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I have looked into Cantor's diagonal argument, but I am not entirely convinced. Instead of starting with 1 for the natural numbers and working our way up, we could instead try and pair random, infinitely long natural numbers with irrational real numbers, like follows:

  • 97249871263434289... 0.12834798234890899...
  • 29347192834769812... 0.23489712349789878...
  • 42987412938478321... 0.23487912836784798...
  • 43921649873612384... 0.58792834796781823...
  • 49238749213847921... 0.58971238456497213...
  • 98123489712348790... 0.58291739429587199...
  • 45678294218374691... 0.09123498915832837...
  • 69217346876217384... 0.23897123484839759...
  • 52189347981283490... 0.34823948750038273...
  • .
  • .
  • .

Couldn't we just as easily apply the diagonal argument to the natural numbers, and therefore generate a new natural number after completely exhausting our list? Here's another way to look at it:

  • 97249871263434289... 0.0
  • 29347192834769812... 0.1
  • 42987412938478321... 0.2
  • 43921649873612384... 0.3
  • .
  • .
  • .
  • 49238749213847921... 0.8
  • 98123489712348790... 0.9
  • 45678294218374691... 0.10
  • 69217346876217384... 0.11
  • 52189347981283490... 0.12
  • .
  • .
  • .

If we accept Cantor's diagonal argument, doesn't the second argument now prove that there are more natural numbers than real numbers between 0 and 1? If you imagine this infinite list of real numbers, which in fact does include all numbers between 0 and 1 (or else the set [1, 2, 3, 4, 5, 6...] wouldn't contain all the natural numbers), it seems like the only distinction between the two sets are the inclusion of a preceding "0." for the real numbers. Does that magically make the set larger?

Shaan Paymaster
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1 Answers1

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There are no "infinitely long natural numbers". Natural numbers have a finite decimal expansion.

Indeed, if you are looking at "infinite decimal expansions", i.e. possibly infinite strings of the digits 0 to 9, this set is uncountably infinite.

Mees de Vries
  • 27,550
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    ...and if you put a decimal point in front of each of them, you get all the (real) numbers between $0$ and $1$, which is the starting point for the Cantor diagonal argument. – John Hughes Oct 04 '18 at 13:52