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Today I'd like to share with you three fake proofs in increasing order of subtlety,
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and then discuss what each one of them has to tell us about math.
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The first proof is for a formula for the surface area of a sphere,
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and the way that it starts is to subdivide that sphere into vertical slices,
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the way you might chop up an orange or paint a beach ball.
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We then unravel all of those wedge slices from the northern hemisphere,
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so that they poke up like this, and then symmetrically unravel all of those from the
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southern hemisphere below, and now interlace those pieces to get a shape whose area we
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want to figure out.
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The base of this shape came from the circumference of the sphere,
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it's an unraveled equator, so its length is 2 pi times the radius of the sphere,
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and then the other side of this shape came from the height of one of these wedges,
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which is a quarter of a walk around the sphere,
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and so it has a length of pi halves times r.
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The idea is that this is only an approximation,
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the edges might not be perfectly straight, but if we think of the limit as we do finer
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and finer slices of the sphere, this shape whose area we want to know gets closer to
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being a perfect rectangle, one whose area will be pi halves r times 2 pi r,
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or in other words pi squared times r squared.
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The proof is elegant, it translates a hard problem into a situation that's easier to
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understand, it has that element of surprise while still being intuitive, its only fault,
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really, is that it's completely wrong, the true surface area of a sphere is 4 pi r
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squared.
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I originally saw this example thanks to Henry Reich, and to be fair,
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it's not necessarily inconsistent with the 4 pi r squared formula,
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just so long as pi is equal to 4.
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For the next proof I'd like to show you a simple
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argument for the fact that pi is equal to 4.
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We start off with a circle, say with radius 1,
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and we ask how can we figure out its circumference, after all,
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pi is by definition the ratio of this circumference to the diameter of the circle.
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We start off by drawing the square whose side lengths are all tangent to that circle.
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It's not too hard to see that the perimeter of this square is 8.
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Then, and some of you may have seen this before, it's a kind of classic argument,
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the argument proceeds by producing a sequence of curves,
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all of whom also have this perimeter of 8, but which more and more closely
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approximate the circle.
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But the full nuance of this example is not always emphasized.
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First of all, just to make things crystal clear,
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the way each of these iterations works is to fold in each of the corners of
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the previous shape so that they just barely kiss the circle,
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and you can take a moment to convince yourself that in each region where a fold happened,
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the perimeter doesn't change.
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For example, in the upper right here, instead of walking up and then left,
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the new curve goes left and then up.
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And something similar is true at all of the folds of all of the different iterations.
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Wherever the previous iteration went direction A then direction B,
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the new iteration goes direction B then direction A, but no length is lost or gained.
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Some of you might say, well obviously this isn't going to give the true perimeter of the
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circle, because no matter how many iterations you do, when you zoom in,
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it remains jagged, it's not a smooth curve, you're taking these very inefficient steps
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along the circle.
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While that is true, and ultimately the reason things are wrong,
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if you want to appreciate the lesson this example is teaching us,
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the claim of the example is not that any one of these approximations equals the curve,
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it's that the limit of all of the approximations equals our circle.
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And to appreciate the lesson that this example teaches us,
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it's worth taking a moment to be a little more mathematically
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precise about what I mean by the limit of a sequence of curves.
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Let's say we describe the very first shape, this square,
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as a parametric function, something that has an input t and it outputs
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a point in 2d space, so that as t ranges from 0 to 1, it traces that square.
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I'll call that function c0, and likewise we can parameterize the next
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iteration with a function I'll call c1, as the parameter t ranges from 0 up to 1,
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the output of this function traces along that curve.
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This is just so that we can think of these shapes as instead being functions.
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Now I want you to consider a particular value of t, maybe 0.2,
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and then consider the sequence of points that you get by evaluating
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the sequence of functions we have at this particular point.
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Now I want you to consider the limit as n approaches infinity of c sub n of 0.2.
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This limit is a well-defined point in 2d space, in fact that point sits on the circle.
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And there's nothing specific about 0.2, we could do this limiting process for
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any input t, and so I can define a new function that I'll call c infinity,
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which by definition at any input t is whatever this limiting value for all the curves is.
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So here's the point, that limiting function c infinity is the circle,
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it's not an approximation of the circle, it's not some jagged version of the circle,
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it is the genuine smooth circular curve whose perimeter we want to know.
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And what's also true is that the limit of the lengths of all of our curves really is 8,
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because each individual curve really does have a perimeter of 8.
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And there are all sorts of examples throughout calculus when
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we talk about approximating one thing we want to know as a
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limit of a bunch of other things that are easier to understand.
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So the question at the heart here is why exactly
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is it not okay to do that in this example?
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And maybe at this point you step back and say, you know,
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it's just not enough for things to look the same, this is why we need rigor,
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it's why we need proofs, it's why since the days of Euclid mathematicians
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have followed in his footsteps and deduced truths step by step from axioms forward.
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But for this last example I would like to do something that
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doesn't lean as hard on visual intuition and instead give a
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Euclid style proof for the claim that all triangles are isosceles.
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The way this will work is we'll take any particular triangle and make no assumptions
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about it, I'll label its vertices a, b, and c,
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and what I would like to prove for you is that the side length a,
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b is necessarily equal to the side length a, c.
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Now, to be clear, the result is obviously false,
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just in the diagram I've drawn you can visually see that these lengths are
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not equal to each other.
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But I challenge you to see if you can identify
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what's wrong about the proof I'm about to show you.
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Honestly, it's very subtle and three gold stars for anyone who can identify it.
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The first thing I'll do is draw the perpendicular bisector, the line bc,
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so that means this angle here is 90 degrees and this length is by definition
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the same as this length, and we'll label that intersection point d.
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And then next I will draw the angle bisector at a,
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which means by definition this little angle here is the same as this little angle here,
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I'll label both of them alpha, and we'll say that the point where these two
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intersect is p.
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And now, like a lot of Euclid style proofs, we're just going to draw some new lines,
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figure out what things must be equal, and get some conclusions.
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For instance, let's draw the line from p which is perpendicular to the side length ac,
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and we'll label that intersection point e.
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And likewise, we'll draw the line from p down to the other side length ac,
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again it's perpendicular, and we'll label that intersection point f.
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My first claim is that this triangle here, which is afp,
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is the same, or at least congruent, to this triangle over here, aep.
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Essentially this follows from symmetry across that angle bisector.
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More specifically we can say they share a side length,
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and then they both have an angle alpha, and both have an angle 90 degrees.
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So it follows by the side angle angle congruence relation.
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Maybe my drawing is a little bit sloppy, but the
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logic helps us see that they do have to be the same.
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Next I'll draw a line from p down to b, and then from p down to c,
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and I claim that this triangle here is congruent to its reflection across that
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perpendicular bisector.
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Again the symmetry maybe helps make this clear,
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but more rigorously they both have the same base, they both have a 90 degree angle,
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and they both have the same height, so it follows by the side angle side relation.
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So based on that first pair of triangles I'm going to mark this side length here
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as being the same as this side length here, marking them with double tick marks.
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And based on the second triangle relation I'll mark this side length here
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as the same as this line over here, marking them with triple tick marks.
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And so from that we have two more triangles that need to be the same,
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namely this one over here, and the one with corresponding two side lengths over here.
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And the reasoning here is they both have that triple ticked side,
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a double ticked side, and they're both 90 degree triangles.
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So this follows by the side side angle congruence relation.
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And all of those are valid congruence relations,
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I'm not pulling the wool over your eyes with one of those,
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and all of this will basically be enough to show us why AB has to be the same as BC.
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That first pair of triangles implies that the length AF is the same as the length AE,
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those are corresponding sides to each other, I'll just color them in red here,
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and then that last triangle relation guarantees for us that the side
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FB is going to be the same as the side EC.
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I'll kind of color both of those in blue.
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And finally the result we want basically comes from adding up these two equations.
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The length AF plus FB is clearly the same as the total length AB,
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and likewise the length AE plus EC is the same as the total length AC.
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So all in all the side length AB has to be the same as the side length AC,
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and because we made no assumptions about the triangle this implies that any triangle
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is isosceles.
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Actually for that matter since we made no assumptions about the
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specific two sides we chose, it implies that any triangle is equilateral.
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So this leaves us somewhat disturbingly with three different possibilities.
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All triangles really are equilateral, that's just the truth of the universe,
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or you can use Euclid style reasoning to derive false results,
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or there's something wrong in the proof.
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But if there is, where exactly is it?
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So what exactly is going on with these three examples?
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Now the thing that's a little bit troubling about that first
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example with the sphere is that it is very similar in spirit to
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a lot of other famous and supposedly true visual proofs from geometry.
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For example there's a very famous proof about the area of a circle that starts
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off by dividing it into a bunch of little pizza wedges,
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and you take all those wedges and you straighten them out,
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essentially lining up the crust of that pizza,
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and then we take half the wedges and inter-slice them with the other half.
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And the idea is that this might not be a perfect rectangle,
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it's got some bumps and curves, but as you take thinner and thinner slices you get
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something that's closer and closer to a true rectangle,
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and the width of that rectangle comes from half the circumference of the circle,
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which is by definition pi times r, and then the height of that rectangle comes from the
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radius of the circle, r, meaning that the whole area is pi r squared.
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This time the result is valid, but why is it not okay to do what we did with the spheres,
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but somehow it is okay to do this with the pizza slices?
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The main problem with the sphere argument is that when we flatten out all
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of those orange wedges, if we were to do it accurately in a way that
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preserves their area, they don't look like triangles, they should bulge outward.
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And if you want to see this, let's think really critically about just one particular
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one of those wedges on the sphere, and ask yourself how does the width across that wedge,
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this little portion of a line of latitude, vary as you go up and down the wedge?
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In particular, if you consider the angle phi from the z-axis down to a point on
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this wedge as we walk down it, what's the length of that width as a function of phi?
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For those of you curious about the details of these sorts of things,
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you'd start off by drawing this line up here from the z-axis to a point on the wedge,
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its length will be the radius of the sphere r times the sine of this angle.
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That lets us deduce how long the total line of latitude is where we're sitting,
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it'll basically be 2 pi times that radial line, 2 pi r sine of phi,
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and then the width of the wedge that we care about is just some constant proportion of
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that full line of latitude.
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Now the details don't matter too much, the one thing I want
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you to notice is that this is not a linear relationship.
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As you walk from the top of that wedge down to the bottom,
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letting phi range from 0 up to pi halves, the width of the wedge doesn't grow linearly,
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instead it grows according to a sine curve.
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And so when we're unwrapping all of these wedges,
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if we want those widths to be preserved, they should end up
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a little bit chubbier around the base, their side lengths are not linear.
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What this means is when we tried to interlace all of the wedges from the northern
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hemisphere with those from the southern, there's a meaningful amount of overlap
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between those non-linear edges, and we can't wave our hands about a limiting argument,
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this is an overlap that persists as you take finer and finer subdivisions.
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And ultimately it's that overlap that accounts for the difference between
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our false answer with a pi squared from the true answer that has 4 pi.
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It reminds me of one of those rearrangement puzzles where you have a number of
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pieces and just by moving them around you can seemingly create area out of nowhere.
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For example, right now I've arranged all these pieces to form a triangle,
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except it's missing two units of area in the middle.
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Now I want you to focus on the vertices of that triangle, these white dots,
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those don't move, I'm not pulling any trickery with that,
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but I can rearrange all of the pieces back to how they originally were so that those
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two units of area in the middle seem to disappear,
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while the constituent parts remain the same, the triangle that they form remains
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the same, and yet two units of area seem to appear out of nowhere.
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If you've never seen this one before, by the way,
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I highly encourage you to pause and try to think it through,
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it's a very fun little puzzle.
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The answer starts to reveal itself if we carefully draw the edges of this triangle and
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zoom in close enough to see that our pieces don't actually fit inside the triangle,
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they bulge out ever so slightly, or at least arranged like this they bulge out ever so
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slightly.
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When we rearrange them and we zoom back in we can see that they dent
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inward ever so slightly, and that very subtle difference between the
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bulge out and the dent inward accounts for all of the difference in area.
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The slope of the edge of this blue triangle works out to be 5 divided by 2,
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whereas the slope of the edge of this red triangle works out to be 7 divided by 3.
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Those numbers are close enough to look similar as slope,
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but they allow for this denting inward and the bulging outward.
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You have to be wary of lines that are made to look straight when
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you haven't had explicit confirmation that they actually are straight.
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One quick added comment on the sphere, the fundamental issue here is that the geometry
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of a curved surface is fundamentally different from the geometry of flat space.
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The relevant search term here would be Gaussian curvature.
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You can't flatten things out from a sphere without losing geometric information.
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When you see limiting arguments that relate to little pieces on a sphere that
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somehow get flattened out and are reasoned through there,
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those only can work if the limiting pieces that you're talking about get smaller
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in both directions.
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It's only when you zoom in close to curved surface that it appears locally flat.
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The issue with our orange wedge argument is that our pieces never got
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exposed to that local flatness because they only got thin in one direction.
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They maintain the curvature in that other direction.
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Now on the topic of the subtlety of limiting arguments,
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let's turn back to our limit of jagged curves that approaches the smooth circular curve.
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As I said, the limiting curve really is a circle and the
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limiting value for the length of your approximations really is 8.
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Here, the basic issue is that there is no reason to expect that the limit of
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the lengths of the curves is the same as the length of the limits of the curves,
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and in fact this is a nice counter example to show why that's not the case.
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The real point of this example is not the fear that anyone is ever
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going to believe that it shows that pi is equal to 4,
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instead it shows why care is required in other cases where people apply
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limiting arguments.
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For example, this happens all throughout calculus.
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It is the heart of calculus, where say you want to know the area under a given curve.
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The way we typically think about it is to approximate that with a set of rectangles,
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because those are the things we know how to compute the areas of.
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You just take the base times height in each case.
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Now this is a very jagged approximation, but the thought, or I guess the hope,
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is that as you take a finer and finer subdivision into thinner and thinner rectangles,
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the sums of those areas approaches the thing we actually care about.
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If you want to make it rigorous, you have to be explicit about the error between
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these approximations and the true thing we care about, the area under this curve.
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For example, you might start your argument by saying that that
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error has to be strictly less than the area of these red rectangles.
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Essentially, the deviation between the curve and our
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approximating rectangles sits strictly inside that red region.
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And then what you would want to argue is that in this limiting process,
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the cumulative area of all of those red rectangles has to approach zero.
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Now as to the final example, our proof that all triangles are isosceles,
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let me show you what it looks like if I'm a little bit more careful about
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actually constructing the angle bisector rather than just eyeballing it.
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When I do that, the relevant intersection point actually sits outside of the triangle.
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And then from there, if I go through everything that we did in the original argument,
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drawing the relevant perpendicular lines, all of that,
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every triangle that I claimed was congruent really is congruent.
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All of those were genuinely true, and the corresponding lengths of
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those triangles that I claimed were the same really are the same.
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The one place where the proof breaks down is at the very end,
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when I said that the full side length AC was equal to AE plus EC.
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That was only true under the hidden assumption that that point E sat in between them.
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But in reality, for many triangles, that point would sit outside of those two.
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It's pretty subtle, isn't it?
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The point in all of this is that while visual intuition is great,
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and visual proofs often give you a nice way of elucidating what's going on with
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otherwise opaque rigor, visual arguments and snazzy diagrams will never obviate the
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need for critical thinking.
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In math, you cannot escape the need to look out for hidden assumptions and edge cases.
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Thank you.
