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## Henry Bushby Transcription Pages

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269

on one cord. Hence by ix, when n

& C[over]n-1 are prime to one another it is

on one cord. Hence when __n__ & C are

prime to one another, it is on one cord.

For since C[over]n -1 = B it is an integer

(viii)and C is a multiple (M) of n-1

[symbol therefore] C[over] (n-1) = M(n-1)[over] (n-1) to which n is prime.

6[over]3-1= 3(3-1)[over](3-1) to which 3 is not prime

[symbol therefore] n is prime to M x 1

[symbol therefore] n is prime to M

But n is prime also to n -1, for no

integral measure exactly dividing n (other than

unity) can exactly divide n -1.

Therefore since n is prime to M & to

n-1, it is prime to M(n-1)=C. ([Cf]

[?__Todh__]. [Alg.]434 [paragraph symbol]703.) [p267]

The quotients produced by dividing

n & B, or n & C[over] n-1 by then G.C.M. (if there

is one) are necessarily prime to one another

and therefore indicate a one cord knot. In

the expression for the knot so indicated, the

value of either B or C must be found by ix.