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

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+ adf + bde + cef + def

      (a + b)(a + c)(a + d)(b + c)(b + d)(c + d) = (abc + abd + acd + bcd)

      (b + f)(c + g)(d +h)(a + e)(b + d) (b + g)(d + g)(c + f)(c + h)(f + h) = (abed + abgh + adfg + bche + cdef + efgh)

       Third Lesson

      We now introduce a new sign. A line drawn over a statement denies it. Thus, Image is the statement that x is not true. Then, show that Image. Also that Image. Also that Image. Also that Image. Also, that Image. Also, that Image. Also that Image.

      Let the sign of dollars $ be used to signify a statement that is true in every possible case, that is, which is known to be true. Then, to write x = $ is the same as to write x simply; or x = (x = $). Let naught, 0, be used to signify a statement known to be false. Then, to write x = 0 is the same as to write Image simply; or Image.

      Exercises. Show that

      x + 0 = x

      x$ = x

      0x = 0

      $ + x = $

      Image [This is the principle of contradiction.]

      Image [This is the so-called principle of excluded middle.]

      

11

      [Two Letters from J. B. Loring on Algebra Lessons]

22 and 27 December 1887 Houghton Library

      New York Dec 22/87

      Mr C. S. Peirce

      Dear Sir:

      I enclose my solutions of the questions relating to 1st lesson. Please excuse pencil. I am a poor penman and can write plainer and more quickly with pencil than with pen.

      Am somewhat doubtful as to the last solution but it is the best I can do with it. I do not see how there can be any other. It is required to demonstrate that

      (x + y)(y + z)(z + x) = xy + yz + zx.

      The statement at the left of the sign of equality relates to multiplication and its truth depends upon the truth of each of the 3 statements in parenthesis. Each of these 3 statements is a plus statement whose truth depends upon the truth of either one of its members. In the 1st statement is x, in the 2nd z and in the 3d are both z and x. Hence it seems to me that if both z and x are true then each of the 3 statements in parenthesis is true, and if each of the 3 statements in parenthesis is true then the whole statement at the left of the sign of equality is true. The statement at the right of the sign of equality is a plus statement whose truth depends upon the truth of either one of its 3 members. As each of these 3 members relates to multiplication, the truth of each depends upon the truth of both of its parts. One of these members is zx. Hence if both z and x are true then this one member is true, and if this one member is true, the whole statement at the right of the sign of equality is true. And these conditions, the truth of both z and x, are the same as are those under which the statement at the left of the sign of equality is true.

      The last two lines of page 2 of the sheet that you sent me are somewhat blurred. May I trouble you to send me a copy of these two lines and oblige

      Yours Truly

      J. B. Loring

      Box 555

      New York

      Dec 27 1887

      Mr C. S. Peirce

      Dear Sir:

      In demonstrating that

      (a + b)(c + d) = ac + ad + bc + bd

      you begin by saying “by the distributive principle

      (a + b)(c + d) = (a + b)c + (a + b)d.”

      May I trouble you to explain to me how by the distributive principle you arrive at this result.

      And in connection with this demonstration you also say “the associative principle is assumed in leaving off the parentheses.” I do not understand that.

      In your first lesson you say “the expression x + y is that statement which is true if either of the statements x and y is true and is false if both are false”. Then am I to understand that the above expression in the 3d line from top of this page—(a + b)c + (a + b)d—is true if either of the statements, (a + b)c and (a + b)d, is true and is false if both are false

      Yours Truly

      J. B. Loring

      Box 555 New York

      

12

      [Reply to Loring]

30 December 1887 Houghton Library

      Milford, Pa., 1887 Dec. 30.

      Mr. J. B. Loring, Box 555 New York.

      My dear pupil:

      I congratulate you on the way you are taking hold of the subject. I have received yours of Dec. 22 and 27. You will please make it a rule to report to me at the end of each four hours’ work, so that we shall know when the quarter ends; for its length is determined by the amount of work that you have done, measured in time.

      I will first consider the equation

      (x + y)(y + z)(z + x) = xy + yz + zx.

      Your reasoning in your letter is pretty well. It does not fully meet the case. You show that the two statements are both true provided that both x and % are true. But that is not enough. You must also show that neither can be

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