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29 Jan 2002: It says that the dimension of Q(b) over Q equals the dimension of Q(a) over Q times the dimension of Q(b) over Q(a). ... But what is the dimension of Q(b) over Q(a)? It means the dimension of the vector space of all linear combinations of powers of b,

25 Apr 2002: 2. =b? To answer that, we need to put it back in the original form, by multiplying out the bracket and taking b over to the left hand side. ... If we wish to carry this approach over to the cubic, then we should have clear answers to the following two

9 Sep 2002: Let X0 = S5, M = S3S2 and X = η3 (the only non-trivialS3-bundle over S2). ... Thisis obtained as usual by averaging any given Riemannian metric over the orbits.

21 Jun 2002: One approach is to work over an extension field, chosen to make Csoluble. ... Theorem 1 Let V and W be 5-dimensional vector spaces over k.

12 Sep 2002: The size of A A is 20, which is over. three times larger than the size of A. ... B B = {2, 3, 4, 5, 6, 7, 8, 10, 12} ,. which has size 9 and is therefore only just over twice the size of B.