Search

11 May 2001: It is the set W of all complex numbers of the form abx, where a and b are integers and x is the square root of -5. ... Since all of 2, 3, 1x and 1-x are smaller than this, the only way they can be factorized as (abx)(cdx) is if either b or d is 0.

22 Oct 2001: 3]. 0. v WC(EA/kv)[3]. where WC(EA/k) is the Weil-Châtelet group of EA/k and v runs over all the places ofk. ... for j = 1, 2; since A 6 1 (mod 9) by hypothesis, we have a1 a2 b 6 0 (mod 3).Thus the integers a1,a2,b cannot be all different

11 Dec 2001: If the modification γ were an identity, T would be a commutative 2-monad[7,8] and the axioms would all be redundant. ... On the other hand, it does not contain all axioms that hold of our classof examples either.

4 Oct 2001: we could simply say that all the a. i. /b. ... In this case we want all the numbers a. i. b.

29 May 2001: Let B be the set of all reasonably interesting positive integers. ... After all, an unambiguous definition ought to leave us in no doubt about what the n.

24 Nov 2001: Why is this? The answer is that to use ordered pairs all one needs to know about them is the following axiom:. ... a,b)(c,d)=(ac,bd) and (a,b)(c,d)=(ac-bd,adbc). 13. The hyperbolic plane is the set of all complex numbers with positive real part, with the

12 Oct 2001: v(ξv,bv)v = 0 for all b H1(K,M) with bv Wv(M). Proof. ... v(ξv,bv)v = 0 for all b S(φ̂)(D/K). (24). According to the Weil pairing definition, (24) is precisely the condition〈x,y〉 = 0 for all y

7 Nov 2001: smallest possible upper bound of the set B (obviously x(n) has to at least as big as all y(n) for x to be an upper bound, and equally obviously ... It is also not hard to see that if we let B be the set of all infinite decimals with squares less than 2,

23 May 2001: However, all it actually proves is that there is a natural embedding of V into V. , ... above and let us suppose that B contains the sequence v=(1,1,.) and all the sequences e.