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Okay, I am more than aware that my logic has fallen through somewhere so please show me where.

But surely irrational numbers can't exist.

Okay, so an irrational number is infinitely long and never repeats, yeah?

But surely those things can't go hand in hand. Either it can be infinitely long, or never repeat, one would stop the other.

So each time you add a new number you reduce the chances, and there's an infinite number of numbers so you have a 1/infinity chance of getting the right number and remainder/no remainder. But you also have an infinite number of possibilities because you have an infinite number of numbers. Giving you an infinity/infinity chance. Which I'm pretty sure should come out to a probability of 1, or certain.

I know Infinity is a concept, not a number so you technically can't divide. But think about it an infinite number of tries to get something. So surely irrational numbers are impossible, just we can't generate them far enough to find the end/repeat.

Thank you in advance. Also, try to remember I'm 15. I'm good with concepts of stuff and will probably understand but you may have to say what some words mean. Sorry

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  • $\begingroup$ This is because there are different sizes of infinity countable and uncountable. $\endgroup$ – Ziad Fakhoury May 28 '17 at 13:31
  • $\begingroup$ en.wikipedia.org/wiki/Cantor%27s_diagonal_argument $\endgroup$ – Travis Willse May 28 '17 at 13:31
  • $\begingroup$ Perhaps this question is better suited for philosophy.SE? $\endgroup$ – Trevor Gunn May 28 '17 at 13:31
  • $\begingroup$ en.wikipedia.org/wiki/Dedekind_cut $\endgroup$ – velut luna May 28 '17 at 13:32
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    $\begingroup$ You're mixing existence in a mathematical sense and existence in a tangible, physical sense. Mathematical objects do not exist because we build them using finite pieces one step at a time. Mathematical objects exists because we can deduce their existence from axioms, we can prove that $\sqrt2$ is irrational and that $\pi$ is irrational, and these are nice and definable objects that we can talk about quite easily, without referring to their decimal expansion. $\endgroup$ – Asaf Karagila May 28 '17 at 13:45
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You don't need formal definitions: already the ancient greek mathematicians knew that $\sqrt{2}$ is irrational, and they could prove it, so irrational numbers do exist. The chances you (or anybody) can tell us its decimal expansion (all digits!) are zero, indeed, that's where your reasoning is correct.

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So each time you add a new number you reduce the chances.

You make a mistake here.

Consider the irrational number given by $0.1010010001000010000010000001\dots$. Each time you add a $1$ you know with certainty that a $1$ will appear after so many $0$s.

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