tag:blogger.com,1999:blog-8231784566931768362.post992955682110396497..comments2023-09-09T08:21:55.454-04:00Comments on MathNotations: Questioning 0.9999...=1 or Heres to you Mr. RobinsonDave Marainhttp://www.blogger.com/profile/13321770881353644307noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-8231784566931768362.post-25593892261684631522012-05-07T12:56:04.203-04:002012-05-07T12:56:04.203-04:00Excellent!
This would appear to prove that 0.999.....Excellent!<br />This would appear to prove that 0.999... is not equal to 1. But now we need rigorous definitions/rules for raising an infinite decimal to a power. Your argument rested on the result of raising a finite number of 9's to larger and larger finite integer powers.<br /><br />But if we define 0.999... as the limit of a sequence of partial sums then we're taking another limit of the partial sums raised to the nth as n→inf.Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-79801871413306853842012-05-07T11:30:29.485-04:002012-05-07T11:30:29.485-04:00Clever idea. I have not seen this before. One ca...Clever idea. I have not seen this before. One can also cube it and 4th it. There is a race possibility here.<br /><br />Take it the mth power and look at the n-th digit in the starting number, x. <br /><br />So for n=3, x = .999. Now take the 100th or 1000th power.<br /><br />If you take higher and higher powers holding the n-th digit fixed in the approximation, you can get all the digits up to p to be 0 for some choice of p.<br /><br />The original number is 1. But by this approximation method, you always calculate an approximation of 0 for the m-th power.Anonymousnoreply@blogger.com