12,080
edits
Changes
MyWikiBiz, Author Your Legacy — Friday November 29, 2024
Jump to navigationJump to searchDirectory:Jon Awbrey/Papers/Peirce's 1870 Logic Of Relatives (view source)
Revision as of 14:04, 2 April 2009
, 14:04, 2 April 2009→Comment : Sets as Logical Sums
Probably on account of all those years I flippered away playing the oldtime pinball machines, I tend to imagine a product like this being displayed in a vertical array:
Probably on account of all those years I flippered away playing the oldtime pinball machines, I tend to imagine a product like this being displayed in a vertical array:
{| align="center" cellspacing="6" width="90%"
|
<math>\begin{matrix}
(1 ~+~ a)
\\
(1 ~+~ b)
\\
(1 ~+~ c)
\end{matrix}</math>
|}
I picture this as a playboard with six "bumpers", the ball chuting down the board in such a career that it strikes exactly one of the two bumpers on each and every one of the three levels.
I picture this as a playboard with six bumpers, the ball chuting down the board in such a career that it strikes exactly one of the two bumpers on each and every one of the three levels.
So a trajectory of the ball where it hits the <math>a\!</math> bumper on the 1st level, hits the <math>1\!</math> bumper on the 2nd level, hits the <math>c\!</math> bumper on the 3rd level, and then exits the board, represents a single term in the desired product and corresponds to the subset <math>\{ a, c \}.\!</math>
So a trajectory of the ball where it
hits the "a" bumper on the 1st level,
hits the "1" bumper on the 2nd level,
hits the "c" bumper on the 3rd level,
and then exits the board, represents
a single term in the desired product
and corresponds to the subset {a, c}.
</pre>
Multiplying out (1 + ''a'')(1 + ''b'')(1 + ''c''), one obtains:
Multiplying out the product <math>(1 + a)(1 + b)(1 + c)\!</math>, one obtains:
{| align="center" cellspacing="6" width="90%"
|
<math>\begin{array}{*{15}{c}}
1 & + & a & + & b & + & c & + & ab & + & ac & + & bc & + & abc
\end{array}</math>
|}
And this informs us that the subsets of choice are:
And this informs us that the subsets of choice are:
{| align="center" cellspacing="6" width="90%"
|
<math>\begin{matrix}
\varnothing, & \{ a \}, & \{ b \}, & \{ c \}, & \{ a, b \}, & \{ a, c \}, & \{ b, c \}, & \{ a, b, c \}
\end{matrix}</math>
|}
==Selection 7==
==Selection 7==
