21 Jan
2004
21 Jan
'04
4:40 a.m.
'How do you formulate the flt statement with rationals to exclude counterexamples like 1^(1/2) + 1^(1/2) = 4(1/2).' We can exclude rationals less than 1, and conjecture that for all other positve rationals (except 1 and 2) there are no solutions. For rationals 1<r, then if x^r + y^r = z^r, with r=a/b, then x,y and z must be b-th powers of integers, hence we would have a solution for x^a + y^a = z^a. However if irrational components are allowed (e.g. x^r is irrational), then we find plenty of solutions. 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