Writings of Charles S. Peirce: A Chronological Edition, Volume 6. Charles S. Peirce

Чтение книги онлайн.

      In your second letter, you ask how the distributive principle proves that (a + b)(c + d) = (a + b)c + (a + b)d. The formula of the distributive principle of multiplication with respect to addition is x(y + z) = xy + xz. Put x = a + b, y = c, and z = d, and you have the result. Of course, in the general formula, x, y, z, may be any statements; hence, it is legitimate to adopt the equivalents just given.

      The next question is what I mean by saying that the associative principle is assumed in leaving off the parentheses. By the associative principle χ + (y + z) = (x + y) + z; so that we may as well write simply x + y + z, for whether this means that x and y are first to be added and then z added on to them, or that to x is to be added the sum of y and z makes no difference according to the preceding formula. In like manner, in the particular case in which I make the remark and to which your inquiry relates, without the associative principle, I should only reach the statement (ac + bc) + (ad + bd). But by the associative principle, this would be the same as ac + [bc + (ad + bd)] and as ac + [(bc + ad) + bd], and in short, without giving all the equivalents, it obviously makes no difference how the parentheses are put in, so long as the factors of no one term are separated, so that they may as well be dropped altogether. All students have to ask such questions at first.

      The blurred lines are x + 0 = x x$ = x 0x = 0 $ + x = $ [This is called the principle of contradiction.] [This is the so-called principle of excluded middle.] In the last two formulae please observe the second x is in each case negatived.

      I will not send you any further exercise today, as I think you have enough for 8 hours, at least. These things are puzzling, at first.

      Yours very truly,

      

13

      [Additional Exercises in Boolian Algebra]

Summer 1887

Houghton Library

      I saw a man in the street who had a way of looking up and squinnying his eyes which showed me that he was either excessively nearsighted or somewhat foolish. He went up to a post-box at the corner and tried to put some object through the top, until he finally found the slit in the side of the box. He was not foolish enough to account for this conduct, and was certainly not intoxicated; so that he was plainly either excessively near-sighted or else absent-minded. Having put something small into the box he next sat down on the curbstone. Nearsightedness would not account for this, which showed him to be either absent-minded or rather foolish. But absentmindedness would not account for his squinnying his eyes; so that I was puzzled. He might be both absent-minded and somewhat foolish, and if so undoubtedly was under the impression he was in a bob-tailed horse-car, and had deposited a nickel in the box. If he was somewhat foolish and excessively near-sighted, the same conclusion would result. If he was absent-minded and excessively near-sighted, the inference will be the same.

      Let a mean he was absent-minded; f, that he was foolish; n, that he was excessively near-sighted; c, that he thought himself in a horse-car. Then the premises are

(a + f),	he was absent-minded or rather foolish;
(f + n),	he was rather foolish or excessively near-sighted;
(n + a),	he was very near-sighted or else absent-minded;
	if he was both absent-minded and foolish, he thought himself СКАЧАТЬ В начало < 36 37 38 39 40 41 42 43 44 45 > В конец