"For the first time we have available in an intelligible form the writings of one of the greatest philosophers of the past hundred years." —Times Literary Supplement
Writings of Charles S. Peirce: A Chronological Edition, Volume 8, 1890--1892by Charles S. Peirce
Volume 8 of this landmark edition follows Peirce from May 1890 through July 1892a period of turmoil as his career unraveled at the U.S. Coast and Geodetic Survey. The loss of his principal source of income meant the beginning of permanent penury and a lifelong struggle to find gainful employment. His key achievement during these years is his celebrated Monist… See more details below
Volume 8 of this landmark edition follows Peirce from May 1890 through July 1892a period of turmoil as his career unraveled at the U.S. Coast and Geodetic Survey. The loss of his principal source of income meant the beginning of permanent penury and a lifelong struggle to find gainful employment. His key achievement during these years is his celebrated Monist metaphysical project, which consists of five classic articles on evolutionary cosmology. Also included are reviews and essays from The Nation in which Peirce critiques Paul Carus, William James, Auguste Comte, Cesare Lombroso, and Karl Pearson, and takes part in a famous dispute between Francis E. Abbot and Josiah Royce. Peirce's short philosophical essays, studies in non-Euclidean geometry and number theory, and his only known experiment in prose fiction complete his production during these years.
Peirce's 1883-1909 contributions to the Century Dictionary form the content of volume 7 which is forthcoming.
Indiana University Press
"For the first time we have available in an intelligible form the writings of one of the greatest philosophers of the past hundred years." Times Literary Supplement
"Volume 8 of the Writings is a comprehensive and meticulously documented presentation of Peirce’s work during the years in question." —TRANSACTIONS C S PEIRCE SOC
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Writings of Charles S. Peirce Volume 8 1890-1892
A Chronological Edition
By Charles S. Peirce, Nathan Houser, André de Tienne, Jonathan R. Eller, Cornelis de Waal, Albert C. Lewis, Diana Reynolds, Joseph Kaposta, Luise H. Morton, Kelly Tully-Needler, Leah Cummins Guinn
Indiana University PressCopyright © 2010 Peirce Edition Project
All rights reserved.
Familiar Letters about the Art of Reasoning
15 May 1890 Houghton Library
Stagira, May 15, 1890.
My dear Barbara:
The University of Cracow once conferred upon a very good fellow a degree for having taught the philosophical faculty to play cards. I cannot tell you in what year this happened,—perhaps it was 1499. The graduate was Thomas Murner, of whose writings Lessing said that they illustrated all the qualities of the German language; and so they do if those qualities are energy, rudeness, indecency, and a wealth of words suited to unbridled satire and unmannered invective. The diploma of the university is given in his book called Chartiludium, one of the numerous illustrations to which is copied to form the title page of the second book of a renowned encyclopaedia, the Margarita Philosophica. Murner's pack contained 51 cards. There were seven unequal suits; 3 hearts, 4 clubs (or acorns), 8 diamonds (or bells), 8 crowns, 7 scorpions, 8 fish, 6 crabs. The remaining seven cards were jokers, or unattached to suits; for such cards formed a feature of all old packs. The object of Murner's cards was to teach the art of reasoning, and a very successful pedagogical instrument they no doubt proved.
If you will provide yourself, my dear Barbara, with a complete pack of cards with a joker, 53 in all, I will make a little lesson in mathematics go down like castor-oil in milk. Take, if you will be so kind, the 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 of spades, and arrange these ten cards in their proper order. I mean by this that the ace, or 1, is to be at the back of the pack, the 2 next, and so on, the 10 alone showing its face. I call this the "proper order," because I propose always to begin the count of cards in a pack at the back, so that, in the pack of ten cards you have just been so obliging as to arrange, every card is in its proper place, that is the number it bears on its face is equal to the number of its place from the back of the pack. The face-value of the 2nd card is 2, that of the 3rd card, 3, and so on.
Now let us add 3 to the face-value of each card in the pack. How shall we do that without a printing-press? Why, by simply taking 3 cards from the back of the pack of ten and carrying them to the face. The face-value of card number 1 is now 3 + 1, or 4; that of card 2 is 5, and so on up to card 7 which is 10. Card 8 is 1; but 1 and 11 are the same for us. Since we have only ten cards to distinguish, ten different numbers are enough. We, therefore, treat 1, 11, 21, 31, as equal, because we count round and round the 10, thus:
We say 13 and 23 are equal, meaning their remainders after division by 10 are equal. This sort of equality of remainders after division is called congruence by mathematicians and they write it with three lines, thus
13 [equivalent to] 23 (mod. 10).
The number 10 is said to be the modulus, that is, the divisor, or the smallest number congruent to zero, or the number of numbers in the cycle.
Instead of ten cards you may take the whole suit of thirteen, and then, imagining a system of numeration in which the base is thirteen and in which we count
1 2 3 4 5 6 7 8 9 10 Jack Queen King
we have a similar result. Fourteen, or king-ace, is congruent with 1; fifteen, or king-two, with 2, etc.
It makes no difference how many cards there are in a pack. To cut it, when arranged in its proper order, and transpose the two parts, is to add a constant amount to the face-value of every card. So much for addition.
Now how shall we multiply? Suppose we have the pack of ten in its proper order, and wish to multiply the face-value of the cards by 3. We deal out the cards one by one from first to last, into 3 piles, laying them face up upon the table. We first take up the pile the 10, or 0, falls upon, then the next pile, last the third. Putting each pile after the first at the back of that last taken.
We now find in place 1 card 3, or 3 times 1;
in place 2 card 6, or 3 times 2;
in place 3 card 9, or 3 times 3;
in place 4 card 2, congruent to 3 times 4;
in place 5 card 5, congruent to 3 times 5;
in place 6 card 8, congruent to 3 times 6;
in place 7 card 1, congruent to 3 times 7;
in place 8 card 4, congruent to 3 times 8;
in place 9 card 7, congruent to 3 times 9;
in place 10 card 0, congruent to 3 times 10.
Take this pack and multiply again by 3. Multiplying by 3 twice is multiplying by 9. But 9 [equivalent to] - 1.
Accordingly we shall now find
in place 1 card – 1 or 9, in place 2 card – 2 or 8, in place 3 card – 3 or 7, etc.
Multiply again by 3, and since 3 × 9 [equivalent to] 7, we shall find
in place 1 card 7 × 1 [equivalent to] 7,
in place 2 card 7 × 2 [equivalent to] 4,
in place 3 card 7 × 3 [equivalent to] 1,
in place 4 card 7 × 4 [equivalent to] 8,
in place 5 card 7 × 5 [equivalent to] 5,
in place 6 card 7 × 6 [equivalent to] 2,
Take a pack of 11 cards. We shall now have,
11 [equivalent to] 0
12 [equivalent to] 1
23 [equivalent to] 1
and, in short, to find what any card will be, having performed the necessary arithmetical operation, we subtract the number in the tens place from the number in the units place, and repay anything we borrow in the addition. Thus, suppose we deal into 5 piles, and take up the piles from left to right putting each one at the back of the pile that was at the left of it. We shall now have
in place 1, since 5 × 1 = 5, card 5;
in place 2, since 5 × 2 = 10, card 10;
in place 3, since 5 × 3 = 15 and 1 from 5 leaves 4, card 4;
in place 4, since 5 × 4 = 20 and 2 from 10 leaves 8, and repaying 1 borrowed we have 9, card 9;
in place 5, since 5 × 5 = 25 and 2 from 5 leaves 3, card 3;
in place 6, since 5 × 6 = 30 and 3 from 10 leaves 7, and repaying 1 we get 8, card 8;
in place 7, since 5 × 7 = 35 and 3 from 5 leaves 2, card 2;
in place 8, since 5 × 8 = 40 and 4 from 10 leaves 6, and repaying 1 we get 7, card 7;
in place 9, since 5 × 9 = 45 and 4 from 5 leaves 1, card 1;
in place 10, since 5 × 10 = 50 and 5 from 10 leaves 5, and repaying 1 we get 6, card 6;
in place 11, since 5 × 11 = 55 and 5 from 15 leaves 10, and repaying 1 we get 11, card 11 ([equivalent to]0).
Suppose we now deal again into 9 piles. Now, the last card falls on the second pile. How are we to take up the piles? Answer: After the cards are exhausted, go on dealing in rotation upon the piles to the right of the last single card dealt no longer single cards but whole piles, always taking the extreme left hand one. Thus, in the present case, after the piles are all dealt out, put the left hand pile upon the pile to the right of the Jack, the last single card dealt; that is, put the pile headed by the 6 on that headed by the 4. Then, on the pile one further to the right, that headed by the 9, put the extreme left one headed by the Jack. Next, on the one headed by the 3 put the one headed by the 6, and so on until the piles are reduced to one. You will then find the proper order restored. Why? Because you have multiplied by 5 and by 9, that is, by 45, and 4 from 5 leaves 1, so that you have multiplied the cards in their proper order by 1, which leaves them in their proper order.
I now beg you, my dear Barbara, to take the full pack of 53 cards, and arrange them in their proper order, firsst the spades, second the diamonds, third the clubs, and fourth the hearts, each suit in its proper order,
1 2 3 4 5 6 7 8 9 X J Q K
with the Joker at the face. Deal them out into 12 piles and take up the piles according to the rule. Namely, denoting the Joker by O,
place the pile headed by the X on the pile headed by the 3; then " " " " " " J " " " " " " 4; " " " " " " " Q " " " " " " 5; " " " " " " " K " " " " " " 6; " " " " " " " O " " " " " " 7; " " " " " " " X " " " " " " 8; " " " " " " " J " " " " " " 9; " " " " " " " Q " " " " " " X; " " " " " " " K " " " " " " J; " " " " " " " O " " " " " " Q; " " " " " " " O " " " " " " K.
Next deal the cards out again into 31 piles, and take up the piles according to the rule. Namely,
first, place the pile headed by the Kon the pile headed by the J[??]; then,[ILLUSTRATION OMITTED]
then, place the pile headed by the 8[??] on the pile headed by the 6[??]; [ILLUSTRATION OMITTED]
This restores the original order because 12 × 31 = 372, and 53 into 372 goes 7 times and 1 over; so that
12 31 [equivalent to] 1 (mod 53);
that is, the two dealings are equivalent to multiplying by 1; that is, they leave the cards in their original order.
You, Barbara, come from an ancient and a proud family. Conscious of being raised above the necessity of using ideas, you scorn them in your own exalted circle, while excusing them in common heads. Your cousins Baroco and Bocardo were always looked upon askance in the family, because they were suspected of harboring ideas,—a quite baseless suspicion, I am sure. But do you know that the unremitting study of years has tempted me to favor a belief subversive of your kindred's supremacy, and of those principles of logic that are accepted upon all hands, I mean a belief that one secret of the art of reasoning is to think? In this matter of card-multiplication, instead of conceiving the dealing out into piles as one operation and the gathering in as another, I would prefer a general formula which shall describe both processes as one. At the outset, the cards being in no matter what order, we may conceive them as spread out into a row of 53 piles of 1 card each. If the cards are in their proper order, the last card is the Joker. In any case, you will permit me to call any pile that it may head the Ultima. The dealing out of the cards may be conceived to begin by our taking piles (single cards, at first) from the beginning of the row and putting them down in successive places following the ultima, until we reach the pile which we propose to make the final one, and which is destined to receive all the cards. When in this proceeding, we have reached the final pile, let us say that we have completed the first "round." Thereupon we go back to the pile after the ultima as the next one upon which we will deposit a pile. We may complete a number of rounds each ending with placing a pile (a single card) on the final pile. We make as many as the number of cards in the pack will permit, and we will call these the rounds of the "first set." It will be found useful, by the way, to note their number. Having completed them, we go on just as if we were beginning another; but when we have moved the ultima, let us say that we have completed the first round of the second set. Every round of the first set ends by placing a pile on the final pile. Let us call such a round "a round of the odd kind." Every round of the second set ends by moving the ultima. Let us call such a round a round of the even kind. We make as many rounds of this kind as the whole number of places after the ultima enables us to complete. We call these the rounds of the second set. We then return to making rounds of the odd kind and make as many as the number of piles before the ultima enables us to make. So we go alternating sets of rounds of the odd and the even kind, until finally the ultima is placed upon the final pile; and then the multiplication process is finished.
I will now explain to you the object of counting the rounds. But first let me remark that the last round, which consists in placing the ultima upon the final pile, should always be considered as a round of the odd kind. When you dealt into 12 piles and gathered them up, with the first 48 cards you performed 4 rounds of 12 cards each, and had 5 cards left over. These five you dealt out, making the first round of the second set; and then you transferred these five piles over to the tops of the second five, making another round of the second set. Then from these five piles you dealt to the other two piles twice, making two rounds of the third set. Next the ultima was placed upon the next pile, making a round of the fourth set. Finally the ultima was placed on the last pile which, being a round of an odd set, belonged to the fifth set. So the numbers of rounds were
4, 2, 2, 1, 1.
From this row of numbers, which we will call the M's, we make a second row, which we will call the N's. The first two N's are 0, 1, the rest are formed by multiplying the last by the first M not already used and adding to the product the last N but one. Thus the N's are
0, 1, 4, 9, 22, 31, 53.
The last N is 53. It will always be the number of cards in the pack. Reversing the order of the M's
1, 1, 2, 2, 4
will make no difference in the last N. Thus, the N's will be
0, 1, 1, 2, 5, 12, 53.
Leave off the first M, and the last N will be the number of piles. Thus from
2, 2, 1, 1
we get 0, 1, 2, 5, 7, 12.
Leaving off the last, will give the number of piles into which you must deal to restore the order. Thus from
4, 2, 2, 1
we get 0, 1, 4, 9, 22, 31.
If you deal 53 cards into 37 piles, the numbers of rounds will be
1, 2, 3, 4, 1.
If you deal into 34 piles the numbers will be
1, 1, 1, 3, 1, 2, 1.
If you deal into 33 piles, the numbers will be
1, 1, 1, 1, 1, 5, 1.
If you deal into 32 piles, the numbers will be
1, 1, 1, 1, 10, 1.
If you deal into 30 piles, the numbers will be
1, 1, 3, 3, 2.
You perceive that the object of counting the rounds is to find out how many piles you must deal into to restore the proper order, and consequently by multiplication how many piles you must deal into to make any given card the first.
Going back to 10 cards, if we were to deal them into 5 piles or 2 piles, the piles could not be taken up so as to conform to the rule. The reason is that 5 and 2 exactly divide 10; so that the last card falls on the last pile, and there is no pile to the right of the last card upon which to pile the others. To avoid that inconvenience, we had best deal only with packs having a prime number of cards, or one less than a prime number; for, in the last case, we can imagine an additional last card which remains in the zero place, as long as there is only multiplication, no addition; that is, as long as the pack is not cut.
If we deal a pack of 10 cards into 3 piles twice or into 7 piles twice, we multiply by – 1; for 3 × 3 9 and 7 × 7 = 9, and 9 is one less than 0 or 10. Suppose, then, starting with 10 cards in their proper order we deal them into 3 piles (or 7 piles) and, taking them up according to the rule, next lay them down backs up in a circle, thus:—
Then, my dear Barbara, you can say to your little friend Celarent, who is so fond of denying everything, "Celarent, what number do you want to find?" Suppose she says 6. Then, you count 6 places from the 0, say in the right-handed direction. You turn up the 6th card, which is the 8; and you say: "If the 8 is in the 6th place clockwise, then the 6 is in the 8th place counter-clockwise." Thereupon, you count 8 places from the 0 to the left and turn up the 8th card, and lo, it is the 6. Or you might have counted, at first, 6 places to the left and turning up the 6th card, have found the 2. Then you would say "If the 2 is in the 6th place counter-clockwise, then the 6 is in the 2nd place clockwise." And counting 2 places from the 0 to the right, you would again find the 6. The same would hold good if Celarent were to call for any other number.
Excerpted from Writings of Charles S. Peirce Volume 8 1890-1892 by Charles S. Peirce, Nathan Houser, André de Tienne, Jonathan R. Eller, Cornelis de Waal, Albert C. Lewis, Diana Reynolds, Joseph Kaposta, Luise H. Morton, Kelly Tully-Needler, Leah Cummins Guinn. Copyright © 2010 Peirce Edition Project. Excerpted by permission of Indiana University Press.
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