LITMIR.BIZ

      [Letter to Noble on the Nature of Reasoning]

      Milford, Pa., 1887 May 28th.

      MISS MARIE B. D. NOBLE,

      Care Messrs. Gorham, Turner, & Co., New York City.

      My dear pupil:

      I wish to begin by giving you some general idea of the nature of reasoning. All reasoning involves observation. A chemist sets up an apparatus of flasks and tubes, he puts certain substances in the former, he applies heat, and then he watches closely to see what the result will be. The procedure of the mathematician is closely analogous to this. He draws a diagram, for example, conforming to certain general conditions, and then he observes certain relations among the parts of this diagram, over and above those which were used to determine the construction of it. The result that the chemist observes is brought about by Nature, the result that the mathematician observes is brought about by the associations of the mind. But this does not constitute so radical a distinction as it seems at first sight to do; for were the laws of nature not intelligible, that is, were they not such as naturally occur to the human mind, the chemist’s observation could never teach him anything, while on the other hand the power that connects the conditions of the mathematician’s diagram with the relations he observes in it is just as occult and mysterious to us as the power of Nature that brings about the result of the chemical experiment. You do not quite see the truth of this? No, why should you: I have stated it in abstract terms, which give you nothing to observe this fact in, and the mind can see no truth except by observation. To enable you to see it, I will give two instances of simple mathematical reasoning; you will please first see that they are proofs, and then remark the fact that even after you know the proofs you have no direct consciousness of the necessity that binds the conclusion to the conditions supposed. First take this simple proposition in geometry, an interesting one in itself, and probably unknown to you. On a plane, take any point, O, and through it draw 3 straight lines. Take 3 new points, A, B, C, lying severally on these 3 straight lines. Take also 3 other points, A′, B′, C′, lying, A′ in the line through O and A, B′ in the line through O and B, C′ in the line between O and C. Now if we draw the straight lines AB and A′B′, and continue them till they intersect at a point which we may designate as X, if we draw the straight lines BC and B′C′, and continue them until they intersect in a point which we may designate as Y, and if we draw the straight lines CA and C′A′, and continue them till they meet in a point which we may designate as Z, then we shall find that the points X, Y, Z lie on one straight line. More briefly stated, if two triangles are so situated that the lines joining corresponding vertices meet in one point, then the points of intersection of corresponding sides (produced if necessary) lie in one straight line. To prove this, through the point O imagine a line not lying in the plane of the triangles, and on this line we imagine any two points, which we may distinguish as S and S′. Now any 3 points whatever lie in some one plane. Consider, then, the plane OSA. S′ lies on this plane, because it is in the line between S and O, both of which are on the plane. A′ too lies on the plane, because it is in the line through A and O. Then, since the point S′ and A′ lie on the plane the whole line through these points lies on the plane. Thus, the lines SA and S′A′ lie on one plane. But any two lines in one plane cross, if sufficiently produced (unless, indeed, they happen to be parallel, a case not necessary to consider, because easily evaded by shifting S or S′). Consequently, the lines SA and S′A′ (produced, if necessary) intersect in a point which we may designate as A″. Similar reasoning would show that the lines SB and S′B′ intersect in a point which we may designate as B″, and that the lines SC and S′C′ intersect in a point which we may designate as C″ Now the points S, A, and A″ lie in a straight line, for A″ designates the intersection of the line SA with another line. And in like manner, the points S, B, and B″ lie in a straight line. These two lines have the point S in common; they therefore lie in one plane. Consequently, the lines AB and A″B″ lie on this same plane, which we may designate as plane No. 1. Likewise, the points S′, A′, and A″ lie in a straight line, and the points S′, B′, and B″ lie in a straight line, and these two lines intersect at the point S′, so that they lie in one plane, which we may designate as the plane No. 2. Consequently, the lines A′B′ and A″B″ lie in plane No. 2. Now the lines AB and A′B′ lie on the original plane of the two triangles first spoken of, which we may call plane No. 3. These three planes, like any three planes, must meet in some one point, and this point lies in the line common to planes 1 and 3, and also in the line common to planes 2 and 3. It also lies in the line common to planes 1 and 2. But the line common to planes 1 and 3 is AB, and the line common to planes 2 and 3 is A′B′, and the point of intersection of AB and A′B′ is the point we called X. Thus, X lies in the line through A″ and B″. It therefore lies in any plane on which this line lies, and thus in the plane of A″B″C″. In like manner, it could be shown that the points Y and Z are on the plane A″B″C″. But these three points are also on the original plane. They are therefore on the intersection of the original plane with the plane A″B″C″ and thus are on a straight line, which is what we had to prove.

      As another illustration of mathematical reasoning I will prove a proposition in arithmetic, namely, that no power of a whole number whose exponent is one less than a prime number can, on division by that prime, leave a remainder greater than 1. Thus, 5 being a prime number, the series of 4^th powers is 1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, etc., and as these all end with a 1, a 6, a 5, or a 0, it follows that the remainders, after division by 5 will all be either 0 or 1. The general proof is as follows. If the base, or root of the power, is divisible by the prime, then the power itself is so too, and there is no remainder. But if the base is not so divisible, then its product by any number less than the prime will, on division by the prime, leave a remainder different from any other product of the base by a number less than the prime. For if two such products were to leave the same remainder, the difference between them would leave no remainder at all, and would thus be divisible by the prime. But this difference would be equal to the base multiplied by some number less than the prime number. Now if a prime number does not divide either of two numbers, neither

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