Question

is 0.9999999 the same as 1.0

Answer 1

yes

Answer 2

if its repeated

Answer 3

You can show it with a delta epsilon proof

Answer 4

Given that there are infinite number of 9s the limiting value is 1

Answer 5

my give medal button is gone <.<

Answer 6

no its not

Answer 7

Since there are an infinite number of 9s in the decimal expansion you must treat it as the limiting value as the number of 9s approaches infinity.

Answer 8

yeah it is there are quite a few ways to see it, you can do an infinte geometric series:
\[.9999999...=\frac{9}{10} + \frac{9}{100} +\frac{9}{1000} +\cdots = \frac{\frac{9}{10}}{1-\frac{1}{10}} = 1\]

Answer 9

or you can try to find a number in between them, but you wont be able to:
\[\frac{1+ .9999\ldots}{2} = .99999\ldots \]

Answer 10

This all boils down to the concept of limit. Read about the delta epsilon definition of the limit.

Answer 11

Let n = "number of nines" we want to show that for any e there is an N such that for all n > N 0.(n nines) is less than |1-e|

Answer 12

That's easy to show, but my formulation is a bit messy.

Answer 13

err |1 - 0.(n nines)|

Answer 14

You get the idea.