Not clear on what we mean with numbers with infinite digits

Question

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

Answer 1

$2$ is not a finite number in the sense you described: $2$ is $2.0000000\ldots$. (it is also $1.999999\ldots$)

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.