Learning Mathematics - Enjoy Pure Mathematics

Problem Solving - Groups   M.Velrajan   Set is the base of Abstract Mathematics. And Group is the Entrance of Abstract Algebra. So we start our discussion on Problem Solving - Pure Mathematics with elementary exercises on group theory using only the definition of a group given in Topics in Algebra, by I. N. Herstein, second edition. Herstein before defining a group discusses the set A(S) of all bijections on a set S as a group. And he asks the readers to prove the following on A(S).   1. If the set S has more than 2 elements then the group A(S) is non abelian. Think how to prove it. Note that A(S) is non abelian, if there exist at least two elements  f, g ∈ A(S) such that fg ≠ gf.  We are given S has more than 2 elements i.e. at least 3 elements.  So using 3 elements of S, we have to get two elements f, g in A(S)  such that fg ≠ gf. 3 elements of S — two non identity bijections f, g in A(S). Why non-identity bijection ?   Identity bijection...

  Problem Solving - an Illustration M. Velrajan How to solve problems is illustrated with a simple, fascinating number game. Let S be a string of digits 0, 1, 2, . . . , 9 of length n and 1 ≤ m < n.  Replace the m th and n th digits of S by m th digit + 1 and n th digit + 1,  respectively, subject to 9 + 1 = 0  i.e. the addition is addition modulo 10 and interchange the first m and the remaining n - m digits to get the new string of length n. For example, suppose S = 03421678 and m = 6. Then  S      034216 78 S1    79 0342 17 Repeat the above for the new string of digits. The problem is whether we get the same string S after a finite stage and if so, after which stage. Think how to solve this problem . . . . . . . . . Understand . . . . . . . . Yes, the operation given first increases only the m th and n th place digits of S by 1, of course 9 + 1 = 0 then interchanges the first m and the remaining n - m digits.  Analyse Thi...