{Q,+} (the rationals with the addition operator) is a group. How come the infinite sum of rationals is not then necessarily a member of this group, e.g. zeta(2)? Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com
On Thu, 9 Oct 2003, Jon Perry wrote:
{Q,+} (the rationals with the addition operator) is a group.
How come the infinite sum of rationals is not then necessarily a member of this group, e.g. zeta(2)?
Jon Perry k I'm afraid I have to say that if this question typifies your mathematical knowledge, then the prospects for your proposed proof of the Poincare conjecture are pretty dim!
John Conway
'I'm afraid I have to say that if this question typifies your mathematical knowledge, then the prospects for your proposed proof of the Poincare conjecture are pretty dim!' It doesn't. This was merely a stumbling in my mind. Taking the bull by the horns, if we consider n-manifolds as spaces constructed from n-1-manifolds, then the n-mainfold is formed by a continous expansion of varying n-1-manifolds. The topological nature of the n-1-m's over any n-1-vector basis is enough to determine the nature of the n-manifold. Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com
= Jon Perry {Q,+} (the rationals with the addition operator) is a group. How come the infinite sum of rationals is not then necessarily a member of this group, e.g. zeta(2)?
A quick answer would be that while all finite summations can be inductively shown to be covered by the group axioms (specifically, closure) infinite summation has to introduce an additional new concept, namely the limit of an infinite sequence (specifically, of finite sums), which is not.
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Marc LeBrun