The Parents' Review
A Monthly Magazine of Home-Training and Culture
"Education is an atmosphere, a discipline, a life."
Euclid as an Amusement.
by P. G. O'Connell.
It is interesting to a well-developed mind to attempt to discover a formal proof of the truth of any fact, no matter how obvious, and many hold that, for a sufficiently developed mind, it is the best form of exercise. To a well-trained gymnast the grand circle may appear to be the best form of exercise, but he does not prescribe it for beginners, knowing that it would probably kill them. Similarly the exercise I am about to suggest would probably injure an undeveloped mind.
It is believed by some that the reductio ad absurdum proof is applicable to cases which are unsusceptible of other proof. This is not true. On the contrary, every reductio ad absurdum proof is capable of being changed into a rational proof, by the alteration of a very few words. I make the statement boldly, having mentally thus altered every reductio ad absurdum proof that I have ever come across. The exercise I suggest consists, not in discovering a new proof, which might be easy, but in restating the old proof with as few alterations as possible, and yet in such words as do not suggest an attempt to conceive the inconceivable.
I will give a few examples, which the reader need not look at until he has first mentally performed the exercise for himself.
(Euc. I. 6.) Let A B C be any triangle having A B and A C unequal. From the greater B A, cut off a part of B D equal to C A. Join C D. Then the triangles D B C and A C B have the two sides D B, B C, equal to the two sides A C, C B each to each, but are obviously unequal in area, since one is a part of the other. Therefore, by I. 4, the angles A B C and A C B cannot be equal. Hence any two angles of a triangle must be unequal if the sides opposite them be unequal, that is, the angles can only be equal when the sides are equal. Hence, if two angles of a triangle be given equal, the sides opposite them must also be equal. Q.E.D.
(Euc. I. 7.) On the same base A B and on the same side of it, let there be two non-coincident triangles A B C and A B D, having A C equal to A D. Join C D and produce A C and A D to E and F. Then, by I. 5, the angles A C D, A D C are equal and likewise the angles E C D, F D C. Therefore, whether the vertex of one triangle fall within, or without, the other triangle, the angle B C D is not equal to the angle B D C. Therefore, by I. 5, B C is not equal to B D. Wherefore, on the same base, etc. Q.E.D.
The word "non-coincident" is necessary, for the theorem is not true without it. In the proof of I. 4, Euclid deliberately places on the same base and on the same side of it two coincident triangles, having "the sides terminated in one extremity of the base equal, and likewise the sides terminated in the other extremity of the base equal." Thus what really is proved in I. 7 is that, if on the same base and on the same side of it there be two such triangles, they must be coincident. Hence, as soon as we have, in the proof of I. 8, placed the triangles on the same base and on the same side of it, we may say: "Therefore, by I. 7, the triangles are coincident."
(I. 14.) Assume the two straight lines to be not in the same straight line, without pretending to assume that the two adjacent angles are equal to two right angles, Prove, as Euclid does, that they cannot be so. The theorem logically follows.
(I. 19.) Let A B C be a triangle, having the angle A B C greater than the angle A C B. Then by I. 5, A B cannot be equal to A C, and, by I. 18, A B cannot be greater than A C. Hence A B must be less than A C. Q.E.D.
The proof of I. 25 may be altered in a similar way.
(I. 26.) Let A B C and D E F be two triangles, having the angle A B C equal to the angle D E F, and the side B C equal to the side E F, but the sides A B, D E, unequal. From the greater, B A, cut off a part B G equal to E D and join C G. Then, by I. 4, the triangles G B C and D E F are equal in all respects. Therefore, by I. 16, the angle E D F is unequal to the angle B A C, and, by axiom 9, the angle D F E is unequal to the angle A C B. Hence two triangles cannot have two angles of the one equal to two angles of the other, each to each, and a pair of corresponding sides equal, unless the sides containing one pair of equal angles be equal, each to each, in which case the triangles will, by I. 4, be equal in all respects.
Wherefore, if two triangles, etc. Q.E.D.
Before proceeding to I. 27, I must briefly refer to the fatal fascination of Euclid's famous twelfth axiom. It is axiomatic enough to those who understand the meaning of the word "angle," the definition of which assumes an appreciation of the meaning of the word "direction." But, if we assume this, we may define parallel lines as lines in the same direction. From this definition I. 27, 28, and 29 would logically follow. To return to the exercise:--
(I. 27.) If a straight line fall on two other straight lines which either meet, or can be produced to meet, then, by I. 16, the alternate angles cannot be equal. The theorem logically follows.
(I. 29.) If a straight line, falling on two other straight lines do not make the alternate angles equal, we may prove, precisely as Euclid does, that the lines will meet if produced, and are therefore not parallel. It follows that, if they be given parallel, the alternate angles cannot be unequal. The rest of the proposition may be proved in Euclid's own words.
(I. 39.) Prove, as Euclid does, that triangles on the same base and on the same side of it, but not between the same parallels, are unequal. The theorem logically follows.
(III. 1.) Let A B be any chord of a circle and D E the chord which bisects A B at right angles at C. Let G be any point not in D E, or D E produced. Join G A, G B, G C. Then G C, C A are equal to G C, C B, each to each, but the included angles are unequal. Hence, by I. 24, G A is unequal to G B. Thus no point, which is not in D E, can be the centre of the circle. Therefore the centre is the middle point of D E.
(III. 4.) Let two chords of a circle, which do not both pass through the centre O, intersect at C. If necessary, join O C. Then O C cannot be perpendicular to both chords. Therefore it cannot bisect both.
(III. 5 and 6.) If two non-coincident circles have a common point A, they cannot have a common centre. Let O be the centre of one. Join O A and draw the straight line O B C, cutting the circles at B and C. Then O A is equal to one of the lines O B and O C, and therefore unequal to the other. Therefore the centre of one circle is not the centre of the other.
III. 9 is a direct corollary from III. 7.
(III. 10.) Let A, B, C, be any three points on a circle whose centre is O. Then O A, O B, and O C are all equal. Hence, by III. 9, O must be at the centre of any circle passing through A, B, and C. Hence, by III. 5 and 6, any such circle must be coincident with the first.
(III. 11 and 12.) Let two circles touch each other at A. Let O be the centre of the first one, let P be any point on any straight line which does not pass through A, even when produced. Prove, as Euclid does, that P cannot be the centre of the second circle. Thus no straight line which does not pass through A can be the straight line joining the centres. Wherefore, etc. Q.E.D.
(III. 13.) Let O and C be the centres of two circles which touch each other at A. Then A, O and C are in a straight line. (III. 11 or 12). Produce this straight line to cut, at B, the circle whose centre is O. Then O A is equal to O B. Therefore C A is unequal to C B. Therefore the circles do not touch at B, and, by the last two propositions, they cannot touch at any other point. Wherefore two circles canot touch each other at more points than one. Q.E.D.
(III. 18.) Let O be the centre of a circles which touches a straight line at A. Join O A. Let B be any other point in the tangent. Join O B, cutting the circle at C. Then O B is greater than O C, and therefore than O A. Hence the angle O B A is less than the angle of O A B (I. 18), and therefore acute (I. 17). Hence O B is not perpendicular to A B. In like manner it may be shown that no straight line through O, except O A, can be perpendicular to A B. Therefore O A is perpendicular to A B (I. 12).
A similar method may be applied to III. 19, which is, however, a logical corollary to III. 18.
This article has been written at the suggestion of my friend the wrangler. If it serve no other purpose it may, at least, induce some teacher to spare their pupils the attempt to conceive the inconceivable.
Proofread by LNL, June 2020
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