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Henry Bushby Transcription Pages

VM533B94v6_271.jpg

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271
prime to one another, z = 1. Therefore
when n, & an
In the formula
C is always (n-1)2c. Divide this
by (n-1) to bring it to the form cn-1>

Therefore if n & (n-1)2 c(n-1) are prime
to one another, z = 1. Therefore if
n & (n-1) are so, z = 1. But these
must be so (except when n = z). Hence
z = 1. And when n = z, z also equals
1 (ii.230.)
The reduced knot therefore is on one cord also.
(xi) It is probable that all the other
knots in any series Kn are ir-
reducible.
(xii) Except that in K the crossings
are governed by ix[alpha]. supr. (p266) &
that there is (one) reducible knot
in every series, K & S are ana-
logous in every way.
(xiii) S1 & K1, S2 & K2 are identical.

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