Q:

How does Cantor's diagonal argument actually prove that the set of real numbers is larger than that of natural numbers?

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?

A:

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.

