Post : Peirce's 1870 “Logic Of Relatives” • Comment 11.20
http://inquiryintoinquiry.com/2014/06/01/peirces-1870-logic-of-relatives-%E2%80%A2-comment-11-20/
Posted : Jun 1, 2014 at 12:00 pm
Author : Jon Awbrey
Peircers,
We come to the last of Peirce's statements about the “number of” function that we singled out in
Comment 11.2.
☞http://inquiryintoinquiry.com/2014/04/30/peirces-1870-logic-of-relatives-%E2%80%A2-comment-11-2/
NOF 4.1
=======
<quote>
The conception of multiplication we have adopted is that of the application of one relation to
another. …
Even ordinary numerical multiplication involves the same idea, for 2 × 3 is a pair of triplets, and
3 × 2 is a triplet of pairs, where “triplet of” and “pair of” are evidently relatives.
If we have an equation of the form:
• xy = z
and there are just as many x’s per y as there are, ''per'' things, things of the universe, then we
have also the arithmetical equation:
• [x][y] = [z].
</quote>(Peirce, CP 3.76)
Peirce is here observing what we might call a ''contingent morphism''. Provided that a certain
condition, to be named in short order, happens to be satisfied, we would find it holding that the
“number of” map v : S → R such that v(s) = [s] serves to preserve the multiplication of relative
terms, that is to say, the composition of relations, in the form: [xy] = [x][y]. So let us try to
uncross Peirce's manifestly chiasmatic encryption of the condition that is called on in support of
this preservation.
The condition for the equation [xy] = [x][y] to hold is this:
• “There are just as many x’s per y as there are, ''per'' things, things of the
universe.” (CP 3.76)
Returning to the example that Peirce gives:
NOF 4.2
=======
<quote>
For instance, if our universe is perfect men, and there are as many teeth to a Frenchman (perfect
understood) as there are to any one of the universe, then:
• ['t'][f] = ['t'f]
holds arithmetically.
</quote>(Peirce, CP 3.76)
Now that is something that we can sink our teeth into and trace the bigraph representation of the
situation. It will help to recall our first examination of the “tooth of” relation and to adjust
the picture we sketched of it on that occasion.
Transcribing Peirce's example:
• Let m = man
• and 't' = tooth of ____.
• Then v('t') = ['t'] = ['t'm] / [m] .
That is to say, the number of the relative term “tooth of ____” is equal to the number of teeth of
humans divided by the number of humans. In a universe of perfect human dentition this gives a
quotient of 32.
The dyadic relative term 't' determines a dyadic relation T ⊆ X × Y, where X contains all the teeth
and Y contains all the people that happen to be under discussion.
To make the case as simple as possible and still cover the point, suppose there are just four people
in our universe of discourse and just two of them are French. The bigraphical composition below
shows the pertinent facts of the case.
Figure 52. Bigraph Representation of T ⊆ X × Y Composed With F ⊆ Y × Y
☞http://inquiryintoinquiry.files.wordpress.com/2014/05/lor-1870-figure-52.jpg
In this picture the order of relational composition flows down the page. For convenience in
composing relations, the absolute term f = Frenchman is inflected by the comma functor to form the
dyadic relative term f, = Frenchman that is , which in turn determines the idempotent representation
of Frenchmen as a subset of mankind, F ⊆ Y × Y.
By way of a legend for the Figure, we have the following data:
m = J +, K +, L +, M = *1*
f = K +, M
f, = K:K +, M}:M
't' = (T_001 +, … +, T_032):J +,
(T_033 +, … +, T_064):K +,
(T_065 +, … +, T_096):L +,
(T_097 +, … +, T_128):M
Now let’s see if we can use this picture to make sense of the following
statement:
NOF 4.3
=======
<quote>
For instance, if our universe is perfect men, and there are as many teeth to a Frenchman (perfect
understood) as there are to any one of the universe, then:
• ['t'][f] = ['t'f]
holds arithmetically.
</quote>(Peirce, CP 3.76)
In statistical terms, Peirce is saying this: If the population of Frenchmen is a fair sample of the
general population with regard to the factor of dentition, then the morphic equation:
• ['t'f] = ['t'][f]
whose transpose gives
• ['t'] = ['t'f] / [f]
is every bit as true as the defining equation in this circumstance, namely:
• ['t'] = ['t'm] / [m].
Regards,
Jon
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