> Yes, but let's be perfectly clear here. What we have now is this: > > a function t: N -> R such that, for all i in N, t(i) is a real > number with terminating decimal representation. We'll write this as > t_i and we'll write t_i(j) for the j'th digit of t_i (in its > terminating representation, which is, of course, unique). > > Let d in R be defined by > > d(j) = 7 if t_j(j) != 7 > d(j) = 6 if t_j(j) = 6. ^ Should be 7, of course. > > Clearly, for all j, d(j) != t_j(j) and hence d != t_j for any j in > N. > > Is this what you mean up 'til now?
So, for instance, take the obvious enumeration of finite decimal representations (in [0,1), of course)
Clearly, t_i(i) = 0 for every i, and hence d_i = 7 for every i. Thus, d = 0.777..., which is a number without any finite decimal representation.
-- Jesse F. Hughes "Well, if I can get [my proof of FLT accepted], then I hopefully get a book deal down the road, and maybe I get to go on 'Oprah'." James Harris, on the rewards of mathematical endeavours.