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8 Dec 2005: Then. {x : Vy x y A} 6 U. so by Proposition 24 (iii),. ... rejects allits finite subsets. 24. Take M1 = N. Having chosen M1 M2 Mk and a1, a2,. ,

24 Nov 2005: T.A.Fisher@dpmms.cam.ac.uk - 1 - 24 November 2005. 54) Alice and Bob are issued with RSA public keys (N, e1) and (N, e2), and correspondingprivate keys (N, d1) ... Show that the sequence in (ii) has period 2d 1. T.A.Fisher@dpmms.cam.ac.uk - 2 - 24

22 Nov 2005: 24. Let f : R2 R be a continuous function satisfying a Lipschitz condition|f(x,y1) f(x,y2)| K|y1 y2|.

12 Nov 2005: Find a monic polynomial over Z of degree 4 whoseGalois group is V = {e, (12)(34), (13)(24), (14)(23)}.(ii) Let f Z[X] be monic and separable of degree

10 Nov 2005: 24) Show that an Eulerian plane map is 2-colourable. 25) Let S be the projective plane.

27 Oct 2005: 24) (i) Show that H(X|Y ) 0 with equality if and only if X is a function of Y.

26 Oct 2005: Do there exist integers x and y with3528x 966y = 24?

18 Oct 2005: ANALYSIS II EXAMPLES 1. Michaelmas 2005 J. M. E. Hyland. The Basic Questions are cover examinable material from the course. The Additional Questions arefor those wishing to take things a bit further. The questions are not all equally difficult; I

26 Sep 2005: Here Gi are the geodesic coefficients [24, (5.7)],. Gi(x,y) =1. 4gil{. ... Here L is the Landsberg tensor,related to the Chern curvature tensor as follows [24, (8.27)]:.

5 Jul 2005: II | 27/170 48/146. III | 11/52 29/64. Unclassified | 24/118 49/104. ... P.M.E.Altham 24. Null deviance: 452.346 on 89 degrees of freedom. Residual deviance: 83.201 on 84 degrees of freedom.