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0 pointsforinti5y ago0 comments

Aren't most transcendental numbers unremarkable?

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4 comments · 1 top-level

grahamlee5y ago· 3 in thread

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.

gnulinux5y ago

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.

jessaustin5y ago

...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....

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jlokier5y ago

> 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?

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