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The thing about this is, if we assign a variable: $$x=10 \cdot 10 \cdot 10 \cdot 10 \cdots$$ and then enclose all but one multiplicand in brackets (as multiplication is associative) and we get: $$x=10 \cdot (10 \cdot 10 \cdot 10 \cdot 10\cdots)$$ We can’t now see that the expression within the bracket is equal to $x$ so we get: $$x=10 \cdot x$$ The value of $x$ could clearly be shown as $0$ because $0=10\cdot 0$. This can’t be right because if we do the same thing with $1$ we can show that every number equals every other: $$x=1\cdot x$$$$4=1 \cdot 1 \cdot 1 \cdot 1 \cdots=3$$

My question is where is the flaw in this logic. Where does this method go wrong. Any insight would be helpful.

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    Well, the expression $x=10x$ assumes there is a value to the expression. Consider this: if $x$ is nonsense, the &10x$ is nonsense. – Thomas Andrews Dec 06 '18 at 06:31
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    In some numeric systems, specifically the $2$-adic numbers, $5$-adic and $10$-adic numbers, the limit does exist and is zero. But these are very specific cases. – Thomas Andrews Dec 06 '18 at 06:34
  • Just because you can write something down doesn't mean it exists. Let $x = $ the man who lives in every house. That means he lives in my house. But I live alone. The that man is me. So I live in your house? – fleablood Dec 06 '18 at 06:40
  • Associativity doesn't hold for infinite products...along with other properties of a group such as closure. – M A Pelto Dec 06 '18 at 07:02

2 Answers2

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I guess the expression is $$ x=\lim_{k\to\infty} 10^k.$$ However, it doesn't make sense to "choose" a value for $x$. This limit diverges to $\infty$. So, your mistake is saying that $x$ is a real number satisfying $x=10x$. It turns out that $x"="\infty$. In particular, the expression above is not a real number, because if $x\in \mathbb{R}$ then for any $r\in \mathbb{R}$, $x>r$. This is impossible.

So, this arithmetic is not meaningful, because arithmetic is not defined (in this context) for $\infty$.

Alekos Robotis
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  • But you see, that $x=\lim\limits_{k\to\infty} 10^k$ is not that same, we are not finding a limit, but an actual value. –  Dec 06 '18 at 08:07
  • There is no actual value. – Alekos Robotis Dec 06 '18 at 08:09
  • Precisely, as shown in other comments and answers. But a limit I see not the same as an infinite product or series. –  Dec 06 '18 at 08:11
  • Also, thank you for the (in this context). As set theory is what I generally work with, this means more than it does to most others... –  Dec 06 '18 at 08:13
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    What is an infinite product if not a limit? – Alekos Robotis Dec 06 '18 at 09:12
  • Literally, an infinite product. –  Dec 07 '18 at 07:19
  • But how do you define such an operation? – Alekos Robotis Dec 07 '18 at 07:20
  • Exactly, there is no limit for the equation (other than $\infty$), and such we shouldn’t define x as a limit. The function, as shown in other answers isn’t a function and so cannot have a value assigned to it. –  Dec 07 '18 at 07:23
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    My point is: how could you define an infinite product of numbers without a limit? – Alekos Robotis Dec 07 '18 at 07:23
  • Exactly, that didn’t cross my mind when writing this question, I was just wondering whether or not something like this hadn’t any meaning whatsoever (in normal arithmetic). –  Dec 07 '18 at 07:25
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First, we have to remind ourselves what it would mean for an infinite product to be equal to anything or to converge to something - much like the infinite summation, it would be the limit of the partial products, right? Well, the partial products for $\prod_{k=1}^\infty 10^k$ are $10, 10^2, 10^3, 10^4, 10^5$... obviously divergent. So, before I even address what you said: no, the product absolutely does not converge to anything.

Now, generally, this is a flaw not unlike with what Ramanujan ran into when showing that the summation $1+2+3+4+... = -1/12$. There's probably a proper formalization of this that someone else can elaborate on in the case of products, but I imagine the idea is the same.

Ramanujan's flaw was that the summation of the natural numbers is divergent. Thus, just "assigning" a variable to the value of the sum, i.e. saying $x = \sum_{k=1}^\infty k$, and then performing manipulations based on that to try to derive a value just is not kosher. The reason is because that summation is divergent - you can check the limit of the partial sums, and they visibly approach $\infty$.

Thus, I imagine an analogous idea holds here: you cannot say $x = \text{some infinite product}$ and perform manipulations as you did if that same product is divergent.

It's like a commenter said - to assume the product has a value and you can assign it to some constant $x$ is nonsense given it clearly has no value, and something reasonable cannot follow from nonsense.

Edit: As noted by a commenter, this is all under the assumptions of us working in the usual topology we normally work with. We could define an alternate topology in which these manipulations for this product make sense. So in a way, you're right - just not in our usual number system. :P

PrincessEev
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    It’s only “obviously divergent” under the usual topology... – Thomas Andrews Dec 06 '18 at 06:34
  • I'm not 100% sure what that means but I imagine that's also an obvious assumption. – PrincessEev Dec 06 '18 at 06:35
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    Well, it means that, when there is a value, the OP is correct that the limit is zero, but in the usual meaning, the value is either undefined or not in the number system. – Thomas Andrews Dec 06 '18 at 06:39
  • Fair enough. I still feel like that being in the usual topology is just an obvious assumption, though, since it wasn't stated otherwise, but you have a point. Actually makes me somewhat curious as to the implications of such alternate topologies ... I should really look into it. – PrincessEev Dec 06 '18 at 06:44