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For example, I know that e, pi, sin (1), Louville's constant etc are just four of the many transcendental numbers, but definitively are any of these the "first", "second" and so on transcendental numbers?

I'm looking for the first 10 transcendental numbers from such a list.

From a puzzle I am trying to solve, I know this much: - the 16th digit of the "2nd" transcendental number is 4 - the 11th digit of the "9th" transcendental number is 1, 2 or 3 - the 23rd digit of the "8th" transcendental number is 0 or 5 - the 12th digit of the "4th" transcendental number is 0 or 9

Thanks! Adrian.

A Tonks
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    There's no way to do this – Akiva Weinberger Mar 12 '17 at 06:01
  • Do you know what the definition of countable is? Do you know that the transcendental are uncountable? – fleablood Mar 12 '17 at 06:16
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    I'm assuming the puzzle doesn't mean the nth transcendental in some universal list, such as the nth prime are the n-th natural number. I assume the puzzle maker simply chose some arbitrary transcendental numbers and list them in an order of her design. It's not a list of all transcentals (which is impossible); it's a list of some transcendentals. For example I could list pi first, then 1/e, then 25 sin 1, then the the number .112123123412345123456...., then the thirty second root ot e, then... – fleablood Mar 12 '17 at 06:25

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There are uncountable many transcendental numbers, so putting them in a list is technically impossible (as a list is defined to be a bijection with $\omega$). However, you can put it in bijection with the ordinal that equals the cardinality of the continuum. There is no canonical way to do this (the natural ordering doesn't work because it's dense and has no endpoints), and any specific example requires the axiom of choice.

I would assume that the puzzle you're doing wants you to fix some ordering and that the problem works for any bijection with the ordinal that equals the cardinality of the continuum.

Stella Biderman
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