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Let $x = 2 \times 2 \times 2 \times 2 \times 2 \times \cdots$

Thus, $x = 2x \Rightarrow x = 0$

So, $2 \times 2 \times 2 \times 2 \times 2 \times \cdots = 0$


This clearly doesn't make any sense. What's wrong then?

Thank you in advance.

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    $\begingroup$ Let $x= 2 \times 2 \times \ldots$ makes no sense. $\endgroup$ – Mauro ALLEGRANZA Aug 31 '17 at 14:27
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    $\begingroup$ Can't we define $x$ as an infinite product? Can we do so as an infinite sum? $\endgroup$ – Gooble Aug 31 '17 at 14:29
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    $\begingroup$ Only if it converges. Defining $x = 1 + 1 + \cdots$, thus $1 + x = 1 + 1 + \cdots = x$, thus $1 = 0$ is equally meaningless. $\endgroup$ – Mees de Vries Aug 31 '17 at 14:30
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    $\begingroup$ If you're going to allow $x$ to represent an infinite quantity, then the equation $x=2x$ is also satisfied by $x=\infty$, so no contradiction here. $\endgroup$ – grand_chat Aug 31 '17 at 14:39
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    $\begingroup$ "Let $x=2\times 2\times\dots$ makes no sense" I would argue that it still does make sense as $x=\lim\limits_{n\to\infty}\prod\limits_{k=1}^n 2$. The difficulty here is that what you are trying to do is essentially $\lim\limits_{n\to\infty}\prod\limits_{k=1}^n 2-\lim\limits_{m\to\infty}\prod\limits_{j=1}^m 2$ which is indeterminate of the form $\infty-\infty$. Just like how $\frac{\infty}{\infty},\frac{0}{0},1^\infty$ have no consistent value to assign to them, so too does $\infty-\infty$ depend on context. $\endgroup$ – JMoravitz Aug 31 '17 at 14:42
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It is possible to have products of an infinite number of factors in a hyperreal number system. However, the infinite number, say $H$, is a specific one not equal to $H+1$. Therefore your $x=\prod_{i=1}^H 2$ is not equal to $2x=\prod_{i=1}^{H+1}2$ and the paradox disappears.

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