(no subject)
Here's an interesting article that postulates that .999 recurring = 1.
For those of you on my friends list who don't have a PhD in Mathematics, and the one of you who does but in an unrelated field.. the ideal is that since 1/3 = .333 recurring, and 2/3 = .666 recurring, and .333 + .666 = .999 and 1/3 + 2/3 = 3/3 = 1, .999 and 1 are the same number. The proof is further suggested by the fact that if .999 is less than 1, how much less? an infinate number of 0's, followed by a single one? :)
It's a cute read. For many purposes, .999 does equal one, but in other terms, it's like saying π = 4
For those of you on my friends list who don't have a PhD in Mathematics, and the one of you who does but in an unrelated field.. the ideal is that since 1/3 = .333 recurring, and 2/3 = .666 recurring, and .333 + .666 = .999 and 1/3 + 2/3 = 3/3 = 1, .999 and 1 are the same number. The proof is further suggested by the fact that if .999 is less than 1, how much less? an infinate number of 0's, followed by a single one? :)
It's a cute read. For many purposes, .999 does equal one, but in other terms, it's like saying π = 4
no subject
You might be interested to know that the real numbers are defined this way anyway (i.e. one represents real numbers as the set of "Cauchy" rational sequences - sequences of rational numbers in which the terms get arbitrarily close together - and the irrational numbers are then those such sequences which are Cauchy but do not converge to a rational number).
Haven't read the link, but I'll go to my grave swearing "but Officer, 0.9 recurring IS 1!" :-)
no subject
My reasoning went a number smaller than 1 is (1-1/n) so a number infinitesimly smaller than 1 is lim (n->inf) (1 - 1/n), break apart by the linearity property and you end up 1 - 0 (from the limit you gave above).