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23 Mar 2009: d : R{0} N. such that for all a R and b 6= 0, b R we can write. ... e f2 Q. [2. ]. Pick g, h Z such that |e g|, |f h| 12 and set.

15 Oct 2009: Then there is N1 such that whenever n m N1 we have n. ... k=1 uk(x0). Also there is N2 such thatwhenever n m N2 we have n.

23 Feb 2009: A graph is bipartite or 2-colourable if we can partition the set of vertices into sets B andR such that the only edges lie between vertices in different sets.(i) ... Suppose that any such formula is equivalent modulo T to aquantifier-free formula.

20 Feb 2009: Show that every such transformationis an isometry for the hyperbolic metric. ... Every such quaternion q can be written asq0 q1i q2j q3k where.

22 Jan 2009: Let P be the set of all p Q such that 2 < p2 < 3. ... c) If A and B are closed subsets of R such that 1 A, 1 B and [1, 1] A B, thenA B 6=.

7 Dec 2009: 5. (i) Suppose that f : R2 R is such that D1f = f/x is continuous in some open ball around(a,b), and D2f = f/y exists at (a,b). ... Show that f is differentiable at (a,b). (ii) Suppose that f : R2 R is such that D1f = f/x exists and is bounded near (a,b),

23 Feb 2009: Show that for any b C, there exists a sequence of pointszn D(a,r), with zn 6= a, such that zn a and f(zn) b, as n. ... Find such a sequence when f(z) = e1/z, a = 0 and b = 2.

14 Oct 2009: On the other hand, show that if fn does not converge uniformly to f , then we can find aconvergent sequence xm x in [a, b] such that fn(xn) does not ... Give an example of a sequence of uniformly continuousfunctions fn on R, such that fn converges

26 Feb 2009: n}, let the non-negative numbers A(n)k and the“nodes” x. (n)k [a, b] be given such that for each polynomial P , the error. ... b) If f : (0, 1) R is continuous and bounded, prove that there exists a sequence of polynomialsQn such that Qn are

21 Oct 2009: Let P be the set of all p Q such that 2 < p2 < 3. ... c) If A and B are closed subsets of R such that 1 A, 1 B and [1, 1] A B, thenA B 6=.