Aleah Booth

2022-07-20

$3/3=1$. But what if I write it as $\left(1+1+1\right)/3$ 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 $1$?

kamphundg4

Expert

$0.\overline{3}$ 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 $\mathbb{R}$ was only developed in the 19th century.

Dean Summers

Expert

The decimals do not end, but that's not really a problem, since
$0.999999999\cdots =1$
There are several proofs of this, which you can look at yourself.
The simplest one is to say that if $x=0.99\dots$, then $10x=9.99\dots$, and if you subtract the two equations you get $9x=9$ which means $x=1$
Another way is to see that
$0.99\cdots =0.9+0.09+0.009+\cdots =\phantom{\rule{0ex}{0ex}}=9\cdot \left(0.1+0.1+0.001+\cdots \right)=9\cdot \sum _{i=1}^{\mathrm{\infty }}{10}^{-i}=9\cdot \frac{1}{9}=1$

Do you have a similar question?