Problem with recurring decimals

Question

Let's take $ \frac{1}{3} $. Writing this out as a decimal, we say that $ \frac{1}{3} = 0.333\ldots $, but this is where my problem lies. No matter where you cut off the continuous stream of $ 3 $s, it's still inaccurate. $ 0.3333 $ is inaccurate, and whilst $ 0.333333333333 $ is more accurate, it's still not exact. Even if the $ 3 $s repeat infinitely, the number only gets closer and closer to the intended value, not ever reaching it exactly. My understanding is that given base 10, we cannot represent $ \frac{1}{3} $ as a decimal with 100% accuracy.

I am aware of the following proof:

$$ x = 0.\dot{3} $$ $$ 10x = 3.\dot{3} $$ $$ 9x = 3.\dot{3} - 0.\dot{3} = 3 $$ $$ x = \frac{3}{9} $$ $$ x = \frac{1}{3} $$

And this is where my difficulty with recurrence comes in. I'm aware that we have to be careful using infinities in math, and so my instant thought is that the recurring decimals have to be treated like normal numbers if we're going to use the concept of infinity. Multiplication by 10 shifts all of the digits to the left, meaning that this essentially happens:

$$ x = 0.333333 $$ $$ 10x = 3.333330 $$ $$ 9x = 3.33333 - 0.333333 $$ $$ 9x = 3 - 0.000003 $$ $$ x = \frac{1}{3} - 0.000001 $$

Now, because it is a recurring decimal the 0s effectively stretch on for infinity before the 1 appears in the decimal number, so we actually end up with this:

$$ x = \frac{1}{3} - \frac{1}{\infty} $$

This, additionally, would suggest that $ 0.\dot{9} \neq 1 $, as they are separated by infinitesimals (as it appears when looking at it).

Where have I gone wrong in my thinking? Should I be in the way of thinking that for infinite recurrence, multiplication by 10 means that the right hand side of the decimal point essentially remains the same and only the left hand side is changed, even when performing arithmetic with the number?

Answer 1

if we say $0.333\cdots = 3\cdot 10^{-1} + 3\cdot 10^{-2} + 3\cdot 10^{-3}+\cdots$

or $0.333\cdots = 3\sum_\limits{n=1}^{\infty} \frac {1}{10}^n$

The decimal expansion is expressed as a convergent geometric series. When we know that the series converges, we can be more comfortable in treating these infinite series as normal algebraic objects.

and so $10\cdot 0.3\cdots = 3\cdot 10\sum_\limits{n=1}^{\infty} \frac {1}{10}^n = 3\sum_\limits{n=0}^{\infty} \frac {1}{10}^n$

Yes you shift all of the decimals over to the left one. But in multiplying by 10, you do not create some sort of $0$ at the end. There is no end to put a zero there. They all shift over one.

You might say that the $3$ in the $10^{-\infty-1}$ place moves into the $10^{-\infty}$ place, but this reasoning feels a little bit sloppy.

The short answer then, is that there is no ifinitessimal that separates

$3.3\cdots$ from $\frac {10}{3}$ or $0.9\cdots$ from $1$

Answer 2

I'm just going to make some comments on the first paragraph of your question. I think these might help a bit.

There were a few statements in the first paragraph that were not correct. I will list some of them below, but please don't take offense to this...

• Writing this out as a decimal, we say that $\frac{1}{3}=0.3$.

We do not say this. We say that $\frac{3}{10} = 0.3$. You might see $\frac{1}{3} = 0.\overline{3}$ written occasionally, and this is to indicate that the $3$'s after the decimal point go on indefinitely.

[EDIT] The OP informed me in the comments that s/he did not intend to write $\frac{1}{3} = 0.3$, but instead wanted $\frac{1}{3} = 0.33\cdots$. Nonetheless, I will still keep the above paragraph in case it is helpful to any one else.

• Even if the $3$s repeat infinitely, the number only gets closer and closer to the intended value, not ever reaching it exactly.

Suppose we consider the number $0.333333\cdots$, where the $3$'s never terminate. Then this number can also be represented as the sum \begin{equation*} \sum_{i=1}^{\infty} \frac{3}{10^i} \end{equation*} and this infinite sum does indeed converge to $\frac{1}{3}$.

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You are correct that, if you starting writing $0.3$ and then $0.33$ and then $0.333$ and did this 1 million times, you would not be 100% accurate in your representation of $\frac{1}{3}$. But we can write $0.\overline{3}$, and this is 100% accurate.