View Full Version : Is pi Infinite 2
pallidin
Mar15-04, 10:33 PM
Thanks, everyone for the original postings.
Question: If a diameter of 1 ascribes a circle of pi, does it it not lead to the conclusion that pi is determinate? After all, is this not a properly constrained system?
Janitor
Mar15-04, 10:53 PM
We've done pi to death. Why don't we ask if e is infinite?
Integral
Mar16-04, 02:05 AM
Or \sqrt 2
cookiemonster
Mar16-04, 02:18 AM
Just for good measure, how about \pi e \sqrt{2}?
cookiemonster
matt grime
Mar16-04, 03:33 AM
But are you sure you've drawn a diameter of exactly one inch? Might it not be .9999999999999999 inches? YOu sure? How are you going to check? And when did infinite start to mean indeterminate, and in what sense are you using that word?
HallsofIvy
Mar16-04, 05:40 AM
Originally posted by pallidin
Thanks, everyone for the original postings.
Question: If a diameter of 1 ascribes a circle of pi, does it it not lead to the conclusion that pi is determinate? After all, is this not a properly constrained system?
Quite frankly, what does this mean? "If a diameter of ascribes a circle of pi". I confess I don't know the word "ascribes" and I don't know what "a circle of pi" could mean.
In any case, "does it not lead to a conclusion that pi is determinate?" puzzles me. What do you mean by "determinate". pi is, of course, a perfectly we determined number. We know exactly what pi is.
Michael D. Sewell
Mar16-04, 02:41 PM
Enough pi! pi can be described and defined in a very short paragraph, that's all the attention that it deserves.
pallidin
Mar16-04, 03:33 PM
OK, point taken! Sorry for beating a dead horse.
Issue is now closed!
Organic
Mar17-04, 03:13 AM
you have correctly identified though, that it is possible to derive a set from pi which is infinite, but that set is not equal to pi because a set can never be equal to a number, a set can only be equal to another set.
There is no limit to a content of a set if it can be compared with N members, therefore set b, which is the binary representation of Pi, is a legal set:
b = {'1_1','1_2','1_3','1_4','0_1','0_2',...}
N = { 1 , 2 , 3 , 4 , 5 , 6 ,...}
Moderators, how much longer does this thread need to be left in the General Math section? The answer to the original question was given many pages ago.
No chroot, please clrealy show why b is not a legal set.
Also please explain why circle's perimeter/diameter(=Pi) representations are ignored by Standard Math?
matt grime
Mar17-04, 04:38 AM
Chroot was making the point that you never acknowledge. Namely that things are only equal when they are equal. The set (b_n) n in N of the elements of the decimal expansion of pi is NOT equal to pi, nor is it equal to the SET of rational approximations.
A set is a set is a set.
Where does chroot say anthing about legality of sets or otherwise?
But then you're happy to equate the empty set and an inequality, so why should we expect you to follow basic conventions and definitions? Just try and remember when other people write things using mathematical terms they are (in all probability) using them in the proper sense, not in whatever way you happen to think it might be.
Organic
Mar17-04, 05:58 AM
Chroot was making the point that you never acknowledge. Namely that things are only equal when they are equal. The set (b_n) n in N of the elements of the decimal expansion of pi is NOT equal to pi, nor is it equal to the SET of rational approximations.
By Standard Math Pi is one of infinitely many irrational numbers of the real line.
Pi is interesting because it is the result of the circle's perimeter/diameter, which is the most symmetrical geometric form, and symmetry is maybe one of the most important concepts of Math language.
Standard Math ignores the verity of structural representation possibilities of Pi and taking care only to its quantitative value that determinates its exact place in the real line.
So I am asking again, please explain why circle's perimeter/diameter(=Pi) representations are ignored by Standard Math?
So I am asking again, please explain why circle's perimeter/diameter(=Pi) representations are ignored by Standard Math?
Because the "structural representation possibilities" (whatever those may be) are irrelevant when one is concerned with the value.
Organic
Mar17-04, 06:32 AM
Then why structural representation possibilities are not valueable to Standard Math?
*dun-dun* *tcssshhh*
Bah, where is a drumset smiley when you need it?
KingNothing
Mar18-04, 04:41 PM
Originally posted by Michael D. Sewell
Enough pi! pi can be described and defined in a very short paragraph, that's all the attention that it deserves.
I agree. Things like prime numbers or perfect numbers could have much more extensive discussions on them.
Bob3141592
Mar19-04, 09:53 AM
Originally posted by Michael D. Sewell
Enough pi! pi can be described and defined in a very short paragraph, that's all the attention that it deserves.
So what's wrong with pi?
Now, if I can only get an infinitely long screen name....
Michael D. Sewell
Mar20-04, 03:48 AM
Originally posted by Bob3141592
So what's wrong with pi?
Nothing. After all these years it hasn't failed me yet.
-Mike
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