What is wrong with pi?

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Yeah, the circle will always curve, no matter how far in you zoom.

It's perfectly possible for a curved line to have a rational length. The proof that pi is irrational isn't that easy.

yeah, I mean if we assume pi is irrational. If we don't, then the circle would get pixelated/not-round if you zoomed in enough. But because every point on the circle is exactly the same distance from the center, then it would be extremely weird for the circle to be pixelated. i can't imagine a pixelated/not-round circle with every point still the same exact distance from the center. So probably you have to assume pi is irrational, as long as it's derived from a theoretical perfect circle, with theoretical perfect, exact measurements (it's extremely unlikely we will ever have either of those things).

That quasi round thing on LED or Video display is not a circle.

A circle is a set of points in a Euclidean plane. It is purely abstract and does not exist in the physical (real) world.

ruveyn

that's what I mean when I say "theoretical perfect circle". If it's not perfect, it's not a circle. And like you said, it's just a set of points on a plane with no thickness--basically no physical presence--whatsoever. I think.
But it's a little easier to pretend circles do exist, even if they don't, because what are you going to call that piece of paper you cut out to make a paper snowman in kindergarten? a quasi-circle? too much of a mouthful for my tastes. And chances are, if you started talking about paper quasi-circle snowmen to other people, they would probably start looking at you funny and back away slowly, thinking that, really, they shouldn't have even started talking to you in the first place. Or something like that.

If the universe could produce a printer that could printer a circle on paper at increasingly high resolutions, and yet there was still a way to view microscopically the increasingly higher resolutions, the circle would continue to be a polygon (look at a circle on YOUR printer with a microscope!).

There are many rigorous ways to define pi, which can all be proven equivalent, and none of them rely on the idea that circles "really exist". You can prove that pi is irrational using any of these definitions. So no matter what your opinion is about whether circles "really exist" or not, it is still a fact that pi is irrational.

For example, we can define pi to be the infinite sum



See? No circles involved.

I suppose your right about that. But if there were no circles, we humans would not have discovered pi till alot later, or not at all. And I like explaining things visually.



yes, but it's only possible for a circle's circumference to be rational if the diameter is irrational. Correct me if I'm wrong, but I think there's no such thing as a circle with both a rational circumference and a rational diameter.

r = 0

OMG Species5618! you're a genius!! !!

The only problem is, I'll never be able to see a circle with r=0, as long as I live. Not enough to tell it's a circle. It would be so cool to see... but then again, it would just be a point. Not very exciting.

why is π pronounced pie (pai) and not pi (pee)….

in greek it is pronounced pi (pee).


it is a bit silly not to pronounce a letter correct (some countries do pronounce it correct though!). that's the only part in π that is flawed…

Take the surface of a sphere as your 2D space (i.e. not Euclidian).
Define a circle as the locus of points equidistant (that is, along great circles) from a given point.
There are now "circles" which have rational ratios of their circumference to their radius.

E.g. on a sphere radius 2/π, the "circles" of unit radius all have circumference two, and hence the ratio of their circumference to their radius is four. (And the ratio varies with radius - with circles of radius two having circumference zero.)

A 2D space with positive curvature throughout always has a ratio of circumference to diameter of less than π. With negative curvature, always greater.

(I think I'm being totally rigorous, but I might need to modify that to "continuous 2D space". It can be extended to 3D spaces, and above, with suitable modifications - such as that the ratio of the surface area of a sphere to its radius squared being one side or other of 4π.)

NB. Seeing as we probably don't exist in a Euclidean space, anyway, then our "real" circles do not define π at all, to start with.

Where did you hear that, aspi-rant (nice profile pic, by the way)? My mom teaches latin/ancient greek, and she told me pi is phonetically exactly like pie... but that could just be so it's different from "p".

then your mom is wrong…

ask her to pronounce phi (Φ greek name Φαι) … is it fee or fai

then let her read:

http://www.goldennumber.net/pronouncing-phi/

the same goes for π.

π (παι) is not pronounced pai as it seems when written in greek… αι pronounces as a long e (ee)

do some googling and find out more.



[youtube]http://www.youtube.com/watch?v=1FyEWbwBarQ[/youtube]

Last edited by aspi-rant on 22 Jul 2012, 8:31 am, edited 1 time in total.

To lau: yeah! But it's weird to talk about circles on non-euclidian planes, cause then you have to go all riemannian...

All euclid's laws get messed up on non-euclidian planes. A triangle an have angles that sum greater (on positive curvature) or less than (on negative curvature) 180 degrees. If you an even call them angles... cause they're all curved and stuff.

Curiouser and curiouser. We write "aspie" as such, and yet pronounce it as "ass-pee", not "ass-pie".

The Greek letter π is certainly pronounced as "pie", although the pi, as in "A flat jade disc with a small hole in the middle, used in ancient China for symbolic or religious purposes", is pronounced as either "pee" or "bee".

In any case, this means that your profile picture, aspi-rant, reading as it does as "ASπ" needs must be pronounced as "ass-pie".

(Note that the "ass" I have used above is the English "ass": "A well-known quadruped of the horse kind...", pronounced as in "lass". It is not the "vulgar and dialect spelling and pronunciation of arse, now chiefly U.S."