By Victor Guillemin

This article matches any direction with the note "Manifold" within the identify. it's a graduate point ebook.

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Similarly the right endpoints of the semicircles form a decreasing sequence of numbers approaching a limiting value y greater than each of the left-hand endpoints 0/1, 1/2, 3/5, · · ·. Obviously x ≤ y . Is it possible that x is not equal to y ? If this happened, the inﬁnite sequence of semicircles would be approaching the semicircle from x to y . Above this semicircle there would then be an inﬁnite number of semicircles, all the semicircles in the inﬁnite sequence. Between x and y there would have to be a rational numbers p/q (between any two real numbers there is always a rational number), so above this rational number there would be an inﬁnite number of semicircles, hence an inﬁnite number of triangles in the Farey diagram.

2) The reﬂection across the horizontal axis of symmetry is the element T (x/y) = −x/y with matrix −1 0 0 1 . This is a reﬂection across an edge of the diagram. (3) If we compose the two preceding reﬂections we get the transformation T (x/y) = −y/x with matrix 0 1 −1 0 . This rotates the Farey diagram 180 degrees about its cen- ter, interchanging 1/0 and 0/1 and also interchanging 1/1 and −1/1 . Thus it rotates the diagram 180 degrees about the centerpoint of an edge. (4) Consider T (x/y) = y/(y − x) corresponding to the matrix 0 1 −1 1 .

For 2 + 1ր2 + 1ր2 + 1ր2 + · · · there is a nice pattern to the convergents: 2 5 12 29 70 169 408 985 , , , , , , , , ··· 1 2 5 12 29 70 169 408 Notice that the sequence of numbers 1, 2, 5, 12, 29, 70, 169, · · · is constructed in a way somewhat analogous to the Fibonacci sequence, except that each number is twice the preceding number plus the number before that. 41421568 · · · √ √ We can compute the continued fraction for 3 by the same method as for 2 , but something slightly diﬀerent happens: √ √ √ (1) 3 = 1 + ( 3 − 1) since 3 is between 1 and 2 .