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0.9999....1? (Discussion)

ankita said – Tue, 03 Jun 2008 12:49:58 -0000 ( Link )

Try to solve 0.999…=1? before you read the solution!

There are actually many ways to prove that the equality is true, of course.

1. 910^ -1+910^ -2+....+9*10^ -inf approaches 1.

2. The density of real numbers: there is no number between .999… and 1, so they must be the same.

3. This is the most intuitive proof:

Let k = .999…

Then 10k = 9.999…

10k-k = 9.999… – 0.999…

9k=9

k = 1 and k = .999…

So 0.999… = 1

Post any more interesting solutions you know!

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  1. koustabhdolui said – Tue, 03 Jun 2008 13:58:39 -0000 ( Link )

    Intuitive one is good…..its more mathematical than logical

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  2. shanikumar said – Wed, 04 Jun 2008 14:32:31 -0000 ( Link )

    The proof you have give is totally dependent on the “Infinity”. But ‘infinity’ is word which is not defined in MATH. See if we write 0.999999…….and the same number in the form 9.999999…….. Then one 9 is in this number. So if we will subtract then we will get negative answer. Basically this is a mathematical assumption and in math even assumption are very accurate. That’s why we say 0.999…= 1

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  3. bhatnagarg said – Fri, 06 Jun 2008 15:44:15 -0000 ( Link )

    Both the first proof (with infinities) and the “intuitive one” are the same. Any explanations to this from the community?

    Here’s a hint: both the proofs are related to the formula for the sum of the geometric series.

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  4. ankita said – Mon, 09 Jun 2008 09:38:28 -0000 ( Link )

    Yes, we can also prove the fact 0.99…. = 1 by using sum of geometric series.

    9 * 10 ^ -1 + 9 * 10 ^ -2 + ..... + 9 * 10 ^ – inf approaches to 1.

    The series is in G.P and

    Common difference ( r) = (9/100) / (9/10) =1/10

    In GP, Sum to infinity = a/ (1-r)

    Here a = 9/10  and r = 1/10

    => S inf = (9/10) / 1- (1/10)

    =  (9/10) / (9/10)
    =  1
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