Math game

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 **

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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• I’m pretty sure he meant Aleph(n+1)=2^Aleph(n). So Aleph(null+1)=Aleph(1)=2^Aleph(0)

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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… ðŸ˜‰

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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 ðŸ˜³

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