- 27 Nov '04 01:11If something is uninteresting, then no one can possibly have interest in it. However, it is conceivably possible that there might be a bet between two people that a computer program that produces a random number will or will not produce any natural number. If there is money riding on this, then whatever the number is, the people involved might have interest in it. Therefore no natural number is entirely uninteresting.
- 28 Nov '04 00:20

I disagree. Consider the grains of sand on the surface of a beach*Originally posted by THUDandBLUNDER***If there are any uninteresting natural numbers, then there must be a smallest one. But this would make it interesting.**

So the next smallest number is the smallest uninteresting number; hence it is also interesting, etc.

Similar reasoning to the Unexpected Hanging puzzle.

Take the leftmost of the uninteresting grains, it's interesting because it is the leftmost. If two grains tie for leftmost uninteresting grain they are both interesting. Hence there are no uninteresting grains of sand on the beach surface.

Except we know that there are uninteresting grains of sand on the beach. Hence we must assume that there are also uninteresting natural numbers. - 28 Nov '04 14:24

First of all we need to define 'uninteresting'. I think a good definition is that no person could ever possibly have interest in an uninteresting thing. However, it''s not for me to decide - it's for Acolyte to define, since he is asking for the proof.*Originally posted by iamatiger***I disagree. Consider the grains of sand on the surface of a beach**

Take the leftmost of the uninteresting grains, it's interesting because it is the leftmost. If two grains tie for leftmost uninteresting grain they are both interesting. Hence there are no uninteresting grains of sand on the beach surface.

Except we know that there are uninteresting g ...[text shortened]... ns of sand on the beach. Hence we must assume that there are also uninteresting natural numbers.

**Except we know that there are uninteresting grains of sand on the beach.**

It depends on the definition of uninteresting. I don't know that there are uninteresting grains of sand on the beach. I know there are grains of sand that I am not interested in at any particular time...but at this moment I am interested in every single grain of sand on a beach, because any one of them might be the uninteresting one - and that would make it interesting. - 28 Nov '04 14:54

Both of those functions are odd, so their integrals over the whole real line are 0.*Originally posted by Acolyte***Prove that there are no uninteresting natural numbers.**

Let f(x) = [e^-(e^|x|)]*x . What is the integral of f between -infinity and infinity?

How about the integral of g(x) = [e^(e^-|x|)]*x ?

If some natural number n is interesting, then n+1, being one more than an interesting number, is interesting as well. To finish the proof, we need to establish that 1 is interesting. The proof of this is trivial and uninteresting and will be left as an exercise. - 28 Nov '04 18:40

should a differentiation be made between interesting numbers and potentially interesting numbers? the numbers that the program generates are not interesting until they are generated, so this doesnt prove that every number is interesting, just that it's possible for each number to*Originally posted by AThousandYoung***If something is uninteresting, then no one can possibly have interest in it. However, it is conceivably possible that there might be a bet between two people that a computer program that produces a random number will or will not produce any natural number. If there is money riding on this, then whatever the number is, the people involved might have interest in it. Therefore no natural number is entirely uninteresting.***become*interesting. i think.

marcussucrammarcussucrammarcus - 28 Nov '04 22:55

g tends to -infinity as x-->-infinity and +infinity as x-->+infinity.*Originally posted by Acolyte***Really? What does g tend to?**

INTEGRAL(-inf to inf)g(x)dx = INTEGRAL(-inf to 0)g(x)dx + INTEGRAL(0 to inf)g(x)dx = INTEGRAL(0 to inf)g(x)dx - INTEGRAL(0 to inf)g(x)dx because g(-x) = -g(x). The last bit is just lim(a-->inf) 'g(a)-'g(a) = 0. Since the only properties used were the oddness of g (which implies that g(0) = 0) and the fact that g doesn' do any funny oscillating, this works for any similar functions (eg x^3).

- 29 Nov '04 14:03

INTEGRAL(-inf to inf)g(x)dx = INTEGRAL(-inf to 0)g(x)dx + INTEGRAL(0 to inf)g(x)dx*Originally posted by royalchicken***g tends to -infinity as x-->-infinity and +infinity as x-->+infinity.**

INTEGRAL(-inf to inf)g(x)dx = INTEGRAL(-inf to 0)g(x)dx + INTEGRAL(0 to inf)g(x)dx = INTEGRAL(0 to inf)g(x)dx - INTEGRAL(0 to inf)g(x)dx because g(-x) = -g(x). The last bit is just lim(a-->inf) 'g(a)-'g(a) = 0. Since the only properties used were the oddness of g (which implies ...[text shortened]... fact that g doesn' do any funny oscillating, this works for any similar functions (eg x^3).

..and here's where it falls down: the RHS is 'infinity - infinity', which doesn't make a lot of sense. You have to make the limiting process clear. Here's another function:

h(x) = 2g(2x) if x>0; h(x) = g(x) otherwise.

h integrated over [-2n,n] is 0 for any n (and this follows directly from the oddness and 'measurability' of g), giving us a sequence of intervals tending to R, and yet I know of no sensible definition of integration that would make h integrate to 0 over the whole real line. What's special about g that means its integral (in some sense) is 0? - 29 Nov '04 15:41

I don't understand the first and second sentences. The definition of the integral of g over (-infinity,infinity) is the limit as a --> infinity of the integral of g over any interval [-a,a], which is 'g(a)-'g(a) = 0. I see why the same does not go for h, because 2n grows faster than n, so I think I see why one can't just take limits as I did, but:*Originally posted by Acolyte***INTEGRAL(-inf to inf)g(x)dx = INTEGRAL(-inf to 0)g(x)dx + INTEGRAL(0 to inf)g(x)dx**

..and here's where it falls down: the RHS is 'infinity - infinity', which doesn't make a lot of sense. You have to make the limiting process clear. Here's another function:

h(x) = 2g(2x) if x>0; h(x) = g(x) otherwise.

h integrated over [-2n,n] is 0 for any n (a ...[text shortened]... over the whole real line. What's special about g that means its integral (in some sense) is 0?

1. What do you mean by 'measurable'?

2. Why can't we say that integrals over the real line of odd functions which tend to some limit as x-->+-infinity are zero, and look at other cases separately? - 29 Nov '04 16:16In a purely philosophical answer, as I have no grasp of mathematics, I would say that there can be no such thing as an uninteresting number. To measure the properties of, or to assign a value to anything, is to give it importance, the mere fact that we could decide upon a number means it carries interest. To merely talk about whether or not such a number exists is to answer the question. By looking for something uninteresting we are giving all possible answers an interest and therefore defusing the search.
- 29 Nov '04 16:39

The definition you give of the integral is not the standard one. Usually, the integral over R is defined as the limit as m and n go to infinity of the integral over [-m,n], provided this limit exists.*Originally posted by royalchicken***I don't understand the first and second sentences. The definition of the integral of g over (-infinity,infinity) is the limit as a --> infinity of the integral of g over any interval [-a,a], which is 'g(a)-'g(a) = 0. I see why the same does not go for h, because 2n grows faster than n, so I think I see why one can't just take limits as I did, but: ...[text shortened]... tions which tend to some limit as x-->+-infinity are zero, and look at other cases separately?**

From an analytical point-of-view it's a bit arbitrary to talk about the integral over R being the limit of the integral over [-n,n], as there's nothing special about 0, and an odd function g(x) is no more special than a translation of it, g(x-c), which won't be odd in general. Nevertheless, the limit you defined is given a name, the**Cauchy principle value**of the integral, because it's useful some contexts, eg in conjunction with Cauchy's theorem, where the line -infinity to infinity is part of the limit of some loop in C you're integrating round, and the way you form the loop allows you to take the limit symmetrically about one of the axes.

'measurable' = Lebesgue measurable (which means that a Lebesgue integral can be defined), but then this won't mean much if you haven't done Lebesgue measure and integrals. Suffice it to say that the proof we were given that non-measurable functions to R even exist required Zorn's Lemma, an alternative formulation of the Axiom of Choice.