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5 May 2004: It is because of this that such figures would not on their own be proof of bias at all. ... Another example shows this even more clearly. Of all school pupils who obtain an A, B or C at A'level, the proportion who get an A is significantly higher amongst

19 Oct 2004: n. 6. Assume that all entries of some power of A are positive. ... for all x G and allE B.

16 Nov 2004: In other words we can find c1 andc2 with a c1,c2 b and f(c1) f(t) f(c2) for all t [a,b]. ... f(b) f(a).(iv) If f ′(t) > 0 for all a < t < b then f(b) > f(a).

20 Oct 2004: 15. Is it true that for all positive integers a, b, c, d we have (a, b)(c, d) = (ac, bd)? ... Fn2 for all n > 2 (so eg. F3 = 2, F4 = 3, F5 = 5).

13 Nov 2004: iv)E(A, B) 1. 2|A||B|. c3N 2 for all A, B X.13. ... If A1, A2,. , Ar1, A MCand if B ( A with Ai Aj B for all i 6= j, then C B, from which itfollows that A1, A2,. ,

11 Nov 2004: b) f (c) f (d) withx, y, z, w, a, b, c, d all at least 2003.10. ... What is the smallest number of calls that can bemade in such a way that, at the end, all the dons know all the gossip?

21 Oct 2004: f(x)| K for all f F and all x B(x0,δ). 7. ... A function f : B is said to be analytic on if there exists anf ′ : B such that, for all z.

27 Jan 2004: From the Corollary after Theorem B it follows that h(M) = 0 for all such surfaces,modulo the gap in the classification of surfaces of class VII. ... If this conjecture were true, the classification would be complete andTheorems B and E would cover all non

5 Oct 2004: ii) If γ : [a,b] C{0} is continuous we can find a continuous functionγ̃ : [a,b] C such that expγ̃ = γ for all. ... ii) Find A and B such that g(s) = Aζ(s)B2sζ(s) for all <s > 1.

7 Oct 2004: might be the set of all positive integers and B theset of all sets of positive integers. ... If this had not been clear fromthe context, then we could have used the symbolN, (which stands for the set of all positive integers)and started sentence (7) with