually deduced geometric rules which permit us to assign a definite sign to any segment, any triangle, any tetrahedron, if the vertices are given in a definite order. Great advantages are thus gained over the ordinary elementary geometry which considers length and contents as absolute magnitudes. Indeed, we can mention general simple theorems where elementary geometry must distinguish numerous cases according to the appearance of the figure.

Let me begin with a very primitive example, the *ratio of the segments* made by three points on a line, say the *x* axis. Denoting the three points by 1, 2, and 4 (see **Fig. 5**), as is convenient in view of what is to follow, we see that the ratio in question will be given by the formula *S* = (*x*1 - *x*2)/(*x*1 - *x*4), and it is clear that this quotient is positive or negative according as the point 1 lies outside or inside the segment (2, 4). If, as is customary in elementary presentations, we give only the absolute value | *S* | = | *x*1 - *x*2 |/| *x*1 - *x*4|, we must always either refer to the figure, or state in words whether we have in mind an inside or an outside point, which is, of course, more complicated. *The introduction of the sign thus takes account of the different possible orders of the points on the line,* a fact to which we shall often have to refer in the course of these lectures.

If we now add a fourth point 3, we can set up the *cross ratio* of the four points, that is,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

This expression has again a definite sign, and we see at once that *D*< 0 when the pair of points 1 and 3, on the one hand, and the pair 2 and 4, on the other hand, mutually separate one another; and that *D* > 0 in the opposite case, i.e., when 1 and 3 lie both outside or both inside the segment 2, 4. (See **Figs. 6** and **7**.) Thus there are always two essentially different arrangements which yield the same absolute value *D.* If this absolute value alone is given, we must give the arrangement also. For example, if we define harmonic points by the equation *D* = 1, as is still the custom, unfortunately, in the schools, one must include in the definition the demand that the two pairs of points separate each other, whereas in our plan the *one* statement *D* = -1 is sufficient. This practice of taking account of the sign is especially useful in *projective* geometry, in which, as you know, the cross ratio plays a leading role. There we have the familiar theorem that four points on a line have the same cross ratio as the four points which arise when we project the given points from a center upon another line (perspective). If we now consider the cross ratio as a relative magnitude, affected by a sign, the converse of this theorem holds without exception: If each of two sets of four points lies on one of two lines, and if they have the same cross ratio, they can be derived one from the other by projection, either single or repeated. For example, in **Fig. 8**, the sets 1, 2, 3, 4, and 1", 2", 3", 4" are in perspective to 1', 2', 3', 4' if we use the centers *P* and *P'*. If, however, we know only the absolute value of *D,* the corresponding theorem does not hold in this simple form; we should have to make a special assumption about the arrangement of the points.

We have a more fruitful field if we consider *applications of our triangle formula.* Let us first select any point 0 in the interior of a triangle (1, 2, 3) and let us join 0 to each of the vertices (see **Fig. 9**). Then the sum of the areas of the three partial triangles, thought of in the elementary sense as absolute magnitudes, is equal to the area of the original triangle. Thus we may write | (1, 2, 3) | = | (0, 2, 3) | + | (0, 3, 1) | + | (0, 1, 2) |. The figure shows that, in all the triangles, the order of the vertices, as they appear in the above equation, is counterclockwise. Hence the areas (1, 2, 3), (0, 2, 3), (0, 3, 1), (0, 1, 2), are all positive in the sense of our general definition, so that we may write our formula in the form

(1, 2, 3) = (0, 2, 3) + (0, 3, 1) + (0, 1, 2).

Now I assert *that the same formula also holds when* 0 *lies outside the triangle, and, further, when* 0, 1, 2, 3 *are any four points whatever in the plane.* If we take **Fig. 10**, for example, we see that the boundaries of (0, 2, 3) and (0, 3, 1) are traversed in counterclockwise sense, but that of (0, 1, 2) is traversed in the clockwise sense, so that our formula for the absolute areas would give | (1, 2, 3) | = | (0, 2, 3) | + | (0, 3, 1) | - | (0, 1, 2) |. The figure verifies the correctness of this equation.

We shall give a general proof of our theorem by means of the *analytic definition,* whereby we shall recognize in our formula a well known theorem of the theory of determinants. For convenience, let us take the point *O* as our origin *x* = 0, *y* = 0, which is obviously no essential specialization, and let us substitute for each of the four triangle areas the appropriate determinant. Then, omitting everywhere the factor , we are to prove that, for arbitrary *x*1, ..., *y*3, the following relation holds:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

The value of each of the determinants on the right will remain unchanged if we replace the second and third 1 of the last column by zeros, since these elements enter only those minors which are multiplied by zero when we develop according to the top row. If we now make a cyclic interchange of rows in the last two determinants, which is permissible in determinants of the third, or, in fact, of any odd order, we can write our equation in the following form:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

But this is an identity, for on the right there are only the minors of the last column of the first determinant, so that we have merely the well known development of this determinant according to the elements of a column. Thus, at one stroke, we have proved our theorem for all possible positions of the four points.

We can generalize this formula so that it will give the *area of any polygon.* Imagine that you had, say, the following problem in surveying: To determine the area of a rectilinear field after having measured the coordinates of the corners 1, 2, ..., *n* - 1, *n* (see **Fig. 11**). One who is not accustomed to operate with signs would then sketch the shape of the polygon, divide it up into triangles by drawing diagonals, perhaps, and then according to the particular shape of the field, paying especial regard to whether some of the angles are re-entrant, find the area as the sum or difference of the areas of the partial triangles. However, we can give at once a general formula which will give the correct result quite mechanically without any necessity of looking at the figure: If *O* is any point in the plane, say the origin, then the area of our polygon, the boundary being traversed in the sense 1, 2, ..., *n,* will be (1, 2, 3, ..., *n*) = (0, 1, 2) + (0, 2, 3) + ... + (0, *n* 1, *n*) + (0, *n,* 1), whereby each triangle is to be taken with the sign determined by the sense in which the circuit about it is made. *The formula yields the area of the polygon positively or negatively according as the circuit of the polygon in the sense* 1, 2, ..., *n is counterclockwise or not.* It will suffice to write this formula. You yourselves can easily supply the proof.

Instead of pursuing this example further, I prefer to take up some especially interesting cases, which, to be sure, could not arise in surveying, namely, cases of *polygons which overlap themselves* as in the adjoining quadrilateral (see **Fig. 12**). If we wish here to talk at all about definite area, it can only be the value which our formula yields. Let us consider what this value means geometrically. At the outset we notice that this must be independent of the particular location of the point *O.* Let us place *O,* as conveniently as possible, at the point where the overlappings cross. Then the triangles (0, 1, 2) and (0, 3, 4) will be zero and there remains:

(1, 2, 3, 4) = (0, 2, 3) + (0, 4, 1).

The first triangle has negative area, the second positive area; hence the area of our overlapping quadrilateral, if we prescribe a circuit in the sense (1, 2, 3, 4), is equal to the absolute value of the area of the part (0, 4, 1) that was traversed in counterclockwise sense, *diminished* by that of the part (0, 2, 3) that was traversed in clockwise sense.

As a second example, let us examine the adjoined *star pentagon* (see **Fig. 13**). If we take *O* in the middle part, all the partial triangles in the sum

(0, 1, 2) + (0, 2, 3) + ... + (0, 5, 1)

are traversed in the positive sense; their sum covers the five-cornered central part of the figure twice, and each of the five tips once. If we again consider a positive circuit around our polygon (1, 2, 3, 4, 5, 1), we see that every part of the boundary is traversed counterclockwise and that, in particular, we have passed twice around the portion of the polygon which is doubly counted in the area, but only once around the remaining portions.

From these two examples we can infer the following *general rule: For any rectilinear polygon with arbitrary overlappings, our formula yields, as total area, the algebraic sum of the separate partial areas bounded by the polygonal line, whereby each of these partial areas is counted as often as we pass around its boundary when the circuit* (1, 2, 3, ..., *n,* 1) *is made once, this counting to be made positively or negatively according as we pass around the partial area in counterclockwise or clockwise sense.* You will have no difficulty in establishing the truth of this general theorem.

*(Continues...)*

Excerpted fromElementary Mathematics From An Advanced Standpoint GeometrybyFelix Klein, E. R. Hedrick, C. A. Noble. Copyright © 2004 Dover Publications, Inc.. Excerpted by permission of Dover Publications, Inc..

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## Table of Contents

I. Line-Segment, Area, Volume, as Relative Magnitudes

II. The Grassmann Determinant Principle for the Plane

III. The Grassmann Principle for Space

IV. Classification of the Elementary Configurations of Space According to their Behavior under Transformation of Rectangular Coordinates

V. Derivative Manifolds

2. Geometric Transformations

I. Affine Transformations

II. Projective Transformations

III. Higher Point Transformations

IV. Transformations with Change of Space Element

V. Theory of the Imaginary

3. Systematic Discussion of Geometry and Its Foundations

I. The Systematic Discussion

II. Foundations of Geometry

Indexes

## Reading Group Guide

1. The Simplest Geometric Manifolds

I. Line-Segment, Area, Volume, as Relative Magnitudes

II. The Grassmann Determinant Principle for the Plane

III. The Grassmann Principle for Space

IV. Classification of the Elementary Configurations of Space According to their Behavior under Transformation of Rectangular Coordinates

V. Derivative Manifolds

2. Geometric Transformations

I. Affine Transformations

II. Projective Transformations

III. Higher Point Transformations

IV. Transformations with Change of Space Element

V. Theory of the Imaginary

3. Systematic Discussion of Geometry and Its Foundations

I. The Systematic Discussion

II. Foundations of Geometry

Indexes

## Interviews

1. The Simplest Geometric Manifolds

I. Line-Segment, Area, Volume, as Relative Magnitudes

II. The Grassmann Determinant Principle for the Plane

III. The Grassmann Principle for Space

IV. Classification of the Elementary Configurations of Space According to their Behavior under Transformation of Rectangular Coordinates

V. Derivative Manifolds

2. Geometric Transformations

I. Affine Transformations

II. Projective Transformations

III. Higher Point Transformations

IV. Transformations with Change of Space Element

V. Theory of the Imaginary

3. Systematic Discussion of Geometry and Its Foundations

I. The Systematic Discussion

II. Foundations of Geometry

Indexes

## Recipe

I. Line-Segment, Area, Volume, as Relative Magnitudes

II. The Grassmann Determinant Principle for the Plane

III. The Grassmann Principle for Space

IV. Classification of the Elementary Configurations of Space According to their Behavior under Transformation of Rectangular Coordinates

V. Derivative Manifolds

2. Geometric Transformations

I. Affine Transformations

II. Projective Transformations

III. Higher Point Transformations

IV. Transformations with Change of Space Element

V. Theory of the Imaginary

3. Systematic Discussion of Geometry and Its Foundations

I. The Systematic Discussion

II. Foundations of Geometry

Indexes