12 Jan 10
Originally posted by FabianFnasSorry I should have been more precise, I meant I believe there are infinite number of INTEGER solutions such that a^2 + b^2 + c^2 = d^2
Not according to my math book, but never mind. My problem was about a hollow sphere. Or else, I think the problem would be much more easy.
(3/13, 4/13, 12/13) gives a correct result. Thanks.
Seems that you're correct - there are probably infinite number of triplets where a, b and c belongs to Q, and each not equal to zero, and a^2+b^2+c^2=1. 1 is enough because we're talking about a unit sphere.
But is there a proof for this?
(like 3^2 + 4^2 + 12 ^2 = 13^2)
so that points on sphere are of the form (a/d . b/d , c/ d)
(like (3/13, 4/13, 12/13) )
Sorry for imprecision!! 😀
13 Jan 10
Originally posted by Palynkadefinitely true! i just thought that an algebraic interpretation might be more accessible if one didn't know about stereographic projection from the plane -> sphere.
Yes, the easiest one comes from stereographic projection, like I said.
though, i'm not sure that the question is necessarily "solved." even without thinking about the implications of "physics sight" with questions of the size of a photon and the size of a point, it seems that the truly interesting mathematical "line of sight" question has more to do with the density of the rationals on the real line.
yes, there are infinitely many rationals in the reals, and similarly infinitely many rational points on the surface of the sphere in question. but it is a denumerable infinity... much smaller than the infinity of the continuum, which exists as a result of the addition of the irrationals to the rationals in defining the real line. i would say, in fact, that there are many more points on the surface of the sphere that can NOT be expressed by solely rational numbers. but does a single point allow "sight?" or do we need an interval, (or in this case more accurately a region) that would constitute a "hole" in the sphere? i think this is more in the spirit of the original problem, and we need to either prove or disprove the existence of a whole sufficiently "large" to allow "sight." ... perhaps an epsilon/delta proof of some part anyone? i'm too out of it right now to really contemplate but this seems right up the alley of some of the other math guys here! 🙂
13 Jan 10
1 edit
Originally posted by AetheraelOh, I agree entirely. But that question was for the other thread. 🙂
definitely true! i just thought that an algebraic interpretation might be more accessible if one didn't know about stereographic projection from the plane -> sphere.
though, i'm not sure that the question is necessarily "solved." even without thinking about the implications of "physics sight" with questions of the size of a photon and the size of a poi plate but this seems right up the alley of some of the other math guys here! 🙂
To answer it, you need to think about whether the rationals are locally compact in the topology of R (important point!), which they are not. To prove this you'd have to realize that, in the topology of R, any neighbourhood of a rational number contains an infinite number of reals so it's not a subspace of Q. You can prove this by contradiction. Assume that it is compact and show then that any compact subspace contains irrationals, which implies that it cannot be a subspace of Q. Contradiction.
Regarding density, it may seem confusing that the rationals are a dense subset of R. This is because every element of R is always arbitrarily close to an element of Q.
So local compactness (in the topology of X) and density (in X) are quite different things.
13 Jan 10
Originally posted by AetheraelI can disprove the existance of any hole of size epsilon (from here on E) 🙂
perhaps an epsilon/delta proof of some part anyone? i'm too out of it right now to really contemplate but this seems right up the alley of some of the other math guys here! 🙂
Given an interval (x, x+E) you can always find a rational inside that interval.
Take a rational interval around the given interval. Take the middle of that interval. If that middle is in (x, x+E) you're done.
If not, take the half that contains (x,x+E)
If you continue in this manner you will eventually end up with a point in (x,x+E)
The 3-dimensional variant goes quite similar 🙂
13 Jan 10
Originally posted by TheMaster37Yep, that's the proof that the rationals are a dense subset of the reals. Any element of R is arbitrarily close to an element of Q.
I can disprove the existance of any hole of size epsilon (from here on E) 🙂
Given an interval (x, x+E) you can always find a rational inside that interval.
Take a rational interval around the given interval. Take the middle of that interval. If that middle is in (x, x+E) you're done.
If not, take the half that contains (x,x+E)
If you continue ...[text shortened]... ll eventually end up with a point in (x,x+E)
The 3-dimensional variant goes quite similar 🙂