Math game

Untitled

Thanks to Secret Name for this submission!

Secret Name says:

I made this comic and wanted to share it with you. Tell me what you think about it in the comments!

** Note: Some posts on Math-Fail are user-submitted and NOT verified by the admin of the site before publication. If you find this post to be distasteful, non-math related, ?or something worse?, then definitely leave a comment letting me know. Thanks very much! Mike **

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11 Comments.

  1. lol, I thought that game was easy.

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  2. Wouldn’t saying Aleph-(null+one) solve the problem?

    Similarly you could say Aleph-(n+1) for any Aleph-n, right?

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    • Yes, you could. The trick is making quick questions so when you say X he quickly says X+1 and he doesn’t think about it.

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    • Mark: That doesn’t work. What if I say Aleph(Aleph(null)). You say Aleph(Aleph(null)+1) and it’s the same number.

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  3. See, that is why the logician will say

    2^x

    where x is the absolute value of whatever you just said.

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  4. actually, you don’t need the absolute value part, just 2^(1000!), 2^(mole^mole), 2^(aleph_null) will be enough

    it follows from Cantor-Bernstein, yes?

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  5. Well, as every cardinal number is also an ordinal number, and as I can use ordinal arithmetic with ordinal numbers, the “trick” of the comic falls apart: “Aleph_0 + 1” is by definition the least ordinal number that is greater than Aleph_0.

    Aleph_0 + 1 is not a cardinality, but only a greater NUMBER was required.

    Your trick would work if I said “1 + Aleph_0”, as that IS the same ordinal number as Aleph_0. But who would say that…?!

    But Yung Hei is absolutely right, too, as his answer ALWAYS produces even a greater CARDINAL number. But for producing greater ORDINAL numbers, the procedure of the comic is completely satisfying.

    So I guess, the moral of it all is this:

    Don’t try to fuck with logicians… 😉

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  6. What if x=0.5, in that case 2>2^0.5 ?

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  7. Nice math-fail by myself there – maybe I should read a little more carefully 😳

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