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I am confused on a rather simplistic question.
1/3 = 0.333333333333 to infinity.
So it has infinite digits. How is it possible to multiply such a number with another one and get a finite number?
6/3 = 6*1/3 = 6*(1/3) = 2

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  • $\begingroup$ You proved it is possible... $\endgroup$
    – Git Gud
    Sep 7, 2014 at 9:10
  • $\begingroup$ @GitGud:Yes but I didn't ask if it is possible.But how is it possible. $\endgroup$
    – Jim
    Sep 7, 2014 at 9:11
  • $\begingroup$ How? By the proof you've given. $\endgroup$
    – Git Gud
    Sep 7, 2014 at 9:12
  • $\begingroup$ @GitGud:I was trying to find an intuitive explanation of this.Perhaps the number with infinite digits has properties I am not aware of $\endgroup$
    – Jim
    Sep 7, 2014 at 9:14
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    $\begingroup$ There is always a way to multiply them. By the way, 2 = 2.0000000000000.... $\endgroup$
    – Tunococ
    Sep 7, 2014 at 9:16

5 Answers 5

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There is more than one way of representing any given number. Obviously if it is possible to represent it as a fraction $\frac{a}{b}$ with $a,b$ not too big that is a convenient choice. However, it is not hard to prove (the famous Cantor diagonalisation proof) that not all real numbers can be represented that way, so we want a system which always works. Hence the decimal system. Sometimes fractions (more usually called rational numbers) can be represented as finite decimals, eg 1/4 = 0.25, but often they cannot, so the decimal representation of 1/3 is not finite.

Incidentally, I am sure you understand that $0.333\dots$ means the limit of the sequence of rational numbers: 0.3, 0.33, 0.333, 0.3333 etc.

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  • $\begingroup$ ...means the limit of the sequence of rational numbers: I am ashamed to say I don't understand this part TBH $\endgroup$
    – Jim
    Sep 7, 2014 at 9:29
  • $\begingroup$ Well, the representation 0.333... would not be much use if we had to use all the digits. But we can get as close as we want to the exact value with a finite number. If you delete all the digits after the $n$th, then you cannot be causing an error of more than $9(10^{-n-1}+10^{-n-2}+10^{-n-3}+...)$. Although that is an infinite series, it has a finite sum which is not hard to find and which tends to zero as $n$ becomes larger and larger. So if we fix the accuracy we want, say within $0.01\%$ of the true value, we can achieve that with only finitely many digits. $\endgroup$
    – almagest
    Sep 7, 2014 at 9:34
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Your number, $0.333..$ Is actualy an infinite sum,

$$ \sum_{n=1}^{\infty} 3\left (\frac{1}{10} \right)^n $$

This is a geometric series it has a finite value $1/3$. If a series has a finite value (i.e. it converges) then when you multiply it by a constant, (in your case $6$) its sum is also multiplied by that constant.

Also $0.999..$ is a geometric series, and when you sum it up you get 1.

When you multiply two numbers with infinite number of digits, you are actually multiplying two series. Series can also be multiplied and the value (sum) of the multiplication is the same as the product of the sums of the series if both series converge (have a finite sum), and one of the series converges absolutely.

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I'll give it a shot. I think you'll agree that $6\cdot0.3333=1.9998$, and $0.3333$ is a pretty good approximation to $1/3$. If you want a better approximation, say $6\cdot0.33333333=1.99999998$ instead, and so forth. That answer is almost $2$, but not quite. The difference is $0.00000002$. But as you add digits to $0.333\ldots$, you also get to add more nines to the answer, and you get closer and closer to $2$. In fact, you only need to add an ever smaller number to get $2$.

So the difficulty comes if you go all the way and consider $1.99999\ldots$. We imagine there should be a digit $8$ at the end, but it's been pushed off to infinity. If were to add a $2$ to this vanished $8$, still infinitely far to the right, you can imagine the digit $8$ turning into $0$, and you need to carry a $1$ to the place before it. That's added to a $9$, and the process repeats ad infinitum until all the $9$s are turned into zeros, and the final carry changes the $1$ in front of the decimal point to $2$. Final result: $2.0000\ldots$. The $2$ we added was infinitely far to the right, so its value is actually nothing.

So the above is an attempt at aiding intuition. It's not proper mathematics, though I suppose you could turn it into mathematics (in the sense that one can define the real numbers using infinite decimal expansions). But it is more common to develop the notion of real numbers without any reference to infinite decimal representations at all. Then what happens with decimal expansions ending in infinitely many nines is just an oddity, but not in any way fatal to the theory.

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  • $\begingroup$ Yes but 6/3=2 not 1.9998. So how come with 2 different ways we get exactly 2 and we approach 2? $\endgroup$
    – Jim
    Sep 7, 2014 at 9:32
  • $\begingroup$ One way to put it: The real line as usually conceived lacks infinitesimals. If some number is closer to $2$ than any positive distance, then that number is $2$. $\endgroup$ Sep 7, 2014 at 9:35
  • $\begingroup$ Another way: I added something like $0.00\ldots2$, where the dots denote infinitely many zeros. This does not make any sense, really, but if you try to make sense of it anyway, it could be a very naïve way of writing an infinitesimal, or in the world of real numbers, a complicated way to write zero. And of course, adding zero to a number does not change it. $\endgroup$ Sep 7, 2014 at 9:40
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$2$ is not a finite number in the sense you described: $2$ is $2.0000000\ldots$. (it is also $1.999999\ldots$)

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  • $\begingroup$ What do you mean by the 1.99999999999...? That 2 = 1.99999999.....? $\endgroup$
    – Jim
    Sep 7, 2014 at 9:25
  • $\begingroup$ Yep. $2$ is one of the numbers that have two different decimal representations. In fact, every decimal that ends in repeated $0$'s is equal to another decimal that ends in repeated $9$'s. $\endgroup$
    – user14972
    Sep 7, 2014 at 9:26
  • $\begingroup$ But then why doesn't 1.99999999... equal to 1.88888888.... and then 2 equals 1.8888888888 for 1.8888.... and 1.777777....> $\endgroup$
    – Jim
    Sep 7, 2014 at 9:27
  • $\begingroup$ For the same reason $2$ doesn't equal $3$: it's simply the way we defined numbers and how to represent them with decimal notation. It's not too different from the fact that both fractions $\frac{1}{2}$ and $\frac{3}{6}$ are two different notations for the same number. There are various ways to motivate why we should have defined things that way -- e.g. converting a repeating decimal into an infinite geometric series, or noting that if you subtract $1.999\ldots$ from $2.000\ldots$ you get a zero in every place: i.e. $0.000\ldots$. $\endgroup$
    – user14972
    Sep 7, 2014 at 9:32
  • $\begingroup$ But 1/2=3/6=0.5 While 6/3=2 vs 6*(1/3) = 1.999999.. =~ 2 $\endgroup$
    – Jim
    Sep 7, 2014 at 9:34
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Let me clear some things up. First off, the infinite expansion of $\frac{1}{3}$ may be represented as one third of the way through the interval of the real line, $[0,1]$. By finite what you mean is a number isomorphic to an element of $\mathbb{N}$. This set is the set of counting numbers $\{1,2,3,4,...\}$. When we talk about the real line however, we often times forget that the elements of the real line are not the "exact" same as their original. For example $1\in \mathbb{N}$ isn't actually the same as $1 \in \mathbb{R}$. But it behaves exactly the same because it is isomorphic to $1\in \mathbb{N}$! So this is where confusion begins. There is no infinitesimal distance in $\mathbb{N}$. However, there is in $\mathbb{R}$. So when we write $3*0.3333333333...=0.999999...$, we notice that $0.9999999...$ behaves, in every sense of the word, exactly like the number $1\in \mathbb{R}$. This is where the confusion lies. Because at this point we may prove that they act in every way the same! And therefore must be the same. This is riddled with underlying math formalities.

I must include that there truly is no intuitive concept of the infinite, small or large.

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