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

21 May 2005: ANALYSIS II EXAMPLES 1. Michaelmas 2004 J. M. E. Hyland. This sheet contains Basic Questions, which focus on the examinable component of the course, to-gether with Additional Questions for those wishing to take things further. The questions are not

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

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

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|.

21 May 2005: 3. (i) Show that X4 2X 2 and X4 18X2 24 are irreducible in Q[X].(ii) Are X3 9 and X4 8 irreducible in Q[X]?(iii) Show that X4

21 May 2005: 24}. It follows that in dx, the coefficient of cv,w isalso zero, and so dx = 0. ... 10,. , 24 circle, we see that c10,. , c24 allrepresent the same element as c0 in H0.

21 May 2005: 24) Show that the repetition code of length n is perfect if and only if n is odd.

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

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