# Thread: Why does Pi go on for infinity?

1. ## Why does Pi go on for infinity?

Is there no possible way that it has a final digit?

2. Originally Posted by anthmoo
Is there no possible way that it has a final digit?
No it is irrational (in fact worse than that it is not even algebraic). If it
did terminate it would of course be rational, but its not so it doesn't

RonL

3. Originally Posted by CaptainBlack
No it is irrational (in fact worse than that it is not even algebraic). If it
did terminate it would of course be rational, but its not so it doesn't

RonL

Can it be proven that it is irrational or not algebraic?

Geometrically, if we were drawing a circle, the ends must touch? (Even at the very small value level?)

4. Hence giving birth to my theory that there is no such thing as a closed circle

5. Originally Posted by anthmoo
Can it be proven that it is irrational
Yes, Lambert 1776.
He even proved a stronger result,
$\displaystyle \pi ^2$ is irrational.
not algebraic?
Yes, Loiville 1882 (I believe).
Geometrically, if we were drawing a circle, the ends must touch? (Even at the very small value level?)
It was not proven geometrical. It was proven based on an integration that produces $\displaystyle \pi$ as a result.

6. Here's my little conjecture (since I don't know how to prove it and excel says it doesn't work, but after 200 terms I don't know how accurate excel can be)

$\displaystyle \lim_{n\to\infty}\frac{\sqrt8}{n}\cdot\sum_{a=0}^n \sqrt{n^2-a^2+a-\sqrt{a^2\left(a-1\right)^2+2an^2(1-a)+n^2(n^2-1)}}=\pi$

This shows that the closer you get to infinity, the more decimal places will be in pi, but what if you reach infinity (which you obviously can't do)? You would get infinite decimal places, and it would equal pi...

7. Originally Posted by Quick
You would get infinite decimal places, and it would equal pi...
He wants to know why it doth not have repeating decimal places.

8. Originally Posted by ThePerfectHacker
Yes, Lambert 1776.
He even proved a stronger result,
$\displaystyle \pi ^2$ is irrational.

Yes, Loiville 1882 (I believe).

It was not proven geometrical. It was proven based on an integration that produces $\displaystyle \pi$ as a result.
Is there any proof geometrically?

Does this mean that the ends of a circle do not touch at all? (Apart from at infinity)

Originally Posted by Quick
Here's my little conjecture (since I don't know how to prove it and excel says it doesn't work, but after 200 terms I don't know how accurate excel can be)

$\displaystyle \lim_{n\to\infty}\frac{\sqrt8}{n}\cdot\sum_{a=0}^n \sqrt{n^2-a^2+a-\sqrt{a^2\left(a-1\right)^2+2an^2(1-a)+n^2(n^2-1)}}=\pi$

This shows that the closer you get to infinity, the more decimal places will be in pi, but what if you reach infinity (which you obviously can't do)? You would get infinite decimal places, and it would equal pi...

How can you calculate that in excel??

The best I can get for calculating pi is doing the sum of about 100000 inverse squares!

9. Originally Posted by anthmoo
Can it be proven that it is irrational or not algebraic?
A quite nice proof can be found here

RonL

10. Originally Posted by anthmoo
Does this mean that the ends of a circle do not touch at all? (Apart from at infinity)
What do you mean, "apart from at infinity"? Of course a circle is "closed", just as a right triangle with sides 1,1,sqrt(2) has corners (they "touch").

11. Originally Posted by TD!
What do you mean, "apart from at infinity"? Of course a circle is "closed", just as a right triangle with sides 1,1,sqrt(2) has corners (they "touch").
If the diameter of the circle is 1 then it's circumference is Pi.

If Pi has infinite decimal places then it must never be closed circle as there must be a space somewhere on the circle that is infinitely small and can not be filled.

What I'm thinking is usually hard to understand so think of it like this...

Pretend your drawing a circle with an incredibly fine pencil and your drawing of the circle is literally perfect.

The diameter of the circle is 1cm so the circumference is Pi. You start at a point and draw 3cm. The circle is not yet complete as there is 0.14159..cm left so you draw 0.1cm but there's still 0.0415926...cm left to draw! You get closer and closer but since pi has infinite decimal places (that aren't all 0s) you will never reach the starting point of the circle!

I hope thats easy enough to follow!

12. It's clear what you mean, but mathematically there's no problem and practically, there's no difference with other numbers.

Take a diameter of 1/pi, then you have to draw the circle with circumference 1. You have your fine pencil and you start drawing, already at 0.9, then 0.95, then 0.9997, then... Stopping at exactly 1, isn't fysically/practially "easier" than stopping at pi.

The circle is closed, because pi is what it is, it's not 3.14 and not 3.141592653, but pi.

13. Originally Posted by TD!
It's clear what you mean, but mathematically there's no problem and practically, there's no difference with other numbers.

Take a diameter of 1/pi, then you have to draw the circle with circumference 1. You have your fine pencil and you start drawing, already at 0.9, then 0.95, then 0.9997, then... Stopping at exactly 1, isn't fysically/practially "easier" than stopping at pi.

The circle is closed, because pi is what it is, it's not 3.14 and not 3.141592653, but pi.
Very well put, thankyou!

My mistake was assuming pi would be drawn from a meaurable point of view when really "it is what it is".

So in theory - no problem. Pi has its value and thats the value in the ratio from diameter to circumference. However, on paper it can be a problem as if you draw a perfect closed circle (ie. no errors in drawing it whatsoever) there will always be an infinitesimally small gap!

14. Why still the gap? The paper doesn't know about real numbers, nor about our concept of 'meters'.
The only problem which we practically encounter is our inability to draw so perfectly.

For us, it's not harder/easier to draw a perfect 3-4-5 (5² = 3²+4²) right triangle, than to draw a 1,1,sqrt(2) (sqrt(2)² = 1²+2²) right triangle, although this last one has a side which has an irrational number as length!

15. Originally Posted by anthmoo
Is there any proof geometrically?
I never seen one, the best I can do is as follows...

A number $\displaystyle a$ is said to be constructable when we can using Euclidean toys (striaghtedge and compass) construct $\displaystyle |a|$. This is an algebra question but the important fact about the set of all constructable numbers is that $\displaystyle \mathbb{Q} \subset \mathbb{F}$. Meaning the it contains all rational numbers. Thus, what you need to show is that you cannot constuct $\displaystyle \pi$ and hence show it is irrational (note, not transcendental! this does not show this). The problem with this approach is that these concepts was purposely created to simplify construcability problems but we are going backwards meaning from this approach making more difficult.

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