mathblab:

question-why-not:

mathblab:

awakeneedsasleep:

the-poetic-mathematician:

awakeneedsasleep:

mathblab:

All Numbers Are Interesting

Proof:

By way of contraction, suppose there exists some collection of non-interesting numbers. According to the well-ordering principle of nonnegative integers, this set has a minimal element, x. However this makes x interesting: a contradiction. Therefore, all numbers are interesting. Q.E.D.

Doesn’t this mean that all non-negative numbers are interesting?

You could do the same with negative numbers, but the greatest element is interesting. And then zero is the only non negative and non positive number, so it’s interesting as well. The proof is incomplete, but easily finished.

My issue is less with the integers, but more with non-ordered fields. I think this implies that there are non-interesting complex numbers, for example.

Yes I was of course not being completely technical. The idea was to not make it overly complicated and use “number” to mean what most people typically think of when they talk about interesting numbers, their favorite number, number theory, etc: these being non-negative interegers or natural numbers.

However, you raise a good point when thinking about the extension of this to the integers, real numbers, complex numbers, etc. As @the-poetic-mathematician mentioned, it is easily extended to integers. But, we run into issues with the reals; there is no well-ordering of the reals and even worse, for complex numbers, there is no usual ordering at all (Complex numbers are not comparable in a less then, greater than way. At best, they can be partially ordered via complex modulus, i.e., absolute value.)

At the same time, just because we cannot use well-ordering doesn’t mean we can’t fix the proof for reals, complex, etc. So there still may be hope for all numbers.

Let F be a non-empty set, and let M be the subset of F containing all utterly uninteresting members of F.

Let {I_n} (n is a natural number going from 1 to infinity) be a countably infinite series of subsets of M, chosen so that these subsets can be ordered by some method into {I_n} so that I_n+1 >= I_n for all n. (Iff M is a finite set, then there exists some natural number N for all n>N, I_n=I_N, and iff M is the empty set, then for all natural numbers n, I_n is also the empty set)

Let us assume in contradiction that I_n isn’t empty, and that there ARE some utterly uninteresting members in F:

Because I_n is ordered, we must get that for any natural n, I_1 <= I_n; therefore the members I_1 are the FIRST utterly uninteresting members of F under the ordering method we have chosen, and that makes them at least SOMEWHAT interesting.

That runs IN CONTRADICTION to our definition of I_n, according to which there can’t be ANY interesting members in I_1, and therefore the assumption of I_n being non-empty must be false. ->

-> for all n, I_n must be the empty set ->

-> M is also the empty set ->

-> There are no utterly uninteresting members in F.

Q.E.D.

Hmmm. I think we may need to consider the case when l_1 = l_2 = … = l_n = … right?

Also we should be careful about claiming everything is interesting; not everything can be put into a set, e.g., the collection of all groups is not a set 😦 So your proof would not apply to all groups, just F := any set of groups.