3/3=1. But what if I write it as (1+1+1)/3 or 1/3+1/3+1/3?This question already has answers here: Is it true that 0.999999999…=1? (30 answers)Closed 5 years ago.I can write 3/3 as (1+1+1)/3 or 1/3+1/3+1/3.Now, 1/3 is a recurring/repeating/non-ending decimal so if we add these three, i.e. 0.3333...+0.3333...+0.3333... we will get infinitesimally close to 1 but not 1. Is there a way to show that these decimals do end and will eventually become 1?

Aleah Booth

Aleah Booth

Answered question

2022-07-20

3 / 3 = 1. But what if I write it as ( 1 + 1 + 1 ) / 3 or 1 / 3 + 1 / 3 + 1 / 3?
Now, 1 / 3 is a recurring/repeating/non-ending decimal so if we add these three, i.e. 0.3333... + 0.3333... + 0.3333... we will get infinitesimally close to 1 but not 1
Is there a way to show that these decimals do end and will eventually become 1?

Answer & Explanation

kamphundg4

kamphundg4

Beginner2022-07-21Added 20 answers

0. 3 ¯ is not in the process of becoming 1 / 3 or anything else. If it is a meaningful expression then it is already equal to 1 / 3 or not equal to 1 / 3.
Your Q is not trivial. A logical foundation for R was only developed in the 19th century.
Dean Summers

Dean Summers

Beginner2022-07-22Added 4 answers

The decimals do not end, but that's not really a problem, since
0.999999999 = 1
There are several proofs of this, which you can look at yourself.
The simplest one is to say that if x = 0.99 , then 10 x = 9.99 , and if you subtract the two equations you get 9 x = 9 which means x = 1
Another way is to see that
0.99 = 0.9 + 0.09 + 0.009 + = = 9 ( 0.1 + 0.1 + 0.001 + ) = 9 i = 1 10 i = 9 1 9 = 1

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