In an infinitely repeating pattern of three integers, said integers being 1, 2, and 2, could one say that there are more twos than ones?
Another manner of phrasing this question: "is a sum count of even numbers equal in size to a sum count of all whole numbers because all infinities are equally infinite?"
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So in a countable subset there are more twos but, in an infinite set, measurements cannot be made. Did I read that right?
Correct-ish. There are as many 1's as 2's if the set is infinite.
Thanks, I intuited that answer but I really don't trust myself when it comes to math on that scale.
That's in terms of cardinality, but in terms of measure, it depends on which measure you use (and although I've never had a formal course about it I'm sure it can give infinitely many different answers).
Pull the ones out of the pattern and stack them in their own.
You are left with:
>An infinite string of 2s
>An infinite string of 1s
If the original pattern was infinite, then you have infinite of each, and you can't tell if there is more of one or the other.
>There are as many 1's as 2's if the set is infinite.
>If the original pattern was infinite, then you have infinite of each, and you can't tell if there is more of one or the other.
Brainlets pls leave.
what exactly does the fact that real numbers have a greater cardinality than integers, have to do with this case, brainlet?
shut the fuck up, reading wikipedia does not make you understand shit, especially when you're completely fucking wrong
retard, the anons obviously know cantor's classic argument if they're able to build bijections between countable sets