The Mayans had it right.
The world is ending.
I have proof.
Not strictly scientific (as if that matters in Canada anymore, anyway), my proof lies more in pure abstract mathematics.
And luckily, I know very little about abstract mathematics. which makes my proof all the more compelling.
It's all based on my theory that the universe runs on a Base Three paradigm.
A little background may be needed here (very little, I'm afraid, because, as I already stated, I really don't understand this stuff, except to know that it proves my theory. or conjecture, really).
Our standard mathematical model works on Base Ten.
It's probably related to the fact that we have a total of 10 fingers (if you count thumbs, naturally). Our totality of standard arithmetic involves only ten digits: 1, 2, 3, 4, 5, 6, 7, 8, 9, and 0.
I put the 0 last, because in Base Ten there are really only nine digits plus a zero which is kind of not really a digit, but a place-holder that marks off the 10th digit that doesn't exist each time you get past the 9, or at the end of multiple 9s at the end of each order of magnitude.
Confused? Yeah. But just wait - I can get far more confusing than that.
Confusion is an important part of my proof that the world is going to end, just as the Mayans predicted, on Dec. 21, 2012.
Confusion has always been an important part of such proofs.
It's an essential factor in proving that my proof is, in fact, a proof, and not just another deranged fantasy aimed at scaring people into paying attention to me, or buying my book, or purchasing my survival gear, or selling me their house cheap, or revering me as a sage, in case it turns out I really am right, after all.
To illustrate more clearly the concept of arithmetic bases, non-quantum computers - the ones nearly all of us use every day - operate on binary code, or Base Two, with only 1s and 0s.
Ancient Babylonians, on the other hand, ran their calculations through Base 60, with a mind-boggling set of 60 digits, with no zero.
I don't know what arithmetic base the Mayans used. It doesn't matter, because I'm only using someone else's confusing interpretation of their calendar to determine the End of the World, and then switching to our completely unrelated calendar to prove my point. because other-my es wise my proof doesn't work.
And then, what would be the point?
So here goes my proof that the world will end on Dec. 21, 2012.
With the universe based on Base Three, the basic universal arithmetic has only three digits: 1, 2, and the place-holder 0.
December is our 12th month.
In descendent numerical form, Dec. 21 this year will be 2012-12-21, so the day after the end of the world would be 20121222.
Translate 20121222 from Base Three into Base Ten, and you get 4,832.
And remarkably, those Base Ten digits add up to 17, a prime number, evenly indivisible by any number other than one and itself.
The day of the actual calamity is 20121221, or 4,831 in Base Ten - which adds up to 16, which is almost a magic number, by comparison to 17, as it is divisible by 2, by 2x2, and by 2x2x2.
The message is clear - Dec. 21 is the last day with significant interaction between itself and previous dates (smaller numbers), while Dec. 22 is a prime - brand new - start.
Hey folks, it doesn't get any simpler than that.
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