the tree

Being finite and bound (that is, if you go far enough you get back where you started) is pretty much the accepted view. Of course there's the added complication that it's expanding as well. And while there are only two dimensions on the surface of the Earth, making it easy to conclude that it's finite and bound on both of them, with the universe as a whole we're talking 11 or something.

For a rough understanding, a real number is something that you can point to on a number line. Or, for the sake of staying on topic, on an axis.

Yes many can. Also some can't.

A matrix where the entries in each row are real, positive and add up to 1. In other words, each row represents a discrete probability distribution.

Okay, I'm just going to be tearing my hair out now.

The only relevant result Google gives is this thread. Did you think to try using Google before announcing to the world that it'd work?

Yeah both of those answers are correct.

Boxxy probably disapproves of your false idols. I'm calling IBTL now

*you're.

Seriously? You want us to match that? 392699/125000=3.141592 there you go. Not exactly challenging. Where you really expecting anyone to not be able to produce a rational approximation of pi? Actually a lot of the people on this forum are kids, since we're discussing pretty basic math here, that's not too surprising. Now, do you think these 'many people' are laughing at the people who know a little maths? Or do you think they are the laughing at the person who brought up numerology in a science forum? Which would you say is more likely? No. He does get the credit for that equation though. I think your comprehension might improve if you try to read a little slower, and more carefully. Don't get ahead of yourself. I'm just quoting this because it's funny to read. Anyone still counting the index?

I think it's worth assuming that in fact, they used maths. As Klaynos said, the numbers weren't just picked out of the air. Also, Newton holds the credit for: [math]\pi = 2 \sum_{k=0}^\infty\frac{k!}{(2k+1)!!}[/math] which a whole lot more impressive than a single ratio. In, y'know, the real world: -10.86902388503848+14.01061653862827 = 3.141592653589790000000000000000000000.. It's fairly obvious, when you think about it, that you're not going to have more decimal places in the sum than the in the terms being added.

I think it's worth noting that all three suggestions so far do lead to the same correct result. The lesson to be learned perhaps, is that if L'Hopital needs to be applied five times then maybe it's worth looking for a quicker approach.

What's there to think about it? It's a well known equality.

"that's not a car, it's a sedan", it's still a car.

I think part of the problem is a lack of an intuitive meaning for xy when y isn't an integer. Multiplying a number by itself i*pi times just doesn't make any sense. I think the trick is kind of to view it as just another function.

Try making these substitutions: [math]\left( \frac{a}{a+b} \right) =x^2 , \, \left( \frac{b}{b+c} \right)=y^2 , \, \left( \frac{c}{c+a}\right)=z^2[/math] It should be fairly easy from there.

Not really, you need very specific programs to plot fractals. If you think of your oridinary plot of say [imath]y=f(x)[/imath], then you're highlighting a subset of [imath]\mathbb{R}^2[/imath], specifically [imath]\{ \left( x,y \right) \in \mathbb{R}^2 \, | \, f(x)=y \}[/imath], which is fairly easy to generate - it'll roughly be something like: for x in range y=f(x) Image[x,y](color)=black loop Even for a calculator trying to plot [imath]f(x,y)=0[/imath] the process will amount to (but be more efficient than): for x in xrange for y in yrange if f(x,y)=0 Image[x,y](color)=black endif loop loop But checking to see if a number in the Mandlebrot set takes a lot more than just some arithmetic function, in short, it's just not what graphing calculators are for. The Wikipedia article on the Mandlebrot set has a pseudocode implementation for plotting a Mandlebrot, but it's basically very roughly checking each number one by one.

Two hyperbola.

what.

I never made such an assertion. The question was about numbers for which a proof exists.

The idea of a statement (such as 'x is(not) rational') being true but unprovable is getting a little Gödelesque for the sake of this thread, I would contend that truth and provability are one and the same though many would disagree and this certainly very far off of topic.

I know, my point was that a proof must exist, known or otherwise, for every computable number.

If a proof does not exist (in the ethereally mathematical sense of existence), then technically speaking it would be neither rational nor irrational, I suppose some non-computable numbers fall between the gaps in that sense.

No, no, no - there are proofs and detailed explanations for pi being irrational, you don't just have to accept it.

Should? There's no particular criterion to pick between two monoids unless they are for something specific.