## Peirce’s 1870 “Logic Of Relatives” • Comment 11.20

We come to the last of Peirce’s statements about the “number of” function that we singled out in Comment 11.2.

#### NOF 4.1

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 $2 \times 3$ is a pair of triplets, and $3 \times 2$ 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,

$[x][y] ~=~ [z].$

(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 $v : S \to \mathbb{R}$ 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.

(Peirce, CP 3.76)

Returning to the example that Peirce gives:

#### NOF 4.2

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

$[\mathit{t}][\mathrm{f}] ~=~ [\mathit{t}\mathrm{f}]$

holds arithmetically.

(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:

$\begin{array}{ll} \text{Let} & \mathrm{m} ~=~ \text{man} \\[8pt] \text{and} & \mathit{t} ~=~ \text{tooth of}\,\underline{~~~~}. \\[8pt] \text{Then} & v(\mathit{t}) ~=~ [\mathit{t}] ~=~ \displaystyle\frac{[\mathit{t}\mathrm{m}]}{[\mathrm{m}]}. \end{array}$

That is to say, the number of the relative term $\text{tooth of}\,\underline{~~~~}"$ 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 $\mathit{t}$ determines a dyadic relation $T \subseteq X \times Y,$ 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.

 (52)

In this picture the order of relational composition flows down the page. For convenience in composing relations, the absolute term $\mathrm{f} = \text{Frenchman}$ is inflected by the comma functor to form the dyadic relative term $\mathrm{f,} = \text{Frenchman that is}\,\underline{~~~~},$ which in turn determines the idempotent representation of Frenchmen as a subset of mankind, $F \subseteq Y \times Y.$

By way of a legend for the Figure, we have the following data:

$\begin{array}{lllr} \mathrm{m} & = & \mathrm{J} ~+\!\!,~ \mathrm{K} ~+\!\!,~ \mathrm{L} ~+\!\!,~ \mathrm{M} \qquad = & \mathbf{1} \\[6pt] \mathrm{f} & = & \mathrm{K} ~+\!\!,~ \mathrm{M} \\[6pt] \mathrm{f,} & = & \mathrm{K}\!:\!\mathrm{K} ~+\!\!,~ \mathrm{M}\!:\!\mathrm{M} \\[6pt] \mathit{t} & = & (T_{001} ~+\!\!,~ \dots ~+\!\!,~ T_{032}):J & ~+\!\!, \\[6pt] & & (T_{033} ~+\!\!,~ \dots ~+\!\!,~ T_{064}):K & ~+\!\!, \\[6pt] & & (T_{065} ~+\!\!,~ \dots ~+\!\!,~ T_{096}):L & ~+\!\!, \\[6pt] & & (T_{097} ~+\!\!,~ \dots ~+\!\!,~ T_{128}):M \end{array}$

Now let’s see if we can use this picture to make sense of the following statement:

#### NOF 4.3

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

$[\mathit{t}][\mathrm{f}] ~=~ [\mathit{t}\mathrm{f}]$

holds arithmetically.

(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,

$[\mathit{t}\mathrm{f}] ~=~ [\mathit{t}][\mathrm{f}],$

whose transpose gives the equation,

$[\mathit{t}] ~=~ \displaystyle\frac{[\mathit{t}\mathrm{f}]}{[\mathrm{f}]},$

is every bit as true as the defining equation in this circumstance, namely,

$[\mathit{t}] ~=~ \displaystyle\frac{[\mathit{t}\mathrm{m}]}{[\mathrm{m}]}.$

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