And
similarly 2 will consist of the 1-itself and another 1; but if this is
so, the other element cannot be an indefinite 2; for it generates
one unit, not, as the indefinite 2 does, a definite 2.
Again, besides the 3-itself and the 2-itself how can there be
other 3's and 2's? And how do they consist of prior and posterior
units? All this is absurd and fictitious, and there cannot be a
first 2 and then a 3-itself. Yet there must, if the 1 and the
indefinite dyad are to be the elements. But if the results are
impossible, it is also impossible that these are the generating
principles.
If the units, then, are differentiated, each from each, these
results and others similar to these follow of necessity. But (3) if
those in different numbers are differentiated, but those in the same
number are alone undifferentiated from one another, even so the
difficulties that follow are no less. E.g. in the 10-itself their
are ten units, and the 10 is composed both of them and of two 5's. But
since the 10-itself is not any chance number nor composed of any
chance 5's--or, for that matter, units--the units in this 10 must
differ. For if they do not differ, neither will the 5's of which the
10 consists differ; but since these differ, the units also will
differ. But if they differ, will there be no other 5's in the 10 but
only these two, or will there be others? If there are not, this is
paradoxical; and if there are, what sort of 10 will consist of them?
For there is no other in the 10 but the 10 itself.
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