20060717, 15:53  #1 
Bronze Medalist
Jan 2004
Mumbai,India
4004_{8} Posts 
A proof with a hole in it?
A proof with a hole: p equals 2
Forget all that business about p being an irrational number with infinitely many decimal places: I can prove conclusively that p is equal to 2. First of all, let's recall that p is defined to be the ratio between the circumference and the diameter of a circle, which is the same regardless of the size of the circle. So, using the notation of the diagram below, we have p = c/d. Now let's start with a circle of circumference 2, and only consider one half of it, as shown in the figure. Since it's exactly one half of the circle, the length of this semicircle is 1. Now let's divide in half the diameter d of the circle, and draw a new, smaller semicircle on each of the two halves. Since the ratio between diameter and circumference is the same for any circle, you can work out that the two smaller semicircles — which are built on half the diameter of the larger one — have circumference half that of the larger one. In other words, the length of each of the two smaller semicircles is 1/2. Now continue in the same manner: divide the original diameter d into 4 equal pieces and draw on each of them a semicircle of length 1/4; then divide it into 8 equal pieces and draw on each of them a semicircle of length 1/8, etc, etc. After n steps you have 2n semicircles, each of length 1/2n. Obviously, the semicircles get smaller and smaller at each stage, and after a great number of steps, your string of semicircles will hardly be distinguishable from the straight line which forms the diameter of the largest circle. The string of semicircles approximates the diameter d, and the approximation gets better and better the more steps you take. This means that the lengths of the semicircles all added up approximate d. In fact, d is the limit of this sum as the number of steps n tends to infinity: d = limn→∞ 2n×1/2n = 1. We know that the circumference c of the large circle is 2, so p = c/d = 2/1 = 2, which proves my claim. Or have I made a mistake? Mally  . 
20060717, 17:38  #2 
Aug 2005
Brazil
2×181 Posts 
I don't think a curved line can ever approximate a straight line. All seems to be correct, but if you take the limit equation 2n×1/2n, you'll see that it's true for every nonnegative n (I'm just too lazy to consider negative ones) which is nonsense. The amount of line that goes above the diameter for each semicircumference, in fact, becomes negligible along the way, and there is a limit that can be taken, but what you're doing is dividing the discrepancy between 1 and the diameter into smaller and smaller segments, that when considered together form the whole difference between 1 and the diameter.
I mean, you're just dividing the amount of line that goes beyond the straight line into smaller pieces so you can fool us into thinking on a particular semicircumference (whose excess is, in fact, getting smaller) and ignoring the fact that you're also getting the total number of excesses proportionally up. Last fiddled with by fetofs on 20060717 at 17:41 
20060717, 21:16  #3 
∂^{2}ω=0
Sep 2002
República de California
10110110101000_{2} Posts 
"Kinda sorta looks like" and "has length equal to" a straight line are completely different things. One can do a similar exercise with (say) a line segment of slope one from (0,0) to (1,1), approximated by a "staircase" function having N equalsized steps. The onestep staircase is just two adjacent sides of a unit square, with total length = 2, or sqrt(2) times longer than the diagonal. As we subdivide our staircase into evermore steps the result "looks" more and more like a straight line, but that is not a mathematically precise concept  the mathematically precise view shows that the *integrals* of the two resulting functions (diagonal vs. Nstep staircase) do in fact converge, but the ratio of the curve *lengths* never varies from sqrt(2).
Similar specious confusebysubdividing reasoning underlies the famous Paradoxes of Zeno. Perhaps I'm just a bit Zenophobic. ;) Here's a slightly more subtle apparent paradox related to basic functions and their integrals for your amusement: Consider the function f(x) = 1/x, from x = 1 to infinity. If we rotate the function about the xaxis we obtain a shape resembling an infinitely longstemmed champagne flute. It's easy to show that whereas the volume enclosed by the flute is finite, its surface area is infinite. So in other words, it seems we have a glass which we can fill with a finite volume of champagne, but since the glass has infinite surface area, no amount of champagne (or other beverage of your choosing) will ever suffice to coat the surface of the glass. How can we reconcile these two points of view? Last fiddled with by ewmayer on 20060718 at 19:01 Reason: Corrected f(x) from log(x) to 1/x 
20060718, 03:04  #4  
Jun 2005
2×191 Posts 
Quote:
You can assume a few different things. 1. First, assume the fluid is a continuum, in which case the layer coating the surface can be infinitessimally thin. The very first drop will be drawn into the flute by capillary action to cover *most* of the surface. The rest is a finite volume covering a finite area. This is probably the more appropriate answer from a purely mathematical perspective, but is not realistic. 2. Assume a scenario closer to reality in that the fluid is a collection of point masses whose motion is described as Brownian. There are very few molecules at any given time contained within the very small filament drawn through the champaign flute. However, the frequency of the collisions of a single molecule with the wall becomes very high because the filament is much narrower than the mean free path of the molecule. Therefore, a single molecule can apply a finite pressure to a very large area of the flute. 3. In reality, molecules have a finite size, although very small. Most of the area wouldnt be covered because the shape would discourage molecules from bouncing *into* that narrow filament, so the probability becomes very low that a molecule will find its way past a certain distance. Furthermore, there's a finite distance beyond which the region is smaller than the space that separates molecules of the container material, and smaller than the characteristic length of the water molecule. Drew Last fiddled with by drew on 20060718 at 03:11 

20060718, 04:16  #5  
Jun 2005
2×191 Posts 
Quote:
Last fiddled with by drew on 20060718 at 04:19 

20060718, 06:14  #6 
Dec 2003
Hopefully Near M48
11011011110_{2} Posts 

20060718, 11:21  #7  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
plus maths.
Quote:
I have never claimed to be an original problen setter when there are so many around. I just pick those which to me are tantalising and for the benefit of those members who may be interested. Also I direct them to those who are too lazy to look around or work them out. Yes you are right jinydu and here again you can go thru the whole issue and perhaps join up. I have and dont regret it one bit. So look forward to a few from the mag. BTW you have not given a reason and neither has the mag. Mally 

20060718, 11:44  #8  
Bronze Medalist
Jan 2004
Mumbai,India
4004_{8} Posts 
Apparent paradox.
Quote:
: Thanks Ernst. Simpler is the Koch's curve where the area of a finite triangle has an infinite length of sides. This can easily be demonstrated and proved. .http://www.ecademy.agnesscott.edu/~l...rve/kcurve.htm mally Last fiddled with by mfgoode on 20060718 at 12:02 

20060718, 19:01  #9  
∂^{2}ω=0
Sep 2002
República de California
11688_{10} Posts 
Quote:


20060927, 16:37  #10  
Bronze Medalist
Jan 2004
Mumbai,India
100000000100_{2} Posts 
Limiting diameter
Quote:
This is the feed back from Plus Maths. The solution What's wrong here? Well, it is definitely true that after n steps there will be 2n semicircles, each of length 1/2n. It is also true that the diameter of the largest circle is in some sense a limit of the strings of semicircles: by making n large enough, you can ensure that the nth string of semicircles squeezes as closely to the diameter d as you like. The mistake lies in the next step: the assumption that the lengths of the strings of semicircles tend to the length of the diameter. This is false! What happens in fact is that while at each stage we replace the semicircles involved by smaller ones, these also become more numerous, so that the replacement makes no difference at all to the overall length. Every string of semicircles has length 1, since 2n×1/2n = 1, for any n. We've simply replaced a number of big wriggles by a greater number of smaller wriggles without changing the length. The lesson is, when it comes to things like length, don't be fooled by appearance. A curve may look like a straight line, but it may be so wriggly at a small scale that in fact it's a lot longer than the line. Even worse, a set may look like a decent curve that should have finite length, but may in fact be infinitely long. And what exactly do we mean by "length" anyway? For an example of a curve with infinite length, check out the Koch curve in Plus article Jackon's fractals. And to learn about notions of length read Plus article Measure for measure. Mally 

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