Shouldn't Pi be equal to 4?

i completely agree with you and have no idea of how to disprove this, so i will try put you in contact with someone who can: try this website

http://www.wikihow.com/Prove-That-Pi-Equals-3

or there is several on

http://uk.ask.com/web?q=should+pi+equal+4&search=&...

hope i helped, Jake

"now if we were to cut the corners of the box in such a fashion as they would cut in and touch the circle, the perimeter does not change and is equal to 4. "

This statement is not true.

Proof: There is a triangle being cut away from each corner of the original box. This is a right angle triangle where the two sides that were originally part of the outer square s perimeter are now being replaced by the hypotenuse of the triangle. For ANY triangle, the sum of the lengths of the two shorter sides will ALWAYS be greater than the length of the longest side. Therefore the new perimeter after cutting away the corners will be less than 4.

Specifically, in the case of replacing the square with an octagon by cutting away the four corners, the original perimeter is 4, and the new perimeter is 8/(1+2^1/2) or 3.3137.... As the corners are cut off, each removed corner is a triangle with two old sides replaced with one new side, the single side always being shorter than the sum of the initial two. As the process continues, the perimeter reduces until when there are infinite sides (and therefore no remaining corners) the length of the perimeter is 3.14159...

RE:

Shouldn't Pi be equal to 4?

If Pi can be expressed as the circumference of a circle divided by it's radius then why is pi=3.1415927 when it should be 4???

To prove this take a circle of diameter = 1

draw a square around the circle with the sides touching the circle, the length of each side is = the diameter of the...

I think most people responding are misunderstanding the way in which you are cutting your square, which is understandable since you can't post a picture of it. I believe I understand what you mean, though. It sounds like you are "folding" the corners of each square so each vertex coincides with a circle, then repeating this process with the new corners that appear. While it is true that the perimeter of the manipulated square remains 4 at each iteration of this process, your assumption that this approaches the circumference of the circle is false. It is very similar to this supposed counter proof of Pythagorean's Theorem:

http://www.futilitycloset.com/2009/07/21/pythagora...

A rigorous counter-proof would be a somewhat involved application of a limiting process. However, direct measurements can easily be used to show that pi should indeed be 3.1415... This alone is adequate in disproving your claim. The flaw in your argument is that you are assuming the perimeter of the square approaches the circumference of the circle, simply because it appears to "hug" the circle more and more closely with each iteration. However, this is an assumption that you are making only because it seems to be true. You haven't given a rigorous argument in support of this hypothesis, which you will find is impossible because your entire claim is false to begin with.

I figured this one out yesterday for some of my friends in engineering. It takes a bit of calculus, but it has to do with the calculation of an arc length.

The way you calculate the arc length using the "rectangle approximation", what you are doing is assuming that the arc length is the infinite sum or integral of differential lengths dx and dy, where x and y are the width and height of your rectangles that you are approximating as your circle. While this approach appears correct to the untrained, it is in fact not a good approximation. The differential length you add up or integrate over [dl] from (x,y) to (x + dx, y + dy) is not dx + dy, but rather the square root of the sum of their squares (pythagorean theorem, dl = sqrt(dx^2 + dy^2)). So rather than adding up their corner distances, you add up the diagonals of the corner distances.

The approach of using the perimeter assumes that dl = dx + dy, thus l = integral (dx + dy), which is incorrect. The correct approach is that dl = sqrt(dx^2 + dy^2), thus after a bit of algebra,

l = integral (1 + (dy/dx)^2 )dx, thus when you compute the integral over a circle you end up with pi = 3.14159 etc (although it is not very easy to prove at the level of computing the integral). Q.E.D.

The perimeter may not change if you do it an infinite amount of times, but your approximation perimeter will have unnecessary length from zig-zagging up and down near the perimeter of the circle. Regardless of how close it hugs the circle, on an infinite scale, just like a fractal, it is adding unnecessary length because the square approximation is not curving.

What do you mean "the perimeter does not change" if you cut the corners of the box (square)? Of course it does! Instead of a square, you'd now have an octagon, and the perimeter would be less than 4. If you can the corners again, you'd have a 16-sided polygon, with an even smaller perimeter, and the perimeter would keep getting smaller and smaller until it reached pi (around 3.14).

Your logic makes no sense.

you are correct that you can tuck in the corners up to the circle and still have the same perimeter. You are incorrect that you can do this an infinite number of times and will eventually have a circle. You will have a very odd geometric figure with a perimeter of 4 basically layered outside of a circle.

Use a piece of string to make your square around the circle (roughly) , take it off and put to one side

Now use another piece of string round the circumference of the circle.

Measure the two pieces against one another and you will see they do not match

This is because when you move the corners of the square to go round the circle there will be a small overlap at each corner hence 3.142, not 4

it is just a theory that some one invented and people figgured it was right so now we use it. 3.1415927 is not able to be rounded up to four because the first decimol is 1.