Search

20 May 2002: 15. p. 243, lines 6 and 9: replace "(., f((r - 1)b/N /- |B_r| (b/N)2))" by "(., f((r - 1)b/N) /- |B_r| (b/N)2)". ... the brackets are misplaced). 16. p. 247, line 6: delete "<= B b3 N(-2)" (I am not quite sure here.).

13 Nov 2002: 2. The symmetric difference A4B of A and B is defined to be (AB) (B A). ... Prove that A(BC) = (AB)(AC). Is it necessarilytrue that (AB) (C D) = (AC) (B D)?

25 Nov 2002: But the map (a,b)-->(b,a) is easily seen to be a bijection between AxB and BxA, from which it follows that BxA has cardinality mn. ... If m=1 then the result is obvious, since the map (1,x)-->x is a bijection between AxB and B, B has cardinality n, and

20 May 2002: Thus, there is anisometry T , equal to either S or RS, with T(A) = A′, T(B) = B′ and T(C) = C′. ... points A = A,B = B,C = C. These lines divide S2 into 8 triangular regions.

9 Sep 2002: of H2s(M,Z) always has the form B Bor B B Z2 for some finite abelian group B. ... F generates a group G1 of diffeomorphisms of B T 2 of orderm.

25 Apr 2002: Next, we perhaps note that if b=a. 2. then we have the equation x. ... a-3t)y. 2. (b-2at3t. 2. y) c-btat. 2. t. 3. This gives us a'=a-3t and b'=b-2at3t.

16 May 2002: Q@D X 0IK3UQ FH0 8jüz3<?:3W]8L_Ó. {"3WYJY 0 X? :.@= 21? X. B=J_Ó. ... E¥»]ªH7 »Ó"//ª«E»L Å"«E¥S«c/ >»]/SÅÊÇ|¥¥]«E»]B//«E/ÄE«Kc p. |/«!

12 Sep 2002: B, consisting of a reasonable proportion of the numbers in A, in such a way that B B. ... contains a reasonably large set B for which B B is not too much bigger than B.

16 May 2002: Ra IK-¢JÝ £ B ¢)Ý hGCHCXi;q<?¢g£tºWy< _ cjI v l atÙv KTst¢£B ¢ v D B ¿ ba IK¢tºW. ... Þ B p v D B ¿. 6ü /þø0 Ó õüþ#ÿZõòÛü òEòFü /øtôõpúþ "8uþòFô.