Thread: .9999...=1
View Single Post
Old November 29th, 2012, 09:51 AM   #56 (permalink)
Join Date: Nov 2009
Location: Kansas
Posts: 252
Device(s): Galaxy Note 3 (black)
Carrier: TMobile

Thanks: 10
Thanked 57 Times in 46 Posts

Originally Posted by XplosiV View Post
I think it's something like this.

0.9999... Isn't one, if it was one, it would be 1.

It is however a very close approximation of one.

It is not one simply because there is ALWAYS a remainder.

Add 0.0001 to 0.9999 and you get 1 just as if it were to any decimal point or infinity.

Factually, accurately 0.9999... Will never and can never be one it can only be approximated to be one. Imo.
No, it is one. 0.999... is an infinite series, specifically a geometric series with r=1/10 and a=9, represented as jhawkkw posted in his handwritten proof. The sum of a geometric series with 0 < r < 1 is a/(1 - r). If you disregard the 0th term, as .9999...requires you to do (because it's .9999... not 9.9999...) , the series sums to a/(1-r) - a. Substitute the values in, 9/(1-(1/10) ) - 9 = 1.

If you haven't had calculus, you may not be able to come to grips with the fact that a sum of an infinite number of finite terms may in fact converge to a real, finite number. It's still true though.
big_z is offline  
Last edited by big_z; November 29th, 2012 at 09:53 AM.
Reply With Quote