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I have developed a proof that contradicts the infinite-ness of $\infty$.

Here is my proof:

Let $a$ be equal to $0.00000000000...001$, where there are $\infty$ number of zeroes. We can also say $a$ is the first real number.
Let $b$ be equal to $0.00000000000...001$, where there are $\infty - 1$ number of zeroes.
$b$ is also equal to $10a$.
However, since $\infty = \infty - 1$, $b = a$, which means $10a = a$, which means that $a$ is a variable that never changes, no matter how much it is multiplied by. (I assume it can be divided or subtracted, though.)
Since $a$ is the first real number, all numbers after it must be some multiple of $a$. Since multiplication does nothing to $a$, we can say that all numbers that are $a$ and above, are equal to $a$.
The problem is that we've already proven that these numbers have values differing from $a$.
This means that $a$ can not have infinite zeroes, and thus, $\infty != infinite$.

However, I believe my proof is incorrect. The problem is, I can not locate the point where the proof is wrong.

Can anyone help me find it?

InitializeSahib
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    There is no first real number. For one thing, even at the informal primary school level, no one should ever hint towards an "infinitieth" digit being a valid notation. –  Dec 08 '16 at 22:47
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    By saying $0.00000\dots001$ where there are infinite $0$'s, but it "ends" with $1$, you are assuming $\infty$ is in fact finite. That's circular – user160738 Dec 08 '16 at 22:48
  • Yeah, if there's an infinite amount of zeros, there can't be a next number. – Simply Beautiful Art Dec 08 '16 at 22:48
  • "Let a be equal to 0.00000000000...001, where there are ∞ number of zeroes. " That doesn't make sense and is not a properly defined number. What do you think it could possible equal? I suppose if it equalled anything it'd be $\lim 10^{-n} = 0$. ANd then 10a = a = 0 and ... no problem. – fleablood Dec 08 '16 at 22:52
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    We can also say a is the first real number. Is $a/2$ not a number? – dxiv Dec 08 '16 at 22:56
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    @dxiv It's not a number. It's a free man! – Daniel Fischer Dec 08 '16 at 22:59

3 Answers3

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Your proof is wrong beginning from the first line -- there is no "first real number", you can always choose a real number between zero and another one.

Hasek
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  • But wouldn't a number with infinite zeroes be the first real number? Since it has infinite zeroes, it must be the closest number to 0? Or is there a problem with assuming the number has infinite zeroes in the first place? – InitializeSahib Dec 08 '16 at 22:49
  • $0+\lim_{x\rightarrow \infty} 10^{-x} = 0$ so if you have infinite 0's and a 1 it is the same thing as being 0. – Kitter Catter Dec 08 '16 at 22:52
  • If you define a number with infinite number of zeros after the dot it is in fact equals to zero. The non-zero number may contain only finite number of zeros before some non-zero element like $0.0\ldots01$, but in that case you can always choose another smaller real number – Hasek Dec 08 '16 at 22:54
  • Of course there is a problem with assuming there is a number with infinite zeroes. What could it's value possibly be? Also every decimal 0.abcd.... is simply short hand for $\sum \frac {a_i}{10^n}$. Every digit has a fixed "place". What place is the 1 fixed at if it comes after an infinite number of 0s. And what would be $a/2$. For any $a > 0$ we can find $a > a/2 > 0$. So there is no first. 0.00000......1 simply doesn't make any sense. – fleablood Dec 08 '16 at 23:04
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Well, suppose everything you said up to $10a=a$ made sense some how. You could then notice that

$$10a=a\implies 9a=0\implies a=0$$

Thus, $a=0$, which is the limit

$$a=\lim_{n\to\infty}0.\underbrace{00\dots00}_n1=0$$

Simply Beautiful Art
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You haven't proven that $\infty \ne \infty$. You've proven that any $0.000........1$ can't have an infinity number of zeros and have any non-zero finite value. Which... is fine. It can't.

$0.000.......1$ simply is not a number and has no coherent meaning.

fleablood
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