Math for fun and glory
Rhapsody on the proof of pi = 4
Correction: when I mark where pi is on the graph, I meant pi/2! Note: If this video were supposed to be teaching you, I'd probably have to make it boring and say that in one sense of limits, spoiler alert, you actually do approach a circle and a line, solving the apparent paradox by saying that the invariant of length does not hold over infinity. Luckily I am an artist, and this is a Rhapsody, and instead of "learning," you get to actually think, if you like. Created by Vi Hart.
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- At1:57, are raindrops actually spherical?(19 votes)
- The shape of a drop of water depends on its size. Smaller ones are spherical.
- If √2 (1.4142135623730951) is a common geometric number is √3 (1.7320508075688772) one?(2 votes)
- sqrt 3 has a lot of interesting properties just like sqrt 2. sqrt 3, for example, is the length of the space diagonal of a unit cube. It is also double the ratio of the height of a equilateral triangle to the side length.
Hope this helped!(5 votes)
- Would a infinagon be considered a circle?(6 votes)
- Maybe, it looks like a circle but in some sense it might be, it depends on how you calculate it.(1 vote)
- Lots of my friends say Pi is right. Who should I believe?(2 votes)
- Tau is a million times easier than Pi, so that's why people think Tau is better.(2 votes)
- At1:39won't the diameter change when you "inflate" the circle?(3 votes)
- honestly whenever she says something like "oh no the teacher's walking around" my heart skips a beat(2 votes)
- If pi is 4,then tau is 8,and in other videos Vi wrote tau as 6.28...,so in this video she proved herself wrong.Why did she do that?(2 votes)
- She says "you may think it's 4", not "it is 4".(1 vote)
- Computers have pixels. Does that mean that pi=4 on a computer?(2 votes)
- Yes, for a computer, pi does equal 4. But the computer's pi is not right.(1 vote)
- The peaks aren't zero. They're 1/infinity. Am I correct? (Or maybe 1/(2^Infinity)😉(2 votes)
- The peaks have a height of zero (1/infinity = 0), and each segment also has a length of zero. Note, however, that 0*infinity is undefined, and it does not make sense to try to calculate it.(1 vote)
- So the logic in this video is exactly the same as the logic in the more famous of Zeno's paradoxes, sometimes known as Achilles and the Tortoise. I never used that, so I'll explain it my way.
Say a man in archery class is shooting arrows at a target and every time he shoots he gets 1/2 of the distance between the last arrow and the target. The first time he shoots, his arrow lands halfway to the target, the second time halfway between those, so 3/4 to the target, and so on until infinity. Does the man's arrows ever actually hit the target? Realistically, yes, but technically, no. The arrow infinitely approaches the target and it seems like it hits it, but in a perfect mathematical reality with no gravity or wind or variables in any way, it never does, only gets infinitesimally close to the target without ever actually touching it.
So, the logic here is that the peaks of the triangles and the zigs and zags of the square around the circle infinitely approach their destination, but as the arrow never hit the target, if you keep zooming in infinitely into the fractal of the idea, you'll see that it never actually becomes flat, and the square will be forever lonely because it will never ever actually hug its perfectly circular love :'[(1 vote)
- But -- if the length of the arrow is a fixed constant
the distance between the shooter and the arrow is a fixed constant
and the man/shooter is a fixed constant...
The arrow will eventually have to hit the target.
If I'm understanding your explanation/example -- it seems to suggest/assume that the length of the arrow is not a fixed constant and that the distance is not a fixed constant.
or -- did I miss something?(2 votes)
Say you're me and you're in math class, and you're supposed to be graphing functions as if there were some deep relationship between y and x that your teacher just won't stop gossiping about. But like most gossip, you really don't care about y's unhealthy dependency on x. Really y, get a life. Luckily, your friend passes you a note. You wait until the teacher is facing the board, and sneakily open it up. And it's this proof of pi equals 4 that you've seen like a billion times on the internet. But because you're bored, and at least it's not graphing, you'd take another look. It's like this. Start with a circle of diameter one and circumference pi. Then draw a square around it. The length of a side of the square is 1, so the perimeter of the square is of course, 4. Now, you start approaching a circle by zigzagging this parameter. This shape still has perimeter 4, and we do it again, and it's still 4, and again. And we approach a circle while the perimeter never changes. So therefore, pi equals 4. Obviously, there's something wrong here. The circumference of a circle is pi, but pi is more like 3.14 numbers, and less like 4. So somehow what this process approaches, is something that looks like a circle, but is not a circle. I mean, what does circle mean anyway? There's this loopy line thing, And then there's this solid disk shape. And while they're related and all, they're not actually the same mathematical object. This troll proof is so much fun because, as you repeat this zigzag move, all the way to zig-finity, it approaches the shape of a circle in the disk sense, and it approaches the area of a circle, but it does not approach a circle. It's all crinkled up. And you imagine that if you stuck a straw in this thing and inflated it-- that is, added as much area as you can while keeping the perimeter as 4-- all the infinite wrinkles would smooth out, and you'd get a circle with perimeter 4 and diameter 4 over pi. In fact, you could inflate the square to get the same thing. Probably has something to do with how circles are the shape with the most area possible given their perimeter, and why soap bubbles like to be spheres, and raindrops are actually pretty sphere-ish too. But anyway, you decide the best way to respond to your friend is to try applying the same fake proof process to something else. Maybe you can choose another irrational number, like square root 2. In fact, square root 2 would work great, because it's also a common geometry number. The ratio of the diagonal of a square to its perimeter. I mean, like, if this square has side length 1, then you've got this right triangle. And a squared plus b squared equals c squared. In fact, that's how you decide to start your proof for your friend. Take a right isosceles triangle. Each leg has length 1. So the hypotenuse is square root 2. Put this all in the square of perimeter 4. Now we can approach a triangle in the same way. And each time, the perimeter of the whole shape is still 4. And thus, by the time we get to a triangle, the length of the hypotenuse must be 2. Thus, the square root of 2 is 2. So you pass that over to you friend. Will he notice that you're approaching the area of a triangle without approaching a triangle? That a triangle has three sides, but the shape you end up with is a polygon with infinite sides? An infini-gon? But not a boring one, like how a regular polygon approaches a circle as sides go to infinity, because infini-gons are more fun when there's actual angles between sides. Zigzagging along to make what you've decided to call a zig-fini-gon. You begin to wonder what other zig-fini-gons you can make. Maybe if you started with a star, and zigged all the points in. And did that again and again to infinity. You get something that looks like a pentagon, but it's actually a zig-finite star with the same perimeter as the original star. Or maybe you could have a rule where at each step you zig down only part way. And then your zig-fini-gons will have more texture to it. Maybe you could throw some zags in there too. There's something fractal-ly about it, except the perimeter never changes. You could do that to a triangle, or maybe make a square that turns into a zig-fini-square. Uh oh, teacher's walking around. Better draw some axes, and pretend to be doing math. So you turn the idea sideways, and start at zero. Go to y equals 1 at x equals 1, then back to 0 at x equals 2. The next iteration is like folding this point down to 0. So the function zigzags from 0 to 1/2, to 0 to 1/2, to 0. The next step brings the 1/2's to 0. And now the highest points are at y equals 1/4. Each step brings the highest points down to 0, and the new highest peaks are only half as high. And each step keeps the total length exactly the same. So what happens when you do this to zig-finity? In one way, it approaches this line, the x-axis, y equals 0. If there were any peaks, they'd get folded down to 0. Therefore, there can't be any. Yet, at each step we have twice as many peaks. So how can there be an infinite number of peaks and no peaks? You might reconcile this by saying that all infinite peaks are equally at zero, since all peaks get brought to zero eventually. But if everything's at y equals 0, you have just a line segment of length 2. That doesn't make sense. The length of the zigzag stays the same at each step, and at the beginning it's like two hypotenuse of two right triangles, so 2 square root 2. And 2 does not equal 2 square root 2. Another problem is that only peaks ever get brought to 0, but not all numbers can be peaks. Any fraction of a power of 2 will be a peak at some iteration, but a number like 1/3, or an irrational number, will never be a peak or a zig or a zag. So they must all be between the zigs and the zags. But there can't be any length between zigs and zags, or else that would create a peak that would have been folded down. Somehow it has to be infinitely zigzagged in a way where there's no actual line segments of any length, but only zigs and zags. Yet, there must be an infinite number of numbers between each zig and zag to fit all the irrationals in, and somehow all the pieces of 0 length add up to be something that does have length. You could imagine grabbing the ends and stretching it out, accordion style, into a line of length 2 square root 2. And then I suppose you could crumple it all back down until the whole length 2 square root 2 line is folded up into a single point.