# Pi = 4

• Oct 3rd 2011, 11:32 AM
alexmahone
Pi = 4
I found this on a website. Where's the fallacy?
• Oct 3rd 2011, 12:50 PM
FernandoRevilla
Re: Pi = 4
I don't understand the last step: $\pi=4!=24$ :)
• Oct 3rd 2011, 12:55 PM
alexmahone
Re: Pi = 4
Quote:

Originally Posted by FernandoRevilla
I don't understand the the last step $\pi=4!=24$ :)

That's an exclamation mark, not a factorial sign. But where's the flaw in this "proof"?
• Oct 3rd 2011, 12:57 PM
TheEmptySet
Re: Pi = 4
When Archimedes did his calculuations he used a lower bound as well. Then he used the squeeze theorem to say that the value of Pi must be inbetween the two values. The above diagram only gives you an upper bound on the value of Pi.

@ Fernando haha that is an awsome observation!
• Oct 3rd 2011, 01:03 PM
FernandoRevilla
Re: Pi = 4
Quote:

Originally Posted by alexmahone
That's an exclamation mark, not a factorial sign.

Irony and Internet: bad marriage. :)

Quote:

But where's the flaw in this "proof"?
Hint: Try $4$ as a lower bound of $\pi$ .
• Oct 3rd 2011, 01:05 PM
emakarov
Re: Pi = 4
Quote:

Originally Posted by TheEmptySet
When Archimedes did his calculuations he used a lower bound as well. Then he used the squeeze theorem to say that the value of Pi must be inbetween the two values. The above diagram only gives you an upper bound on the value of Pi.

Can't you have a similar polygonal path whose length is arbitrarily close to 4 inside the circle?
• Oct 3rd 2011, 02:14 PM
Deveno
Re: Pi = 4
the polygonal path (at the limit) is not differentiable...this shows one must be careful as to how you define "length" of a curve. put another way:

length(limit) ≠ limit(length).
• Oct 3rd 2011, 02:21 PM
TheEmptySet
Re: Pi = 4
Quote:

Originally Posted by emakarov
Can't you have a similar polygonal path whose length is arbitrarily close to 4 inside the circle?

I do not think so. With a little bit of calculus you should be able to convice yourself that the side length of an inscribed square with maximum periemter would have length $\dfrac{\sqrt{2}}{2}$ So the starting Perimeter would be $2\sqrt{2} \approx 2.83$

Again this just shows that pi is less than for but bigger than 2.83
• Oct 10th 2011, 11:18 AM
HallsofIvy
Re: Pi = 4
For a similar fallacy, let the broken line from (0, 0) to (1/2, 0) to (1/2, 1/2) to (1, 1/2) to (1, 1) be an "approximation" to the straight line from (0, 0) to (1, 1)- it has length 1/2+ 1/2+ 1/2= 2. A better approximation would be the broken line from (0, 0) to (1/4, 0) to (1/4, 1/4) to (1/2, 1/4) to (1/2, 1/2) to (3/4, 1/2) to (3/4, 3/4) to (1, 3/4) to (1, 1) for a total length of 8(1/4)= 2 again. In general, you have 2n line segments each of length 1/n. "In the limit" this appears to converge to the straight line, which has length $\sqrt{2}$, but the total length is always 2.
• Oct 10th 2011, 11:55 AM
TheChaz
Re: Pi = 4
Quote:

Originally Posted by FernandoRevilla
I don't understand the last step: $\pi=4!=24$ :)

All of this (including the wit) has been thoroughly discussed...
Is value of $\pi$ = 4? - Mathematics - Stack Exchange
• Oct 10th 2011, 01:22 PM
Deveno
Re: Pi = 4
this is true. also on physicsfoums.com, and other places.

this is the nicest "non-technical" explanation i have seen:

when we approximate things in calculus, we always have a way of showing that the uncertainty, or error of our calculations goes to 0, in the limit.

in this case, the length of the approximating curve, the zig-zagging polygon, always stays the same, so the error is always 4-pi.

so, in the limit, all we know is that pi ≤ pi + 4-pi, which is true.