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!

Posted by mathfail
on April 3, 2012

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!

lol, I thought that game was easy.

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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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Please prove that “Aleph(1)=2^Aleph(0)”

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

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Nice math-fail by myself there – maybe I should read a little more carefully ðŸ˜³

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