Lobachevski - Non-Euclidean Geometry

Lobachevski Geometry - Part II

PART I

The following selection consists of two sections. First, we have six more propositions from Book I of Euclid's Elements (Propositions 27-32). These are propositions dealing with parallel lines. With these Euclidean propositions we have placed some pages from Lobachevski's Theory of Parallels. This work discusses Euclid's theory of parallels, finds fault with it, and substitutes another theory for it. In so doing, Lobachevski develops a version of "non-Euclidean geometry."

Euclid: Elements of Geometry

BOOK I

Proposition 27

If a straight line falling on two straight lines make the alternate angles equal to one another, the straight lines will be parallel to one another.

For let the straight line EF falling on the two straight lines AB, CD make the alternate angles AEF, EFD equal to one another;

proposition 27

I say that AB is parallel to CD.
For, if not, AB, CD when produced will meet either in the direction of B, D or towards A, C.
Let them be produced and meet, in the direction of B, D, at G.
Then, in the triangle GEF, the exterior angle AEF is equal to the interior and opposite angle EFG:
which is impossible.
Therefore AB, CD when produced will not meet in the direction of B, D.
Similarly it can be proved that neither will they meet towards A, C.
But straight lines which do not meet in either direction are parallel;
therefore AB is parallel to CD.
Therefore etc.          Q.E.D.

Proposition 28

If a straight line falling on two straight lines make the exterior angle equal to the interior and opposite angle on the same side, or the interior angles on the same side equal to two right angles, the straight lines will be parallel to one another.

For let the straight line EF falling on the two straight lines AB, CD make the exterior angle EGB equal to the interior and opposite angle GHD, or the interior angles on the same
proposition 28
side, namely BGH, GHD, equal to two right angles; I say that AB is parallel to CD.
For, since the angle EGB is equal to the angle GHD, while the angle EGB is equal to the angle AGH,
the angle AGH is also equal to the angle GHD; and they are alternate; therefore AB is parallel to CD.
Again, since the angles BGH, GHD are equal to two right angles, and the angles AGH, BGH are also equal to two right angles,
the angles AGH, BGH are equal to the angles BGH, GHD.
Let the angle BGH be subtracted from each; therefore the remaining angle AGH is equal to the remaining angle GHD; and they are alternate; therefore AB is parallel to CD.
Therefore etc.          Q.E.D.

Proposition 29

A straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle, and the interior angles on the same side equal to two right angles.

For let the straight line EF fall on the parallel straight lines AB, CD; I say that it makes the alternate angles AGH, GHD equal, the exterior angle EGB equal to the interior and opposite angle GHD, and the interior angles on the same side, namely BGH, GHD, equal to two right angles.
proposition 29
For, if the angle AGH is unequal to the angle GHD, one of them is greater.
Let the angle AGH be greater.
Let the angle BGH be added to each; therefore the angles AGH, BGH are greater than the angles BGH, GHD.
But the angles AGH, BGH are equal to two right angles;
therefore the angles BGH, GHD are less than two right angles.
But straight lines produced indefinitely from angles less than two right angles meet;
therefore AB, CD, if produced indefinitely, will meet; but they do not meet, because they are by hypothesis parallel.
Therefore the angle A GH is not unequal to the angle GHD, and is therefore equal to it.
Again, the angle AGH is equal to the angle EGB;
therefore the angle EGB is also equal to the angle GHD.
Let the angle BGH be added to each; therefore the angles EGB, BGH are equal to the angles BGH, GHD.
But the angles EGB, BGH are equal to two right angles;
therefore the angles BGH, GHD are also equal to two right angles.
Therefore etc.          Q.E.D.

Proposition 30

Straight lines parallel to the same straight line are also parallel to one another.

Let each of the straight lines AB, CD be parallel to EF; I say that AB is also parallel to CD.
proposition 30
For let the straight line GK fall upon them.
Then, since the straight line GK has fallen on the parallel straight lines AB, EF, the angle of AGK is equal to the angle GHF.
Again, since the straight line GK has fallen on the parallel straight lines EF, CD, the angle GHF is equal to the angle GKD.
But the angle AGK was also proved equal to the angle GHF; therefore the angle AGK is also equal to the angle GKD;
and they are alternate.

Nicholas Lobachevski: The Theory of Parallels

In geometry I find certain imperfections which I hold to be the reason why this science, apart from transition into analytics, can as yet make no advance from that state in which it has come to us from Euclid.

As belonging to these imperfections, I consider the obscurity in the fundamental concepts of the geometrical magnitudes and in the manner and method of representing the measuring of these magnitudes, and finally the momentous gap in the theory of parallels, to fill which all efforts of mathematicians have been so far in vain.

For this theory Legendre's endeavors have done nothing, since he was forced to leave the only rigid way to turn into a side path and take refuge in auxiliary theorems which he il-logically strove to exhibit as necessary axioms. My first essay on the foundations of geometry I published in the Kasan Messenger for the year 1829. In the hope of having satisfied all requirements, I undertook hereupon a treatment of the whole of this science, and published my work in separate parts in the "Gelehrten Schriften der Universitaet Kasan" for the years 1836, 1837, 1838, under the title "New Elements of Geometry, with a complete Theory of Parallels." The extent of this work perhaps hindered my countrymen from following such a subject, which since Legendre had lost its interest. Yet I am of the opinion that the Theory of Parallels should not lose its claim to the attention of geometers, and therefore I aim to give here the substance of my investigations, remarking beforehand that contrary to the opinion of Legendre, all other imperfections—for example, the definition of a straight line— show themselves foreign here and without any real influence on the Theory of Parallels.

In order not to fatigue my reader with the multitude of those theorems whose proofs present no difficulties, I prefix here only those of which a knowledge is necessary for what follows.

1. A straight line fits upon itself in all its positions. By this I mean that during the revolution of the surface containing it the straight line does not change its place if it goes through two unmoving points in the surface: (i. e., if we turn the surface containing it about two points of the line, the line does not move.)

2. Two straight lines can not intersect in two points.

3. A straight line sufficiently produced both ways must go out beyond all bounds, and in such way cuts a bounded plain into two parts.

4. Two straight lines perpendicular to a third never intersect, how far soever they be produced.

5. A straight line always cuts another in going from one side of it over to the other side: (i. e., one straight line must cut another if it has points on both sides of it.)

6. Vertical angles, where the sides of one are productions of the sides of the other, are equal. This holds of plane rectilineal angles among themselves, as also of plane surface angles: (i. e., dihedral angles.)

7. Two straight lines can not intersect, if a third cuts them at the same angle.

8. In a rectilineal triangle equal sides lie opposite equal angles, and inversely.

9. In a rectilineal triangle, a greater side lies opposite a greater angle. In a right-angled triangle the hypothenuse is greater than either of the other sides, and the two angles adjacent to it are acute.

10. Rectilineal triangles are congruent if they have a side and two angles equal, or two sides and the included angle equal, or two sides and the angle opposite the greater equal, or three sides equal.

11. A straight line which stands at right angles upon two other straight lines not in one plane with it is perpendicular to all straight lines drawn through the common intersection point in the plane of those two.

12. The intersection of a sphere with a plane is a circle.

13. A straight line at right angles to the intersection of two perpendicular planes, and in one, is perpendicular to the other.

14. In a spherical triangle equal sides lie opposite equal angles, and inversely.

15. Spherical triangles are congruent (or symmetrical) if they have two sides and the included angle equal, or a side and the adjacent angles equal.

From here follow the other theorems with their explanations and proofs.

16. All straight lines which in a plane go out from a point can with reference to a given straight line in the same plane, be divided into two classes—into cutting and not-cutting.

The boundary lines of the one and the other class of those lines will be called parallel to the given line.

From the point A (Fig. 1) let fall upon the line BC the perpendicular AD, to which again draw the perpendicular AE.

In the right angle EAD either will all straight lines which go out from the point A meet the line DC, as for example AF, or some of them, like the perpendicular AE, will not meet the line DC. In the uncertainty whether the perpendicular AE is the only line which does not meet DC, we will assume it may be possible that there are still other lines, for example AG, which do not cut DC, how far soever they may be prolonged. In passing over from the cutting lines, as AF, to the not-cutting
figure 1
lines, as AG, we must come upon a line AH, parallel to DC, a boundary line, upon one side of which all lines AG are such as do not meet the line DC, while upon the other side every straight line AF cuts the line DC.

The angle HAD between the parallel HA and the perpendicular AD is called the parallel angle (angle of parallelism), which we will here designate II (p) for AD = p.

If II (p) is a right angle, so will the prolongation AE' of the perpendicular AE likewise be parallel to the prolongation DB of the line DC, in addition to which we remark that in regard to the four right angles, which are made at the point A by the perpendiculars AE and AD, and their prolongations AE' and AD', every straight line which goes out from the point A, either itself or at least its prolongation, lies in one of the two right angles, which are turned toward BC, so that except the parallel EE' all others, if they are sufficiently produced both ways, must intersect the line BC.

If II (p) < π/2 it, then upon the other side of AD, making the same angle DAK=11 (p) will lie also a line AK, parallel to the prolongation DB of the line DC, so that under this assumption we must also make a distinction of sides in parallelism.

All remaining lines or their prolongations within the two right angles turned toward BC pertain to those that intersect, if they lie within the angle HAK = 2 II (p) between the parallels; they pertain on the other hand to the non-intersecting AG, if they lie upon the other sides of the parallels AH and AK, in the opening of the two angles EAH = π/2 — II (p), E'AK = π/2 — II (p), between the parallels and EE' the perpendicular to AD. Upon the other side of the perpendicular EE' will in like manner the prolongations AH' and AK' of the parallels AH and AK likewise be parallel to BC; the remaining lines pertain, if in the angle K'AH', to the intersecting, but if in the angles K'AE, H'AE' to the non-intersecting.

In accordance with this, for the assumption II (p) = π/2, the lines can be only intersecting or parallel; but if we assume that II (p) < π/2, then we must allow two parallels, one on the one and one on the other side; in addition we must distinguish the remaining lines into non-intersecting and intersecting.

For both assumptions it serves as the mark of parallelism that the line becomes intersecting for the smallest deviation toward the side where lies the parallel, so that if AH is parallel to DC, every line AF cuts DC, how small soever the angle HAF may be.

17. A straight line maintains the characteristic of parallelism at all its points.

Given AB (Fig. 2) parallel to CD, to which latter AC is perpendicular. We will consider two points taken at random on the line AB and its production beyond the perpendicular.
figure 2
Let the point E lie on that side of the perpendicular on which AB is looked upon as parallel to CD.
Let fall from the point E a perpendicular EK on CD and so draw EF that it falls within the angle BEK.

Connect the points A and F by a straight line, whose production then (by Theorem 16) must cut CD somewhere in G. Thus we get a triangle ACG, into which the line EF goes; now since this latter, from the construction, can not cut AC, and can not cut AG or EK a second time (Theorem 2), therefore it must meet CD somewhere at H (Theorem 3).

Now let E' be a point on the production of AB and E'K' perpendicular to the production of the line CD; draw the line E'F' making so small an angle A E'F' that it cuts AC somewhere in F'; making the same angle with AB, draw also from A the line AF, whose production will cut CD in G (Theorem 16).

Thus we get a triangle AGC, into which goes the production of the line E'F'; since now this line can not cut AC a second time, and also can not cut AG, since the angle BAG = BE'G' (Theorem 7), therefore must it meet CD somewhere in G'.

Therefore from whatever points E and E' the lines EF and E'F1 go out, and however little they may diverge from the line AB, yet will they always cut CD, to which AB is parallel.

18. Two lines are always mutually parallel.

Let AC be a. perpendicular on CD, to which AB is parallel; if we draw from C the line CE making any acute angle ECD with CD, and let fall from A the perpendicular AF upon CE, we obtain a rightangled triangle ACF, in which AC, being the hypothenuse, is greater than the side AF (Theorem 9).
figure 3
Make AG = AF, and slide the figure EFAB until AF coincides with AG, when AB and FE will take the position AK and GH, such that the angle BAK = FAC, consequently AK must cut the line DC somewhere in K (Theorem 16), thus forming a triangle AKC, on one side of which the perpendicular GH intersects the line AK in L (Theorem 3), and thus determines the distance AL of the intersection point of the lines AB and CE on the line AB from the point A.

Hence it follows that CE will always intersect AB, how small soever may be the angle ECD, consequently CD is parallel to AB (Theorem 16).

19. In a rectilineal triangle the sum of the three angles can not be greater than two right angles.

Suppose in the triangle ABC (Fig. 4) the sum of the three angles is equal to π + a; then choose in case of the inequality of the sides the smallest BC, halve it in D, draw from A through D the line AD and make the prolongation of it, DE, equal to AD, then join the point E to the point C by the straight line EC. In the congruent triangles ADB and CDE, the angle

ABD = DCE, and BAD = DEC (Theorems 6 and 10); whence follows that also in the triangle ACE the sum of the three angles must be equal to π + a; but also the smallest angle BAC (Theorem 9) of the triangle ABC in passing over into the new triangle ACE has been cut up into the two parts EAC and A EC. Continuing this process, continually halving the side opposite the smallest angle, we must finally attain to a triangle in which the sum of the three angles is π + a, but wherein are two angles, each of which in absolute magnitude is less than a/2; since now, however, the third angle can not be greater than π, so must a be either null or negative.

20. If in any rectilineal triangle the sum of the three angles is equal to two right angles, so is this also the case for every other triangle.

If in the rectilineal triangle ABC (Fig. 5) the sum of the three angles = π, then must at least two of its angles, A and C, be acute. Let fall from the vertex of the third angle B upon the opposite side AC the perpendicular p. This will cut the triangle into two right-angled triangles, in each of which the sum of the three angles must also be π, since it can not in either be greater than π, and in their combination not less than π.

figure 6

So we obtain a right-angled triangle with the perpendicular sides p and q, and from this quadrilateral whose opposite sides are equal and whose adjacent sides p and q are at right angles (Fig. 6).

By repetition of this quadrilateral we can make another with sides np and mq, and finally a quadrilateral ABCD with sides at right angles to each other, such that AB = np, AD = mq.
figure 6
DC = np, BC = mq, where m and n are any whole numbers. Such a quadrilateral is divided by the diagonal DB into two congruent right-angled triangles, BAD and BCD, in each of which the sum of the three angles = π.

The numbers n and m can be taken sufficiently great for the right-angled triangle ABC (Fig. 7) whose perpendicular sides AB — np, BC = mq, to enclose within itself another given (right-angled) triangle BDE as soon as the right-angles fit each other.

Drawing the line DC, we obtain right-angled triangles of which every successive two have a side in common.

The triangle ABC is formed by the union of the two triangles ACD and DCB, in neither of which can the sum of the angles be greater than π, consequently it must be equal to π, in order that the sum in the compound triangle may be equal to π.

figure 7

In the same way the triangle BDC consists of the two triangles DEC and DBE, consequently must in DBE the sum of the three angles be equal to π, and in general this must be true for every triangle, since each can be cut into two right-angled triangles.

From this it follows that only two hypotheses are allowable: Either is the sum of the three angles in all rectilineal triangles equal to π, or this sum is in all less than π.

21. From a given point we can always draw a straight line that shall make with a given straight line an angle as small as we choose.

Let fall from the given point A (Fig. 8) upon the given line BC the perpendicular AB; take upon BC at random the point D; draw the line AD; make DE —AD, and draw AE.
figure 8

In the right-angled triangle ABD let the angle ADB = a; then must in the isosceles triangle ADE the angle AED be either a/2 or less (Theorems 8 and 20). Continuing thus we finally attain to such an angle, AEB, as is less than any given angle.

22. If two perpendiculars to the same straight line are parallel to each other, then the sum of the three angles in a rectilineal triangle is equal to two right angles.

Let the lines AB and CD (Fig. 9) be parallel to each other and perpendicular to AC.

Draw from A the lines AE and AF to the points E and F, which are taken on the line CD at any distance FC > EC from the point C.
figure 9
Suppose in the right-angled triangle ACE the sum of the three angles is equal to π — α, in the triangle AEF equal to π — β, then must it in the triangle ACF equal π — α — β, where α and β can not be negative.

Further, let the angle BAF = a, AFC = b, so is α + β = a — b; now by revolving the line AF away from the perpendicular AC we can make the angle a between AF and the parallel AB as small as we choose; so also can we lessen the angle b, consequently the two angles α and β can have no other magnitude than α = 0 and β = 0.

It follows that in all rectilineal triangles the sum of the three angles is either π and at the same time also the parallel angle II (p) = π/2 for every line p, or for all triangles this sum is < π and at the same time also II (p) < π/2.

The first assumption serves as foundation for the ordinary geometry and plane trigonometry.

The second assumption can likewise be admitted without leading to any contradiction in the results, and founds a new geometric science, to which I have given the name Imaginary Geometry, and which I intend here to expound as far as the development of the equations between the sides and angles of the rectilineal and spherical triangle.

Part II


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