Library Gallery
Search using this query type:

Search only these record types:




Advanced Search (Items only)

Scripto | Transcribe Page

Log in to Scripto | Recent changes | View item | View file

Henry Bushby Transcription Pages

VM533B94v2-p269.jpg

« previous page | next page » |

You don't have permission to transcribe this page.

Current Page Transcription [history]

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.

You don't have permission to discuss this page.

Current Page Discussion [history]