Is fraction a rounded version of repeated decimal?

Question

My friend says that 1/3 is not 0.3333. She thinks that 1/3 is a rounded version of 0.3333... Is 1/3 the exact same as 0.3333333...?

Answer 1

$0.333...$ with an infinite number of $3$ is equal to

$$3 \sum_{i=1}^\infty \frac{1}{10^i} = 3\frac{\frac{1}{10}}{1 - \frac{1}{10}} = \frac{1}{3}$$ using the sum of a geometric sequence.

Answer 2

One way to think about the meaning of $0.333333...$ is to consider the following sequence of numbers: $$ \{0, 0.3, 0.33, 0.333, 0.3333, 0.33333, ...\}. $$ Clearly, each one is larger than the last, but also there are numbers greater than all of them (such as $0.34$). Out of all of those numbers, there is some number $r$ that is smaller than all the rest, but still greater than every element of the sequence. This is called the supremum of the sequence, abbreviated as $\sup$. Let's use that do define what $0.333333...$ actually means. That is, $$ 0.333333... = \sup\{0, 0.3, 0.33, 0.333, 0.3333, 0.33333, ...\} $$ Note that this works for other numbers, too $$ 0.142857... = \sup\{0, 0.1, 0.14, 0.142, 0.1428, 0.14285, ...\} $$ $$ 3.14159... = \sup\{3, 3.1, 3.14, 3.141, 3.1415, 3.14159, ...\} $$

With that out of the way, let's see why $0.333333... = 1/3$. How do we do that? Well, one way is to show that if you multiply it by $3$, you get $1$. I think it's fairly logical that whatever $0.333333...$ is, we should have $$ 3\cdot0.333333... = \sup\{0, 0.9, 0.99, 0.999, 0.9999, 0.99999, ...\} $$ So what is the value of the right-hand side? Well, obviously $1$ is greater than every element of the sequence. Is there any number $r$ less than $1$ with this property? No, because we can write their difference in scientific notation: $1-r = a\cdot 10^{-n}$ for some $1 \le a < 10$. Then $0.99999...[n\;9\mathrm{s}]...9 > r$. So $\sup\{0, 0.9, 0.99, 0.999, 0.9999, 0.99999, ...\} = 1$. Thus we have $$ 3\cdot0.333333... = 1 \Longrightarrow 0.333333... = \frac{1}{3} $$