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Let $x = 1 + 1 + 1 + \cdots \infty \textrm{ times}$.

$\therefore \quad x = 1 + x$

$\therefore \quad x - x = 0 = 1$

But this is absurd. I am unable to understand where I have gone wrong. I remember using this kind of analysis for solving problems like this.

Apoorv Potnis
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    Hi Apoorv. $x$ doesnt really equals to $x+1$. The 1st statement is not correct.. – Redsbefall Jan 23 '18 at 05:10
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    $\infty - \infty$ is undefined. – user1101010 Jan 23 '18 at 05:11
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    $x$ is not a real number. You can do normal arithmetic when you are dealing with real numbers – clark Jan 23 '18 at 05:11
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    "I remember using this kind of analysis for solving problems like this" Using such a technique to solve for the value of $x$, one must first make the assumption that $x$ is actually a real number in the first place. If it happens to be true that $x$ is in fact actually a real number, then the logic and result that follows could be valid, however if it turns out that $x$ is not a real number then the result and all of the logic that follows is moot and should be discarded. Similar logic would tell you that $\sum\limits_{n=0}^\infty (-1)^n=1-1+1-1+1-1\dots "=" \frac{1}{2}$ which is false. – JMoravitz Jan 23 '18 at 05:17

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Thank you for pointing out a clear absurdity! There are some ludicrous people on the internet who come across this pseudo-proof as you have and then think that all of mathematics must be false. (You are fortunate to recognize that you have made some error, and that's good.) So the problem with this line of reasoning is that $1 + 1 + \cdots$ is not a number. In fact, the series $\sum\limits_{i=0}^\infty 1$ does not converge to a number. (If you have not yet taken Calculus II or dealt with convergence of series, then...don't worry so much about the convergence details and just recognize that $1 + 1 + \cdots$ does not represent an actual number in the sense of an integer, rational, real, or complex number.)

Hence you cannot add 1 to $x$, because there is no definition of $+$ in this context. You also cannot subtract $x$ from itself, because $x$ is, again, not a number!

Hopefully that answers your question! Feel free to ask for more details, though I'm not sure what more I can give.

Tanner Strunk
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  • To address your question relative to the link you give, notice that the continued fraction (the name of the thing in your link) isn't getting bigger in some unbounded way. Presumably that continued fraction actually represents a number to which it converges or can at least be approximated by some real (ideally rational for all our purposes) number. That is the difference between that problem and the one you have presented in your question. – Tanner Strunk Jan 23 '18 at 05:15
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    Yes! If you have a series which converges to a number, then really that series is just another representation of that number. You can think of the series as just another symbol--just like $y$ or $x$--representing the number to which it converges. For instance, the series $\sum\limits_{i=0}^\infty (1/2)^n$ converges to $1/(1 - 1/2) = 2$, meaning that the series can be used as simply another representation of the number we know as 2. Hence we have $2 = 1 + 1, 2 = 4/2$, and $2 = \sum\limits_{i=0}^\infty (1/2)^n$. – Tanner Strunk Jan 23 '18 at 05:22
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    I missed my chance to reply to a comment of yours on another answer as it was deleted. @TannerStrunk "Also, even the first statement is not correct, because there is no definition of $+$ between $\infty$ and $1$." That is untrue. In many contexts we do define such an operation such as in the context of the Riemann Sphere and the extended complex numbers where $\infty+a$ is defined to be $\infty$ for any finite value of $a$. In those contexts though, $\infty-\infty$ generally remains undefined. – JMoravitz Jan 23 '18 at 05:24
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    Thanks for the prompt reply. I had not expected such a quick response. – Apoorv Potnis Jan 23 '18 at 05:25
  • @JMoravitz, I must admit that I'm being nitpicky at this point, and you obviously know what you're talking about. However, I think that $1 + 1 + 1 + \cdots \neq \infty$ in the context of projective geometry (e.g. for the Complex projective line, the Riemann Sphere). In this case, infinity is representing a limit (maybe not specifically the limit represented by the summation above) or just the point added when compactifying the complex plane, right? At any rate, you're still right that arithmetic is defined in that way for projective geometry using the point at infinity. – Tanner Strunk Jan 23 '18 at 05:30