Cantor’s diagonal argument (please!)
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Imagine a long list of numbers that are between 0 and 1. The list is so long that it goes on forever! But we can still number those numbers; the first is #1. The second is #2. The third is #3. And so on.
We won’t run out of whole numbers for numbering our list, because whole numbers go on forever.
But does this list have all the numbers between 0 and 1 on it? NO! It CAN’T! And I can prove it. All I have to do is find a magic number that’s not on the list.
The magic number has a different digit 1 place after the decimal than the #1 number. So, for example, if #1 was 0.673903240, the magic number starts with 0.3 (because 3 is not 6, so this magic number I am making is not the same as #1 on the list).
The magic number has a different digit 2 places after the decimal than the #2 number. So, for example, if #2 was 0.743603557, the magic number now starts with 0.38 (because 8 is not 4).
The magic number has a different digit 3 places after the decimal than the #3 number. So, for example, if #3 was 0.9745378, the magic number now starts with 0.381 (because 1 is not 4).
And so on, forever… I always pick a digit that is different from the Nth digit than the Nth digit of #N on the list.
The list of numbers goes on forever, but I can still find a magic number that is not on the list. (Actually, I can make as many magic numbers as I like, and none of them will be on the list! Can you see how?)
The numbered list may go on forever, and there may be an endless number of numbers between 0 and 1, but there are more numbers between 0 and 1 than there are on the list! SOME INFINITIES ARE BIGGER THAN OTHERS! ACK!
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