[grahamlee]

No number is unremarkable. Let's construct the set of all unremarkable numbers. Now, let's construct the sequence of those numbers in order. The first member of that sequence has the remarkable property that it is the smallest unremarkable number. That is remarkable, so remove it from the set. By induction, the set must be empty.

[gnulinux]

That doesn't work because real numbers are not enumerable, so you cannot induce over them. That joke "proof" only works for natural numbers and goes like this:

Theorem: all natural numbers are interesting

* Base case: 0 is interesting because it is the smallest natural number, as well as the identity element of + operation.

* Inductive case: Assume the theorem holds for all m, m<n. Take n. If it is not interesting, then n is the smallest non-interesting number. But that's interesting because it's the smallest such number. Therefore it cannot be non-interesting. Therefore theorem holds for n.

By induction, we conclude all natural numbers are interesting. QED.

[jessaustin]

...only works for natural numbers...

That proof also works for the rationals with a suitable ordering. Example: 0, 1, -1, 2, -2, 1/2, -1/2, 3, -3, 1/3, -1/3, 2/3, etc....

[gnulinux]

Yes works for all enumerable set (i.e. all sets that have a bijection with natural numbers).

[jlokier]

> Now, let's construct the sequence of those numbers in order. The first member of that sequence

Not happening. Sets of numbers do not always have a first in order.

Consider the set of unremarkable real numbers > 0 under the regular arithmetic ordering. Which one is first?