My Calc II prof would always do this. He'd start going through a proof like it was critical to the course and we'd all be furiously copying it off the board. Then he'd get to the 1=2 conclusion at the end and laugh at how none of us saw it coming. Then we'd spend time trying to see where the error was (usually a very well hidden "divide by 0" mistake) I'll always remember him. He made calculus actually fun and interesting. The world needs more educators like him

@pik33100

Жыл бұрын

Let a+b=c Then 4a+b=3a+c (added 3a) 4a+4b=3a+3b+c (added 3b) 4a+4b+3c=3a+3b+4c (added 3c) 4a+4b-4c=3a+3b-3c (now substracted 7c from both sides) 4(a+b-c)=3(a+b-c) 4=3

@user-tf2oh3cu7l

Жыл бұрын

@@pik33100 I understand now! Because a+b=c, a+b-c is actually 0, and because every number times 0 equals 0, so 4(a+b-c)=3(a+b-c) is valid. So it's 4.0 = 3.0, which is true. How is it 198?!?!?! Idk but this is a lot for me. Thanks and also, ⛽.

@jonathanhagerty9807

Жыл бұрын

@@user-tf2oh3cu7l exactly right, so you could divide a+b-c from each other because 0/0 is undefined. As a result, the proof given is correct until you try to say 4=3, which makes sense

@user-tf2oh3cu7l

Жыл бұрын

Tanks (i mean literal tanks and thanks)

@Noelciaaa

Жыл бұрын

Damn, I wish I had one like this. Would've been super engaging!

You know you’re a good math communicator when you can lie with proofs.

@MouseGoat

Жыл бұрын

yeah, guy could easily have convinced me with these false proofs. but knowing why they wrong gives a great insight into weird aspects of our world and how things can be folded up into small spaces

@Mutual_Information

Жыл бұрын

@@MouseGoat 100%, pointing out how proofs can be false highlights how precise these tools really are.

@bohanxu6125

Жыл бұрын

your statement lacks rigor. I accidentally lied with proofs before, yet I'm not a good math communicator :] or... your statement is true, I just don't know I'm a good math communicator yet~

@Mutual_Information

Жыл бұрын

@@bohanxu6125 yea it’s not intended to be precise. Maybe I should have said can “lie convincingly”.. but even with that, it’s not serious. He’s not actually lying.

@l.3ok

Жыл бұрын

Yeah Grant Sanderson is indeed an exceptional math communicator; the thing here is that all these false proofs were not invented by him, they've been very well known for a long while.

The pi=4 "method" can also be used to "prove " √ 2 = 2 by approximating a diagonal with finer and finer scale taxicab paths

@Loony-cz3xb

Жыл бұрын

hello. I agree but what do you think about my opinion: in pure mathematics this kind of proove is obviously wrong, because when you shrink and double the starecases to infinity you get 3,1415... (or √ 2 in the other example) BUT in our reality, in OUR universe with a minimum physical length (planck) you can't go to infinity, you cannot iterate to infinity; -> so the starecases have the planck length in horizontally and vertically direction; -> so the aproximation stops at this point -> and this means pi is still 4 (√ 2 = 2) in this universe...

@ThomasFackrell

Жыл бұрын

@@Loony-cz3xbthis is fascinating. However a problem is that for the 45-45-90 triangle, under this Planck taxicab metric the hypotenuse length would be 2, not ~root 2. Any practical doodling with a ruler shows that the Euclidean length cannot be 2 though. So how do we define the length of a curve or line that requires 2+ dimensions to be constructed? Root 2 is incommensurable (not rational) which is another way of saying it cannot be measured in one dimension. So does it exist in one dimension? Is root 2 actually on the number line? No one has found it, so I say no.

@Jayantea

Жыл бұрын

Math people are dumb and smart at the same time

@zern7617

Жыл бұрын

@@Loony-cz3xb square root of 2 cannot be 2. 2 * 2 = 4, simple maths can completely dislodge the entire point of pi = 4. lmfao

@Loony-cz3xb

Жыл бұрын

@Jaaaco i agree, there is no proove.. But...still you can''t build a perfect circle with inifinitive small triangles; in my opinion, when you reach with the short side of the triangel (the arc length) the planck length, nobody could say in which way the line goes between these two edges. You cannot double again the triangles to iterate the formula of PI. So the real path of the arc lengh is unnown. Of course this is only my opinion. :-) Have a nice day.

when he drew the triangle and split the base in "two" i was like "wait that's not even close to the middle"

@kyx5631

Жыл бұрын

Right? Even when eyeballing it, that was too sloppy to happen by accident.

@saurabhkumarsingh3986

Жыл бұрын

And that's exactly where the trick lied

@kipchickensout

Жыл бұрын

@@saurabhkumarsingh3986 yes

@rosepinkskyblue

Жыл бұрын

Angle bisector wasn’t right either, but I thought maybe I was wrong and maybe he did measure it 😭

@joshc5613

6 ай бұрын

I think the bisector for the side was fine but the angle bisector was kinda freehanded with extreme generosity

As a non-math person who does a lot of CGI work, I’m rather proud of myself for getting the first one right. That sort of warping is something we have to watch out for all the time in UV Unwrapping. Yay!

@StormBurnX

Жыл бұрын

As a 3D printing need that deals with a lot of low-poly work I knew from the thumbnail exactly what the problem was! Yay!

@johnson42069

Жыл бұрын

yeah the moment he unwrapped it i thought "wait those are triangles when they're supposed to be curved on the sides"

@kyrawr83

Жыл бұрын

I'm studying geography and I caught that too, unsurprisingly.

@tilenkos2065

Жыл бұрын

@@kyrawr83 CG artists 3D printers 🫱🏻‍🫲🏼 Geographers

@ajaiyp8679

Жыл бұрын

its strange how math connects all the professions and fields together.

This is the video mathematicians want to make after being told convergence proofs are "unnecessary details, it's obvious" for the 1000th time. Great work!

@artey6671

Жыл бұрын

Who are the people that tell mathematicians things like that?

@aguyontheinternet8436

Жыл бұрын

@@artey6671 engineers?

@Etern1tyOne

Жыл бұрын

@@artey6671 I will admit: some physicists 🙈 I try not to, but... sometimes it's hard to resist. Not with convergence, but with some other "formal" proofs. 🙈

@artey6671

Жыл бұрын

@@Etern1tyOne Well, some mathematicians try to avoid conflict. If they are ok with your proof, they might secretly be like "well, he tried".

@TomJones-tx7pb

Жыл бұрын

yup I used to hate physicists who did this. Then I got really good at functional analysis and realized that if you sat inside a good function space you can do this at will. I realized that the physicists were doing correct math, but they were probably oblivious as to why.

I am a seamstress, an knew from my experience taking 2D shapes to make 3D objects that the "triangles" of the sphere should have a curve. It was very exciting learning the maths behind my empirical knowledge, thank you so much!

@haveatyou1

5 ай бұрын

Are you the one that ghosted me on Tinder?

@pommeverte531

5 ай бұрын

​@@haveatyou1I don't use tinder, so no 😅

@haveatyou1

5 ай бұрын

@@pommeverte531 Why flirt then ghost me though 😒?

@pommeverte531

5 ай бұрын

​@@haveatyou1why write when you can't read tho?

@haveatyou1

5 ай бұрын

@@pommeverte531 I thought we got on pretty well.

This was the last video I watched before leaving my aunt's house. This was around 3 months ago, the last conversation I had with her was about my college classes and this math video I watched on KZread (being you). She passed away roughly a month later and every time I watch this video I get reminded of her. I know you'll never read this comment, but your channel does mean a lot to me and you've helped me learn and nurture my curiosity about mathematics. I don't have any sort of deep statement, I just love these videos and they remind me a lot about days gone by. Thanks for making these, keep up the good work.

@Phoenix-nh9kt

Жыл бұрын

Sorry for your loss man.

@JR-White

Жыл бұрын

Wishing you well. My condolences.

@Joe-mv8mq

Жыл бұрын

Thank you both.

@somepunkinthecomments471

Жыл бұрын

Sorry to hear about your loss. I hope you are doing well.

@3blue1brown

Жыл бұрын

I'm very sorry for your loss, and thank you for such kind words.

I got the third one. I'm usually not smart enough to figure out anything that this channel says is "tricky" by myself so I was really satisfied with that

@MasterHigure

Жыл бұрын

You should be. That is indeed a known and famously tricky false proof. P is actually outside the triangle (as long as the triangle isn't isosceles), as is exactly ONE of E and F. (Oh, right, I see now that the second half of the video is going through the errors.)

@bloom945

Жыл бұрын

Same!

@ekisacik

Жыл бұрын

I figured it out by going, "Okay, what does that point P look like in a real isoceles triangle?", before realizing that the perpendicular bisector of the bottom line and the angle bisector of the top angle would be *the same line*. Then, I knew something had to be up with where point P actually lied... and once I understood that it was always outside the triangle, then I finally understood where the proof was wrong.

@petrsmital7340

Жыл бұрын

I'm proud of you, bro

@squeakybunny2776

Жыл бұрын

Me too. Point p was the one thing that "came out of nowhere" and I didn't like how it was assumed to be inside the triangle. Especially because, is the triangle actually would be equilateral there ouldnt exist a point p because the perpendivual bisector and angle bisector lines would overlap. Really happy with myself😊 only thing is I didn't do the math to then conclude it was the last step where it all went down... Missed oppertunity for full points.. I'll take 2 of those gold stars then.. Still happy

One of my favorite ways to determine if a statement is plausible is to draw an extreme example. For the triangle proof, I drew a scalene triangle with one very short side. As soon as I tried to connect P, it was obvious it wasn't going to work because the intersection points existed outside of the original triangle. TL;DR I got three gold stars from Grant and I'm very proud of myself.

@bangscutter

Жыл бұрын

That is actually a good approach to test by setting extreme limits. Like the equations for special relativity in physics, should reduce to classical newtonian mechanics in the limit of speeds much smaller than c.

@Metal_Master_YT

Жыл бұрын

I did the _exact same thing_ wow, and I came to the same conclusion, its a good strat.

@XxZeldaxXXxLinkxX

Жыл бұрын

@@bangscutter Good ol' binomial approximation

@stevencraeynest7729

Жыл бұрын

Exactly! I did the same and it was obvious. Must admit, if I hadn't done that the fake proof does sound convincing

@waroftheworlds2008

Жыл бұрын

Loads of theories fail at 0, 1, or approaching infinity or approaching 0.

The fact that any circle drawn on a TV will have the same edge length as a square or rectangle which perfectly houses that pixel-based circle is honestly mind boggling. It makes perfect mathematical sense, but at the same time I just can’t wrap my head around it! Thank you for this enlightening piece of information - I’ll be sure to spread it to random people next time I’m drunk!

I've never been great at math. I barely passed geometry and algebra, and I've never even touched a calculus textbook. But something about your math videos make me feel like I understand it all, it brings back a sort of childlike curiosity where I don't have to feel bad about making mistakes or being wrong, I just get to sit here and learn things. It's oddly soothing for me and I love your videos :)

I had a SAT prep book way back in the 80s that said whenever the test says "Not drawn to scale", you should immediately redraw it to scale. That rule of thumb has served me well my entire life. Unfortunately, most people will throw out everything they know in the face of a confident deception.

@andrewharrison8436

Жыл бұрын

Yes, nice

@altrocks

Жыл бұрын

Like the infamous telephone wire question. A quick sketch shows the answer while the math just frustrates.

@coloradowestaerialarts1316

Жыл бұрын

I agree. Very good video.

@a_commenter

Жыл бұрын

@@altrocks What question is that?

@reidflemingworldstoughestm1394

Жыл бұрын

@@altrocks Yes and no. It depends on what answer you are asked to provide. If the problem states the wire hangs in a parabolic arc and asks for values of that arc, then the correct answer is independent of the fact that real wires don't hang in parabolic arcs. If on the other hand the solution asked for is to use the real world model despite the parabolic example given, then it's appropriated to go all pedantic on they ass.

A note on the sphere proof explanation: We can also prove that those "triangles" don't have flat sides by looking at their angles. The dividing lines on the sphere are all perpendicular to the "equator" of the sphere, so each of those triangles would have two right angles.

@codegenesis80085

Жыл бұрын

Lol yeah I noticed the same thx for explaining!

@i_love_vi_

Жыл бұрын

Oh yeah

@nenmaster5218

Жыл бұрын

@@i_love_vi_ Science-Fans are Sometimes Dumb. Its ok, i can say that, cause im a Science-Fan myself. But jokes aside: Do you keep yourself updated on local and global Issues, friend? Are you aware of the the possible Food-Shortages as well as the Danger depicted in the GOP-Videos of "Some More News"? Said Channel covers Crops, Hate, War, aaaand mooore, so it’s really valueable-much to keep you updated.

@cyanclouds2127

Жыл бұрын

but isnt that not a problem as for higher no of divisions the central angle would approach zero? just like the circle area proof

@reubenmanzo2054

Жыл бұрын

@@cyanclouds2127 Then your "triangle" is just a single line, which doesn't have area at all.

There’s something so frustrating about knowing you’re right, but not being able to prove it… good job, magic numbers man, that last example really got me.

It's interesting how the Pi listeners in the animation react when something is noteworthy or thought provoking. I noticed myself understanding better when I saw the pi people understanding together. Very kewl

The fun thing about the second "proof", you don't even need to do it with curves. Apply it to a diagonal of a square, and you can "prove" that the square root of 2 equals 2.

@karunk7050

Жыл бұрын

Could you please elaborate?

@FaranAiki

Жыл бұрын

@@karunk7050 Just watch this video: LWPOlZBXtD8. (put it on the KZread link)

@buttonasas

Жыл бұрын

@@karunk7050 Imagine a diamond instead of a circle. And you get perimeter 4×2=4×sqrt(2), which is false.

@slightlybluish3587

Жыл бұрын

@@karunk7050 Take a unit square, ABCD, and draw the diagonal AC. The perimeter is 4. Now, fold the two edges of the square B and D inwards toward the center of the diagonal. The perimeter is still 4. You have now created 4 new "corners." Fold those in as well. The perimeter is still 4. Keep on folding the corners in, infinitely. The perimeter approaches two times the length of the diagonal (because the square is "getting closer" on both sides). The perimeter is still 4, so therefore the diagonal of the circle is 4/2=2. idk why it's wrong

@stephaneduhamel7706

Жыл бұрын

@@karunk7050 You first draw an isosceles right triangle of sidelength 1, by definition its hypothenuse is of length sqrt(2). Then you can do the same manipulation as with the circle, by floding in the corner with the right angle to make it go closer to the hypothenuse. This way, you can construct a sequences of jagged lines of length 2, and the limit of this sequence is the hypothenuse, which would make you think sqrt(2)=2.

This channel is absolutely fantastic. Absolutely LOVE you guys! You explain everything very well, and I especially appreciate explanations for the simple terms that I, being a non-native english speaker, don't know from school! Absolutely fantastic stuff!

Beautiful. A set of correct axioms plus a hidden incorrect assumption leads to an incorrect conclusion. Even if all the pieces fit perfectly together. This just goes further to show that something that is plausible isnt necessarily true.

That’s funny. I knew the sphere example was wrong, I know you can’t flatten sections of a sphere without the pieces looking deformed. But you flattened it out and I said “huh, I must be wrong”. Blind trust I guess.

@mrpedrobraga

Жыл бұрын

I was like that, too, except it was when he said "this is innacurate, but as we take the limit" and there I was with "uh... i guess..."

@ed_iz_ed

Жыл бұрын

he does mention they arent flat, but the same goes for the circle

@freshrockpapa-e7799

Жыл бұрын

@@ed_iz_ed a circle is flat man

@wiggles7976

Жыл бұрын

That's true, a sphere is famously not flat, because there's no isometry from the surface of a sphere to the plane. Though a sphere is a manifold, so locally, it approaches being flat. But, if you are taking a bunch of pieces and laying them next to each other, even though they get smaller and flatter, the error due to the non-flatness could accumulate. That would be my guess as to why the proof is wrong, before watching the rest.

@Felipe-sw8wp

Жыл бұрын

@@ed_iz_ed but the circle IS flat (this is the flat-circle theory and I`m a flatcircler) No, jokes aside, the circle is flat as a surface, it isn't flat as a curve.

A good way to catch yourself on the triangle proof at the end: you’re arguing that a triangle is isosceles, and the argument constructs a perpendicular bisector and an angle bisector and makes use of their intersection. But if you do that argument with an isosceles triangle, those two lines are the same and they don’t have just one intersection. The argument makes itself incoherent. That doesn’t fully explain how the argument is wrong, but it should be a red flag, and that kind of self-consistency check will often help to flag this kind of faulty assumption in practice.

@gregoryfenn1462

Жыл бұрын

Good point! A retrospective sanity check on something you think you've proven goes as long way :D

@fares8005

Жыл бұрын

Good point, but it's also worth noting that many euclidean geometry theorems have a "generic proof" that only works for non-degenerate cases and those special cases (e.g isoceles triangle) are verified separately.

@thomasjunker5415

Жыл бұрын

This was the same conclusion I arrived at as well. As P isn’t well defined if the triangle is in fact isosceles, the entire proof goes out the window by contradiction

@mishaerementchouk

Жыл бұрын

The proof starts to look suspicious from the very beginning, when it suggests that P is inside the triangle. It could be inside or outside. The suggestion is turned into an assumption at the end, when it is concluded that from AF + FB = AE + EC follows AB = AC. This is the only mistake in the proof, since the conclusion holds only if both E and F are inside the triangle, which doesn’t follow from anything. Such transitions from «geometry» to «algebra» are always a weak point because some effort is required to establish signs geometrically: one needs to keep track of the mutual arrangements of the points.

@mishaerementchouk

Жыл бұрын

Oh, it turns out that this is explained at the end of the video.

I actually got that last one, but even then it took me 2 minutes or so, despite me spending essentially all of my life until the age of 19 at math competitions. What I'm saying is: If you construct a false proof cleverly enough, you can fool anyone. Including yourself, sometimes. And not just in mathematics.

@alinpopescu4147

Жыл бұрын

the intersection of the angle bisector and perpendicular bisector is on the circumcircle (since they both pass through the midpoint of arc bc) , which is not even close on the drawing.

@penguin_reader_yt9510

6 ай бұрын

Ima try it on my dad later. Hes an engineer. It should be interesting. Ill report how it goes. Maybe.

@ca1498

5 ай бұрын

@@alinpopescu4147 Nice - I was wondering why P seems to always have to be on the outside. Now I know.

@ca1498

5 ай бұрын

It is surprising how many known mathematicians have made known mistakes. Including Euler. Probably not Gauss, though.

@sleepntsheep1169

4 ай бұрын

​@@penguin_reader_yt9510so, how did it go?

This is why maths is the most interesting subject

@rakinhasan1360

20 күн бұрын

Agreed 💯

as a former geometry teacher, i was very excited when I saw "SSA" used as a reason. if i ever teach again, I will most certainly used this video to engage my students!

@hopechr

Жыл бұрын

the ASS congruence.

@anandixitin

Жыл бұрын

I think that can be said to be RHS congruence.

@doggo6517

Жыл бұрын

I was a sailor on the HMS Congruence

@mnek742

Жыл бұрын

Actually the SSA "congruence" is a false congruence unless a right angle is involved

@balrighty3523

Жыл бұрын

@Murtaza Nek Yep, I was thinking the same thing. The second S in the ASS "congruence" can potentially reflect and create two valid triangles that aren't the same. The only times this is guaranteed to not be the case is when the A in the ASS is obtuse (because the reflection ends up being outside the triangle entirely) or if, as you say, there's a right angle involved (meaning the second S can still reflect, it just doesn't become anything new).

These fake proofs have been boggling in the back my mind for months(specially the p=4 proof) and I'm *very* thankful to finally get porper answers and finally let my mind rest in peace. I'm *also* very thankful that you gave tips on being careful when looking at proofs. You're truely the best Math teacher I've ever seen!

@jansenart0

Жыл бұрын

To me it was more about calculus caring about the area under the curve instead of the length of the curve.

@martinepstein9826

Жыл бұрын

@@jansenart0 Calculus does care about the lengths of curves. Arclength is defined as an integral.

@tychobrahe6379

Жыл бұрын

I have also had that pi=4 proof on my mind lately, and I’ve seen some other people attempt to explain what it gets wrong, but to very unsatisfactory results (like saying the curve is always going to have jagged edges, ignoring that the limit will indeed be the circle). I’m glad 3b1b can come and show how it’s done.

@derekdreery

Жыл бұрын

Another interesting angle for this problem is that functions can converge in different ways. One type of convergence is where every point on the curve converges to the corresponding point on the limit curve (where 'corresponding' means 'the same t for a parametrisation'). Another is where you integrate the area under a curve to get a number, and then those numbers converge to the area under the limit function. Neither of these implies the other and it's fun to think of counterexamples. You don't even need curves, you can find counterexamples for functions that are straight lines joined together.

@chlli

Жыл бұрын

porper

1:29 when i saw this it seems like some part of "unraveling" the slices into flat shapes is an approximation for the actual curvature, and the actual difference between the 2 areas boils down to approximating 4 as pi (or vice versa) in some capacity

The graph overlay really helps to visualize the triangle illusion one, is great

In 5:57, where 3B1B is drawing, the text in the book says next: "The point of rigour is not to destroy all intuition; instead it should be used to destroy bad intuition while clarifying and elevating good intuition. It is only with a combination of both rigorous formalism and good intuition that one can tackle complex mathematical problems" -Terence Tao "When you yourself are responsible for some new application in mathematics in your chosen field, then your reputation, possibly millions of dollars and long delays in the work, and possibly even human lives, may depend on the results you predict. It is then the need for mathematical rigor will become painfully obvious to you." -Richard Hamming

@kaszagryczana2260

Жыл бұрын

You can read

@realperson9951

Жыл бұрын

@@kaszagryczana2260 i cant

@PacoCotero1221

Жыл бұрын

@@kaszagryczana2260 no joke, this may be helpful to blind people

@PacoCotero1221

Жыл бұрын

@ThatOneRookie 🚫🧢

@aureliontroll2341

Жыл бұрын

@@PacoCotero1221no joke blind people probably will never like that video at all because is a video with VISUALS proofs in KZread .

when you started drawing the triangle I immediately noticed "hey that isn't the halfway point" and wondered why you didn't just redo the drawing. Tried drawing it myself and got a point outside the triangle

@benoit-pierredemaine3824

Жыл бұрын

Ah yes, AF=FB clearly does not seem right ...

@redstonulo

Жыл бұрын

yea the middle point was about 1/22 off (1/22 of the distance between B and C)

@totoshampoin

Жыл бұрын

Also expalins why he didn't use manim this time

@redstonulo

Жыл бұрын

@@totoshampoin yea pretty much

@julianw1010

Жыл бұрын

@@totoshampoin Because he cheated?

Hey, just wanted to let you know that my school now has this in the official syllabus for discrete math. Congrats and hope you become even more widespread in popularity!!

I'm honestly just impressed at the graphics/animations in this video, they either took a lot of work or were done using an algorithm

When you showed the first "proof", my immediate thought was: you can't just convert from spherical geometry into euclidean geometry like that, otherwise there would be no need for the Mercator projection

@squirrelcarla

Жыл бұрын

i know next to nothing about math but im a 3d artist so when i saw him do that i immediately went "boy, IF ONLY it was that easy" 🤣

@feynstein1004

Жыл бұрын

I had the same thought. However, that does make me wonder. Why does it work for the 2D proof though? Like, you can take the curved circumference of a circle and bend it into a straight line without problems.

@SioxerNikita

Жыл бұрын

@@feynstein1004 The thing is, you don't do that. He explained that, that is the error, but with smaller and smaller slices it approaches a straight line, and if we take an infitesimally small slice, it is essentially a straight line.

@SgtSupaman

Жыл бұрын

@@feynstein1004 , for the 2-dimensional circle, slicing smaller and smaller vertically is all you need to approach the limit where a curve and a straight line become the same thing, because only one side of those slices is even curved in the first place, and that is what you are making smaller with each successive division. For a 3-dimensional sphere, however, the vertical slices are still just shortening the one side, but now the other sides also have curvature. That's why the video points out that you need to slice both vertically and horizontally to approach something that would be equivalent to flat.

@feynstein1004

Жыл бұрын

@@SgtSupaman Ah okay. That helps a lot. Thanks 😀

The triangle trick reminds me of the infinite chocolate trick. If you cut a chocolate bar on the diagonal of one of the rows, cut off the longest column from the top piece and switch the top pieces, you seemingly get an extra piece, but in reality, the pieces along the diagonal cut get smaller and the missing chocolate in that row accounts for the extra piece.

@atanvardecunambiel8917

Жыл бұрын

To quote the gr8 jan Misali, math prank’d!!!!!!!!

@abhinavdp7376

9 ай бұрын

Isn't it that pizza trick

This seemed a lot of fun also coming up with ways where proofs could be false is a good learning experience. Thank you Grant 🙏🙏

I’ve worked out the exact drawing error in example 3: The point D is too far towards C on the line segment DC, while the angle bisector for angle A leans too much towards AB and away from AC. This is how the lines intersect within the Triangle, and why point P is situated there instead of outside as it’s supposed to be.

Not only is P outside the triangle, it's actually on the circumscribed circle of the triangle. In Germany this fact is called "Südpolsatz" (South pole theorem).

@michamiskiewicz4036

Жыл бұрын

Nice observation! Moreover, the two projections E, F are colinear with the middlepoint D (in particular, one of them has to be inside and the other outside the triangle). That's a special case of Simson's line theorem.

@jay_sensz

Жыл бұрын

Also, the intersection point P doesn't exist at all for an isosceles triangle because the angle bisector and the opposing line segment bisector coincide. But you can pick any point P on the coinciding bisectors that is within the triangle and make the construction valid. In all other cases, P, as well as either E or F lie outside the boundary of the triangle, making the construction invalid.

@caspermadlener4191

Жыл бұрын

The Dutch mathematical olympiad came up with "BOM-lemma", but "zuidpoolstelling" sounds way better. I also just heared that New Zealand also doesn't have a name for this. Thank you, random German Math person!

@ericzhu6620

Жыл бұрын

I think it can be proved using Incenter/Excenter Lemma

@lc1777

Жыл бұрын

@@ericzhu6620 yes

Oh man I stared at the TV for a solid few minutes of pain on that 3rd one before realizing that the intersection point wasn't inside the triangle. Unfortunately I had convinced myself that it lived at point D, so I can only take partial credit for finding the error and immediately replacing it with my own lol. awesome video

@MasterHigure

Жыл бұрын

The point being inside or outside is not really relevant. The relevant part is that exactly one of E and F is outside the triangle, which makes one leg a sum and the other a difference, instead of both being a sum (which allows the two legs to be unequal in length). Of course, the fact that one of them is outside is easier to hide if P is drawn inside.

@hens0w

Жыл бұрын

@@MasterHigure you can show that they amount to the same thing

@MasterHigure

Жыл бұрын

@@hens0w My point is, they could've drawn P outside the triangle and still E and F both inside (or both outside), and the faulty conclusion would still be that the triangle is isosceles. Whether P is drawn inside or outside the triangle is rather irrelevant to the actual crux of this mistaken proof. If you want to use this proof to fool someone, where you draw P is of rather minimal importance. It is the placements of E and F that are crucial.

@hens0w

Жыл бұрын

@@MasterHigure can you give an example, I don't think you can I think if p is inside then so are E and F

@megamaser

Жыл бұрын

Initially I made the same guess but then I realized the same conclusions would follow so I thought for a bit longer and I realized an isosceles triangle from the same angle would require the left segment to shrink or the right segment to stretch, which made it clear that the midpoint would have to fall on the right side of the current midpoint.

thanks to these videos i've been improving my problem solving skills. i was so proud when i figured out what was wrong with the isoselces triangle proof

Found the flaw. Obviously it was because he assumed that it was a triangle. It was actually just a shape with three angles and three sides.

Excellent choice of which pages to use for drawing the diagram for the third "proof"; the quotes are nicely apropos indeed.

@ahsan4306

Жыл бұрын

Wow. I missed that little detail. Or maybe it was just coincidental. Nah! Can't be. GS is a perfectionist.

I'm really glad you took the time to talk about this, since your channel spends most of its time presenting visually intuitive proofs of mathematical things. It's important to acknowledge that while visual intuition is really useful for aiding understanding, it is NOT a substitute for carefully checking every aspect. The devil is in the details often enough, even in the beautifully idealized world of mathematics. I got the triangle one, but probably only because I've seen a similar fake proof before, which purported to prove a 90 degree angle was equal to a 100 degree angle. It really is very sneaky. Actually the most confounding one for me was the sphere one. I couldn't quite put my finger on what the issue was. Thanks for the clear explanation there. I was very glad to see how you explained the pi=4 one. I've seen a lot of bad explanations out there, which make the error seem complicated, when in fact, it's exactly what you said: the limit of the property is not the same thing as the property of the limit. It's probably the most broadly applicable cautionary tale among these, though "make sure your diagrams are at least topologically correct" is pretty broadly applicable too.

@purplegill10

Жыл бұрын

Ironically, it was the other two that got me, but the sphere one was easy for me purely because I had a map projection/sewing phase for most of my late-teen life. The second the lines turned straight sent off alarm bells in my head since, both in sewing and map projection, straight lines on a curved object almost never translate to straight lines when dealing with 2d things. It took many confused pillows and plushies before I finally got around to that in my head.

@rocketsandmore6505

Жыл бұрын

speech op

Similar to the 2nd problem of square approximating the circle, I encountered a similar problem. While working on images and pixels, i realised that for any flat right angle triangle whose non-hypotenuse sides are aligned with the sides of Flat rectangular images, the triangle has its hypotenuse always = sum of sides of it other two sides. It can be seen by similar method of folding the corner of right angle. When i start to think in terms of Calculus and limits, my conclusion was that the thickness for the hypotenuse curves is always more than the sides. (In calculus as we never mention anything about thickness, so it must be that all curves have limit to zero thickness but equal thickness). Interesting thing i realised was that the increase in thickness of hypotenuse is of the same factor as the (sum of length of other two side/ length of hypotenuse ). Now, after watching this video's 2nd explanation, i can understand that the problem lies in the error calculation. Again interestingly, the error will be in the same ratio ( sum of length of side/ length of hypotenuse).

As an Engineer and lifelong lover of math (but working with Finance&Accounting), I am astonished with your skills of explaining so many complex concepts on a simpler way - which makes it so much fun to watch (back in school I never thought I’d be watching a math video in YT just for fun, but you changed that and I’m grateful 😊) Congrats for all you do and wish you all success!

@bleepblop

3 ай бұрын

I love math as well and am wondering about how you like finance/accounting/engineering in reference to math. I am close to college and wondering about all of these fields

Another thing to note about the isosceles triangle proof: In an actual isosceles triangle, the perpendicular bisector of the base *is* the angle bisector for the angle opposite the base. (This is because of SSS congruency between the two triangles you get when you draw this line.) When we assume the triangle ABC is isosceles, point P, the intersection between the perpendicular bisector of the base and the vertex angle bisector, is not well defined because they are the *same* line. They have infinite intersection points. Therefore, to speak of the distance between a point on the triangle and point P is meaningless if the triangle is isosceles (which is what the proof tries to prove)

@arcguardian

Жыл бұрын

That's why I concluded his first step is where the lie went wrong. There's no sense in me over thinking it, a yellow flag in math is usually a red one.

@XyntXII

Жыл бұрын

​@@arcguardian I think you are right, that this is the point. If as he said, the hidden assumption, that would be needed for the proof, was that the point P would have to be inside the triangle, then this would prove, that all triangles where P is inside would be equilateral triangles.

@sihaskumarasingha8148

Жыл бұрын

*breaking bad noises intensify*

@linkhyrule5800

Жыл бұрын

@@XyntXII Indeed. It therefore follows that for all scalene triangles (triangles where no two sides are equal), P must be outside the triangle. As @ Tomáš Slavík discovered while trying to follow along. There's no contradiction there.

@ca1498

5 ай бұрын

The fact that for an isosceles triangle we get one and the same line in no way breaks the proof; it only makes it incomplete. It just means that with those you are encountering a special case, which you have to acknowledge and possibly have to do something different. In this case, you can take any point P on the line inside or outside of the triangle and repeat the rest of the arguments. You will find that the triangle is... drumroll... isosceles! So, the problem is not that it breaks down for a triangle that you already know is isosceles; the problem is that if the triangle is not isosceles (I mean actually AB != AC), then it turns out to be isosceles, because we incorrectly assume that E and F are on the same side of the the BC line.

Thank you for stating that triangle fallacy, and for explaining its flaw. It's one of the few lasting contributions to mathematics by an obscure nineteenth-century mathematician called Charles Dodgson, better known to us as Lewis Carroll.

@vigilantcosmicpenguin8721

Жыл бұрын

"Beware the Jabberwock, my son! The fallacies that seem consistent!"

@MichaelRothwell1

Жыл бұрын

Nice! The origin of the alias Lewis Carroll is rather fun. From Charles Lutwidge Dodgson, Charles translated to Latin is Carolus and Lutwidge translated to Latin is Ludovicus, so (after swapping) we get Ludovicus Carolus, anglicised to Lewis Carroll. I think the Dodo character comes from Dodgson.

@rosiefay7283

Жыл бұрын

@@MichaelRothwell1 You are correct. Mr. Do-Do-Dodgson had a stammer.

At first I was about to make a joke about the first two proofs were very well animated while the third was drawn on paper. Turns out, that was actually required for it to fool us. That said, I'm pretty proud that I got the third one. The longer the drawing went on, the more I was convinced that those lines just wouldn't intercept like that. Fantastic video, love it.

When he said: "they get increasingly subtle", i took that to mean I should get the first one; maybe the second, and that the third one was out of reach. Missed the first, missed the second, felt terrible about myself, and then got the third. Nice.

@SgtSupaman

Жыл бұрын

It just goes to show how considering things more or less difficult is often just one person's perspective, because we can all have different thought processes and approaches (though, of course, a consensus can be reached). For instance, I agree with you that the third one was easier to see through than the second one, but I agree with the video that the first one was the easiest of the three to find the issue.

@wiener_process

Жыл бұрын

I've been through too much courses on mathematical analysis to get fooled by the first two, but the third one gave me more trouble. Until I tried drawing it myself, and then it became pretty clear.

@TlalocTemporal

Жыл бұрын

And still more conversely, I completely missed the first one, saw the second one as soon as the first fold happened, but only got halfway on the third one (saw that p was outside the triangle, but didn't realize what that meant).

I remember, when I was a "child" (14 years old), I asked my math teacher why the length of the diagonal of a square isn't twice the length of one side, with the same reasoning as you did with the circle, bending the corner over and over until, at the limit, there's just a straight line which is the diagonal. This really haunted me for days, I couldn't figure out why that was wrong (I obviously knew about Pythagoras' theorem). He tried to explain to me the way you did, how it is not always possible to assume that f(lim x_n) = lim f(x_n), but he couldn't really give me a formal answer to **why** this doesn't work. By that I mean, yes, we can't say the length of the limit is the limit of the lengths because our example shows it's not true, but it doesn't explain why it's not true. It's only years later, when I learnt about function sequences and their convergences, and also about what's a curve and what's the length of a curve (at least of class C^1 piecewise), that I could fully understand what's going wrong. For the ones who don't know, the length of a curve is given by the integral of it's derivative (in modulus, or norm if you consider it over a normed vector space). If you have a sequence of functions (f_n) that converges uniformly to a limit f, even if they all are of class C^1, the sequence of their derivatives doesn't not necessarily converges to f'. This means the sequence of their lengths doesn't necessarily converges to the length of f. In other words, the length function isn't continuous. Knowing that, it is also quite easy to create a sequence of curves that converges to a circle (or a segment), but with their sequence of length diverging towards infinity (think of the sequence of function (1/n sin(n^2 x)) )

@mystic839

Жыл бұрын

i struggled with this well into adulthood, and still find my heuristics leading me towards these continuous functions that prove things incorrectly. i definitely have to remember that sometimes, they aren't.

@ragnkja

Жыл бұрын

The zig-zag approximation works for area, but not for distance.

@saschabaer3327

Жыл бұрын

There's a generalization of the definition of length for not necessarily differentiable curves that works in a pretty general setting (any continuous function from a compact interval into a metric space I believe): Let I = [a,b] be the interval on which the curve c is defined, and let d be the metric on the target space. Say a subdivision of I is a finite collection of points x0 = a Pick a sequence of subdivisions of I with the property that the n-th subdivision satisfies d(c(xi), c(x{i+1})) One can prove that if c is piecewise differentiable, this definition coincides with the one you gave, but it gives a well-defined answer for some curves that aren't differentiable. It's also incidentally very similar to the procedure that doesn't work, as you can imagine it as approximating the curve with straight line segments - the big differences being that (a) both endpoints of each segment must be on the curve and (b) the segments must get shorter uniformly.

@_4y4m3_ch4n_

Жыл бұрын

so baiscally.... in the fake proof, the curve is continuous, but the approximation isn't, right?

@MK73DS

Жыл бұрын

@@_4y4m3_ch4n_ Not really, what is not continuous is the length function for a curve. This means that the limit of the lengths of a sequence of curves isn't necessarily equal to the length of the limit of this sequence of curves. And this is exactly what's happening there, the limit of the lengths of the curves is 8, but the length of the limit of these curves is 2pi (because the limit of these curves is a circle). In a more extreme case, if you give me any length you want, as large as you want, I can find a curve arbitrarily close to a circle that has a length greater than the one you chose. This means I can find a sequence of curves, which converges to the circle, but their lengths diverges to infinity! (you can do that by oscillating very quickly around the circle) If the length of a curve was a continuous function, this proof would have worked. For example, the area delimited by a curve is continuous, so this means that the area delimited by these modified squares converges to exactly pi (if there was a way to easily calculate the area delimited by each of these curves, this would give us a way to approximate pi). But the length is not a continuous function, and therefore it is incorrect to say that because the curves converges to a circle, their lengths converges to the length of the circle too. Proving that one function is continuous isn't always easy, but proving that one function isn't continuous is often done by finding a sequence that doesn't satisfy the "limit of f(sequence) = f(limit of sequence)". For example, how can I prove the step function isn't continuous (the step function is the function defined by f(x) = 0 if x = 0), I just look at the sequence (-1/n). This sequence converges to 0, but the sequence (f(-1/n)) is the sequence (0,0,0,...) and does not converge to 1 = f(0). Therefore this function is not continuous. What I wanted to do in my first comment, is to give another proof why the length of a curve is not continuous. Proofs by counter-example are great, they are quick ways to prove the non-continuity, but they don't always show why, what was the issue, what's special about that function that makes it not continuous. I wanted (at least, I wanted to try) to give an explanation of why the length function is not continuous, not just a proof that it isn't. Sometimes, proofs and understanding aren't the same things.

"You eat my food?" "No." "Prove it" "Alr let me show you"

I'm happy because although I couldn't prove rigorously, I still had an intuition about that Alpha angle and about the sides of the sphere slices not being a flat line, but a curve of sorts (I do a bunch of UV Mapping, so it helps)!

I'm an engineer and while getting my degree I had my fair share of math. This is the perfect elucidation of why I could have never been a mathematician even tho I can work with math with no problem. I tried really hard to spot the mistakes and, except in the first case, I wasn't able to spot the error in the other two before they were explained. Amazing video!

@pannekook2000

Жыл бұрын

euclidation 💀💀💀

@ionymous6733

Жыл бұрын

Though to any inspiring mathematicians, just because Francesco Vultaggio believes he "could have never been a mathematician" because he didn't get it before the explanations, that doesn't prove you can't, ironically perhaps.

@JM-us3fr

Жыл бұрын

I’m a mathematics grad student, and I only spotted the mistake in the second one. Proofs are very unnatural for our minds, which is why rigor is required. My friends often ask me “How are you so naturally good at proofs?” The reality is no one is “naturally” good at proofs. I had to practice a lot before gaining important intuitions.

@stefanperko

Жыл бұрын

That doesn't mean at all that you couldn't be a good mathematician. The issue with these "proofs" is that they aren't proofs in the first place. By that I mean the logic is so subtle and opaque that it's really hard to find faults with them. They are mostly based on intuition But our initial intuition is often times just wrong. That's why in the end you need to write down the argument in symbols to really convince yourself it works. And then it will usually become very obvious that it simply doesn't or at least that there is a huge logical leap somewhere.

@francescovultaggio2540

Жыл бұрын

I admit I wrote the comment in a way that could be misconstrued quite easily. Not being able to spot the mistakes does not mean that you are not cut to be a mathematician, just like not being able to spot a bug in a code at first reading does not mean that you are not cut for CS. Growing up I noticed that what really makes the difference is not your natural raw talent or intuition but your dedication and passion. The true indication that one can be a mathematician is not necessarily if you found the mistake but how hard you tried, did it stayed in your mind afterwards, will you read more about this kind of puzzles/problems? Honestly I love the channel and the clarity in which he explains difficult concepts but I won't do any of that, mainly because math to me is a mean to an end and not an end in itself. That's the difference btw a mathematician and an engineer/CS/etc. I hope I didn't discouraged the next field medal recipient from getting into math with my previous comment, lol

I'd like to remind you of your series on Differential Equations and request for its completion. It is such a great pleasure to be able to comment on a video which was just released a few minutes ago.

@iankelley9302

Жыл бұрын

I second this!

@iOSMinecraft120

Жыл бұрын

agreed, but perfection requires patience!

Frigging amazing video, esp since visual proof is something that we all enjoy from this channel

When I first learnt about using integration to calculate areas in high school, I tried to calculate the surface area of a sphere and got pi^2 r^2 too. I checked and checked my calculations over and over again and still cannot find any fault in it. It turns out calculating the surface area requires path integral, which is like in this video, actually summing over the curve vs summing over the jagged lines. That day I understood the importance of rigorous limits.

There's a sort of extension to the isosceles triangle proof. Once someone correctly mentions that P is outside the triangle, you draw another inaccurate diagram where the points E and F are also both outside of the triangle. (i.e. On extensions of the sides that they correspond to.) Then you still get the conclusion that the triangle is isosceles, but this time by subtracting at the end instead of adding. In reality, one of E and F lies on a side of the triangle strictly between the two vertices, and the other one lies on an extension of the side that it is on.

When reading The Elements, I often find multiple propositions proving what looks like the same thing, but different cases where some point is on the outside instead of the inside. This shows that Euclid really needed to include those different cases as separate proofs.

I’ve watched this video about 3 times cus it’s just so clear and concise. :)

Th preface in that notebook is probably the truest and most inspirational of all the quotes in it.

For the sphere "triangles", an intuitive way of phrasing the problem is that, proportionally, the overlap stays the same as we approach the limit. It may shrink as we go finer, but so does the main area of our triangles (both their widths are scaled by the same "interval" of approximation while their height remains the same). Thus, the overlap may shrink out of view, but it doesn't shrink away from our problem :P

@JamesChurchill

Жыл бұрын

I struggled with this for many years until I realized it was a failing in my understanding of how a limit works. Any approximation is the true value plus some error. For a limit to work (ie reveal the true value) the successive approximation has to *reduce* the error. But in these examples the error is maintained, so the limit still ends up with the original wrong value.

@black_platypus

Жыл бұрын

@@JamesChurchill Exactly. 3b1b phrases it pretty much the same when he says the error must approach zero in the limit. I just felt like looking at this example, realizing that the width of both the triangles and their overlap are scaled by the same factor, this might serve as an intuitive way of coming to that realization

@WanderTheNomad

Жыл бұрын

I feel like you could probably apply that to life somehow. The errors disappear from view over time, but are still there.

@black_platypus

Жыл бұрын

@@WanderTheNomad whoa... 🤯😁

@Oturan20

Жыл бұрын

@@WanderTheNomad and come back to haunt you in your nightmares!😈

As many have noted, in the third 'proof', P actually lies outside the triangle. However, that's not enough; you can rework it so that it still seems valid. The actual sticking point is that, of points E and F, one of them lies on a side of the triangle while the other one only lies on an extension of a side of the triangle. For example, F is actually on the segment AB, while E is beyond C, relative to A. (I must confess I saw this explained in a book some time ago. It's a really good one.)

@Kerostasis

Жыл бұрын

That explains why I was having trouble. I quickly realized that P could sometimes be outside the triangle, but I couldn't figure out how that actually changed the solution.

@vincentjiang6358

Жыл бұрын

What book?

@tomkerruish2982

Жыл бұрын

@@vincentjiang6358 I'm sorry, it's been at least 40 years. (Also, I was referring to the puzzle as being a really good one. I wish I could remember the book.)

@ahsan4306

Жыл бұрын

@@vincentjiang6358 Probably Mathematics and the Imagination. At least that's where I saw it first

@TheEulerID

Жыл бұрын

It is simply not true that we have to go through the whole process to invalidate a proof. Finding a single flawed step is sufficient. It's then up to the original mathematician to submit a corrected version. The goes for any mathematical proof; if there is a single flawed step, then the proof is invalid and the onus is always on the one making the claim or, "onus probandi", as it is in Latin. In this case we know the "proof" must be flawed from the outset, as we can start with a counter-example. However, it's not always so clear. For example, Andrew Wiles' original published proof of Fermat's last theorem was found to be flawed, even though the conjecture turned out to be true. Andrew Wiles was able to publish a valid proof of Fermat's last theorem a couple of years later when that flaw was corrected.

The first thing I noticed wrong with the triangle proof was trying to find the intersection point (P) between the perpendicular bisector of BC and the bisector of ∠A. If the triangle was really isosceles, they would be the same line, so APD would be collinear and P could be anywhere along AD, which is the triangle's axis of symmetry. I don't think this is the key to the proof being wrong, since the intersection point actually does exist for non-isosceles triangles (or even for isosceles triangles if you pick one of the equal sides instead of the odd one out), but I just thought it was interesting that you're using a point in the proof that would not exist if what you were trying to prove was true.

@sandhuekam14

Жыл бұрын

So true even i think the same and also the perpendicular bisector is not correct its a bit off centre the point must be somewhere else

@MK-13337

3 ай бұрын

Yes, the point P must exist, although for all non isosceles triangles it is outside of it, and the proof falls apart.

Thanks for sharing your knowledge with all of us

Excellent video as always, you managed to verbalise/visualise the whole "limiting process needs to reduce the error" part so succinctly when I've failed to convince a few students of mine about that for a long time now - definitely showing this video to them. Cheers!

@maxthexpfarmer3957

Жыл бұрын

That's why I think finding the area in a way analogous to Darboux integration as Archimedes did is a better way to do it.

@squorsh

Жыл бұрын

You can't forget the delta from your epsilon delta proofs

@mvmlego1212

Жыл бұрын

I wish that he spent an extra minute or so deriving the limit of the error, because otherwise he's basically asking us to trust us that it doesn't approach 0, but it was a great video overall.

I found all the flaws quite quickly, and it only took an entire 3 years of undergraduate mathematics to prepare me for this moment. Great vid!

Why this actually shows a lot of very interesting nuances that they didn't teach me in univesity. the thing about proving that error of the area under the curve goes to zero is something I've never considered.

I was so close to nailing the mistake in the isosceles triangle proof -- my first thought when he intersected the angle bisector and the perpendicular bisector was "how do you know those two lines intersect? They could have been parallel, for all we know!" And I thought about how, in an actual isosceles triangle, the angle bisector of the vertex angle ks also the perpendicular bisector of the base. So once again, I figured "that point can't exist." But I was still thrilled to see the subtlety of the fact that the point COULD exist but it's location in relation to the triangle is everything. Love it!

@anjnagupta5985

2 ай бұрын

SSA doesn't exist

@AKWLMath

Ай бұрын

​@@anjnagupta5985It does for right triangles.

If I'm honest, the second proof was for me much harer than the third one. The first one is quite easy (curvature) while the third one can be disproven by drawing the "proof" correctly. Second one really requires you to understand how limits work, instead, and even with your explanation, I still don't understand the issue that much.

@SimonBuchanNz

Жыл бұрын

Basically the problem is that it assumes that the perimeter is the same between the two curves (the true circle and the limit of folding the square), but that's not a guarantee unless you have the third part of the proof showing that the limit of the error between them is 0. For the classic example of the area under a curve, you can prove that with the upper bound as well as the lower bound forming a rectangle with area greater than the error, and showing that the limit of the area of the error bounding rectangle is 0, and thus the limit of the error must also be 0 (handwaving detail about discontinuities and turning points)

@rdaysky

Жыл бұрын

You could try to apply the same faulty reasoning to claim that the length of a straight line from (0, 0) to (1, 1) is 2. The problem is that the error doesn’t go to zero as you add more segments. For each individual right triangle the error does go to zero but only because the triangles become smaller; the error would need to go to zero faster than 1/n because you’ll be adding n of those error terms, where n is the number of the triangles. It’s similar to why 1+½+⅓+¼+⅕+… doesn’t add up to a finite number, the addends do get closer and closer to zero but not fast enough.

@n0ame1u1

Жыл бұрын

Calculating the arc length of a function (which is how we would find the lengths of these curves) is not so simple. It involves an infinite sum (which in some cases can be simplified into an integral). And the limit of an infinite sum is not in general the same as the infinite sum of a limit. There are some conditions that need to be true for that to be the case, which must not be true here.

@Michael-sq5ju

Жыл бұрын

I think the difference in difficulties for most people is that almost all pure math students will have taken a real analysis course, where arguments like the second one are discusssed in great detail, whereas fewer pure math students know about curvature (real analysis is often a first year course, whereas courses covering geometry for pure math majors are typically later) and fewer pure math students will remember Euclidean geometry from their high school days.

@jeffsamuelson7221

Жыл бұрын

It is actually even possible to find a sequence of curves which hug the circle whose lengths blow up to infinity. What is not possible is that a sequence of approximating curves exists whose lengths converge to a smaller value than the true length (in technical terms, arc length is lower semicontinuous)

9:06 just wanted to point out, the step that turned out to be flawed was obfuscated by that favorite word among math professors, "clearly". three gold stars to you Grant if that was intentional!

You got me pretty good in the third "proof" there. For some reason I was convinced that assuming FP = EP was the error, although this actually is correct.

A lot of similar fallacies are to be found in the classic 'Riddles in Mathematics' by Eugene P. Northrop, first published in 1960 - and still a very good and enjoyable book.

If one wants to know more rigorously what happens in the second "proof", it essentially means that the uniform limit of curves, that is, the limit in the C^0 topology, does not necessarily preserve lengths of curves. This is indicative of the fact that "length" is not an innate property of curves that are only continuous, as many continuous curves do not necessarily have a well defined length (fractal curves for example). You would need what is defined as a rectifiable curve in order to talk about it's length. If one considers, on the other hand, curves that are piecewise smooth and they have a well defined limit in the C^1 topology, that is, their derivatives (ergo their velocities) also have a limit, then length is in fact preserved. The jagged curves approximating the circle do not have their velocities converging uniformly to the "velocity" (or the tangent velocities) of a circular path, hence it does not converge in the C^1 topology. Even more thoroughly, one could imagine the most general space on which a C^1 topology makes sense, and one would have to start dealing with Sobolev spaces from functional analysis. I find this remarkable in that it shows that even though our intuition fails, mathematics finds a way to rescue a significant and meaninguful portion of it on a maybe more subtle or advanced context, but nethertheless a correct and rigorous one.

@WarDaft

Жыл бұрын

I've seen this one before and to be frank it's always baffled me that it baffles anyone. Behaviour at the limit is not behaviour approaching the limit unless you explicitly force it to be so. Consider lim n->2 of n. At no point is n equal to 2, but the limit is. Why should any given derived property be conserved if you haven't proven that it is? Proving what is conserved is why we use them in the first place and should be the first thing anyone learns about them!

@hybmnzz2658

Жыл бұрын

@@WarDaft yeah I always think about the sequence 1,1/2,1/3,... which shows positive things don't have to converge to a positive quantity. But somehow, if you show a simple example to someone they won't be so convinced.

@londonl.5892

Жыл бұрын

@@hybmnzz2658 The 1, 1/2, 1/3 one makes more sense because they do seem to approach something non-positive because each one is getting smaller. Here, each perimeter in the sequence is 8. And it seems really bizarre that a sequence of 8s doesn't have a limit of 8. Like, if you do lim(len(curve)), that's 8s all the way down and at the limit. But if you do len(lim(curve)), for some reason, each curve individually has a len of 8, but the limit doesn't? That's very, very odd. I still don't understand. The explanation above helps a little bit, but it's still not quite clear.

@cr10001

Жыл бұрын

@@londonl.5892 Not a mathematician! But it seems to me that, with the second example, as you zoom in and add more and smaller right-angle 'kinks' to the rectilinear line, the ratio of [length of line] to [length of circle perimeter] doesn't get any better. Therefore the approximation isn't tending towards the correct value. (Whereas, comparing it with integration, the finer you subdivide it, the closer the area under the rectangles gets to the curve.) Similarly in the first example - in the case of the sphere the 'triangles' all bulge in the middle, and as you narrow them down the degree of 'bulginess' stays the same. Whereas with the circle, the triangles all have straight sides, with a tiny curve at the top, and as the triangles reduce in width that degree of 'bulginess' reduces (as a proportion of the triangles' area).

@amarbapat8599

Жыл бұрын

@@cr10001 congratulations Your intuition has led you to discover the very primitive form of Convergence criterias. I am not being sarcastic! Not being from a mathematical background but still having that kind of intuition is crazy amazing.

This video was honestly genius. You communicated everything so well, and it really helped my understanding in the basis of proofs. Thanks!

That was so much fun. I got that last one, but couldn't get the circle's perimeter. Shows us how important understanding definition of basics in maths.

As you stated. PI = 4, non-Euclidean geometry is a broad mathematical discipline that deals with numerous types of geometry. One of the primary topics covered in this is a type of geometry known as "hyperbolic geometry," which is effectively the "opposite" of the normal geometry you are familiar with. This implies that when lines, angles, and forms are represented in hyperbolic space, they take on different qualities than when they are represented in conventional geometry. This leads to a distinct conception of geometry in which the same points might be given dramatically different features and behaviors. This assertion holds true only within the context of a non-Euclidean space known as the Poincare Disk of One Dimension. This is referred to as a Hyperbolic Space. In various respects, hyperbolic space varies from ordinary Euclidean space. One important distinction is that parallel lines in Euclidean space will ultimately converge in hyperbolic space. The Poincare Disk of One Dimension is unique among hyperbolic spaces in that it is not endless, as you may return to your origin by traversing around in a "circle." The term Poincare refers to the French mathematician Henri Poincare, who popularized the concept of hyperbolic geometry in his 1882 work, Le science pure, Ou la mathématique non-Euclidian. This book is widely regarded as one of the most important mathematical publications in history, with much of its content still in use today. So, how is PI equal to 4? Euclidean space is a space where parallel lines remain parallel. Hyperbolic space is a space where parallel lines converge. Because of this unique property, hyperbolic space contains different geometric definitions in places where Euclidean space contains a "circle" or a "sphere". What this means is that Euclidean measurements of area and volume are not applicable to hyperbolic space, and thus they will make no sense due to the unique behavior of the lines. This fact is what means that hyperbolic space can be represented by the equation pi = 4, but Euclidean space cannot.

This reminds me of a ‘problem’ we have when teaching calculus: How to explain why you can calculate a volume by integrating the area of the cross-section but you can’t calculate a surface area by integrating the perimeter of the cross-section.

@shre6619

Жыл бұрын

I never thought about this before, can u explain more

@matheusalmeidadamata

Жыл бұрын

I would also like to understand why it is not possible to do this. Thanks!

@johnsmith-gq5jw

Жыл бұрын

@@matheusalmeidadamata ​ @Shre You can see the problem if you try to find the length of a the diagonal of a triangle by integrating the width of cross sections. For example, the length from (0,0) to (1,1) would be \int_0^1 1dx = 1. It might also make sense to compare it do what is should be. You might recall that ds=\sqrt{1+[f'(x)]^2} dx.

@ladyravendale1

Жыл бұрын

If I remember correctly from my calc 3 course, this happens because of missing information. Integrating in multiple dimensions is (I'm fairly sure, at least) generizable to a series of nested integrals of the value 1. When integrating just a perimeter of a shape with the value 1 you are given no information about the shape inside the perimeter that you are trying to get the surface area of. If I remember correctly, you need to integrate sqrt(1+ fx^2 + fy^2) or ||ru x rv|| depending on parametrization, which gives you the information in the form of partial derivatives to actually know what's happening inside the perimeter.

@Victor-sw4ne

Жыл бұрын

Consider a sphere, for simplicity, and a cross-section which does not pass through it’s center. While the volume of a cylinder whose base is this cross-section is a good approximation for the volume of a thin ‘slice’ of the sphere around this section, the lateral area of the cylinder is not a good approximation for the surface area of the slice, since the surface of the sphere meets the cross-section plane at an angle. That’s what creates this problem. Basically, you would have to consider this ‘angle’ due to the change in the perimeter, and, in the case of the sphere (or actually any surface of revolution), think of a sum of areas of cone trunks rather than cylinders. In the general case, this leads to you having to considering an integral involving partial derivatives, which carry information about the direction of the tangent planes.

Curiously, the one that got me was the second one, because since I knew they were fake visual proofs I was very careful with the assumptions of shapes and sizes. But the second one also preys on our hidden assumption that 2D limits are about converging areas, when in this case it's about lengths, and the length of the curve doesn't approach the length of the circle even if the area between them goes to zero. Very cool

@slightlybluish3587

Жыл бұрын

How can that be? I’m a bit lost, because if the area between them approaches 0 then you have the same line

@andrewnazario2253

Жыл бұрын

im confused also by this :/ can you explain?

@slightlybluish3587

Жыл бұрын

​@@andrewnazario2253 i've discovered that it can be true. Imagine you have a straight line going from point A to point B. Now imagine another line that starts from point A, goes midway to B, then up 10 units (arbitrary), and then down 10 units before heading towards B. There is no area because it didn't have any horizontal movement, but you end up with a tiny spike that is longer than the original straight line. I find it a little complicated to apply it to this problem, but oh well :P

@Herio7

Жыл бұрын

@@slightlybluish3587 I think its better to visualize 2D things with 2D things. I mean your example isn't wrong but operating on 0 area is a bit unintuitive. It would be easier to visualize and compare simple and similar shapes say square 10x10 (area = 10, circumference=40) and long pole like rectangle 1x100, area is the same but circumference is 202.

@slightlybluish3587

Жыл бұрын

@@Herio7 but in this specific problem the part that is troubling is the fact that the area *between* these two different lines (the perimeter of the circle and the zig-zagged perimeter) approaches zero, even as their perimeter is different; rephrased, the shapes are getting more alike and approaching the same area, while still having different perimeters. In your example the shapes are obviously different aren't they? I mean it's obvious that they have the same area but different perimeter, but in this problem what's confusing is how the two different shapes seemingly get closer to being identical while still having different perimeters but the same area.

Very enjoyable video. Especially the second part (pi = 4). Playing with infinity (and limits and the oddities they bring) again.

I was screaming throughout this entire video. It was awesome. Thank you.

The third proof sounded fishy when I saw the intersection of the perpendicular bisector and the angle bisector and thought "Where should that intersection point really be?" The answer I thought was that on a true Isosceles triangle, that intersection lies directly on the unequal side. Then after seeing the explanation, it clicked further that the moment one side elongates or shortens bit to make a scalene, that intersection point begins to lie outside the triangle.

@diabl2master

Жыл бұрын

In fact, for an isosceles triangle, the perpendicular bisector and the angle bisector are the same line. So they don't have a single point of intersection.

@Theimtheimtheim

Жыл бұрын

Inrerestingly, these two lines always intersect on the circumference of the triangle, thus if they are not equal, P will always lie outside of the triangle.

Oh man, he gave the answers to all the easy questions but left out discussion of the most challenging and intriguing curiosities @16:55. I hope you do please cover these in the future because I have ALWAYS been curious about these questions ever since I learnt calculus. And I've never found a calculus books that has discussed these very questions either, ever!

Wow wow wow, I was completely lost as to why those proofs were wrong until you laid it out. This is such a good video, thank you!

Thank you very much for making such full blown videos years ago!

Such an excellent video! The clarity in your explanations as well as the production quality never cease to amaze me! Bravo!

I think I'll use the triangle "proof" as a collaborative exercise for my discussion in geometry on Monday, it ties in nicely with our exercises in thinking about the different ways to present diagrams and the importance of construction.

Magnificent, this is why in math and science we have to check if the steps we are taking are valid in that mathematical frame work and we should give a second thought on the claim by edutech that learning should be visual

This is a nice video, but I noticed a subtle mistake at 17:21. You are proving that triangle BPD is congruent to triangle CPD, but there is written "CPE" instead

I finished my master's degree in math a few years ago. During my undergraduate this channel was an immense source of inspiration when I was finding topics tough and continues to be a wonderful reminder of what my life was like during that time. Thank you.

This is a really important topic! As a science animator, I always do my best to depict everything accurately and honestly, and it's great to see this in action amongst the community. Much love 💛

@everythingisalllies2141

Жыл бұрын

you must have a problem being honest when trying to make the lies of Einstein's Relativity look convincing.

this vid was soooo good that it gave me chills in some parts

World is so wired where you spend money on education institutions where you should learn new thing but here i am after clearing the exams barely now learning from KZread. As a game dev its very necessary to have a good knowledge about maths. Also its interesting and fun here. Great job.

This video elucidates why mathematical rigour is necessary for proofs so well. In high school, I loved geometry and for almost every problem I encountered, I would think of some visual interpretation and try to solve it that way. It was simply more fun! But then in college math, all the subtleties and nuances of even the simplest of proofs became so important. The ignorant me used to think it unnecessary to state obvious lemmas at every corner and eventually, I almost lost the love I once had for mathematics. But these examples help me appreciate the need for such cautious methods and rekindle my passion for maths. Thank you 3b1b! ❤

That video would have been so useful to me a few weeks ago! But hey, at least it gave me the opportunity to think by myself instead of referring to your videos at the slightest obstacle. But thanks to you now I know why my approximation of a sphere based on cylindres was giving me π²R² and not something else! I really like how your videos make you understand the topic but also the quirks that come with it so as to leave the viewer prepared for anything it has to throw at it. In a nutshell, great video as always!

@pavelkalugin4537

Жыл бұрын

I made that mistake too sometime ago :) I calculated ∫ 2π√(1 - x²)dx for x ∈ [-1; 1] and got the same π² answer. It was only after I had looked at the extreme case of cylinder being close to the upper sphere that I realized the error wasn't limited there. I guess, as always, special intuition is needed here.

3Blue1Brown, I love your videos! But something really impressive is your animations and I'm really curious to know how you do it. What software do you use?

My geometry prof went through the third example in the first class as a cautionary tale not to rely too much on pictures when constructing proofs (which was especially relevant when we covered hyperbolic geometry) I had actually watched this video before but unfortunately in the moment I couldn't remember what the trick was

In your proof, you gave it away with the "I made no assumption" lines. You actually did assume that the point P, the point of intersection of Perp. Bisector and Angular Bisector of angle A was inside of the triangle. It could very well be outside the triangle too.

@micuhh

Жыл бұрын

More accurately, for this reason you cannot assume that |AC| = |AE| + |EC| as E may not even be on line segment AC

@pantoffelkrieger8418

Жыл бұрын

It is also very easy to show that exactly one of E and F lie on the side and one is outside unless they are equal to B and C (apart from using the contradiction that ABC would be isosceles): It is well known that P is on the circumcircle, specifically, the midpoint of arc BC. Then D,E,F are the Simson line of P with respect to ABC, so they are collinear

@ThePharphis

Жыл бұрын

I thought the giveaway was that he clearly didn't measure the angle before bisecting it

@person8064

Жыл бұрын

The perpendicular bisector of BC doesnt look like its bisecting.

@kerstinhoffmann2343

Жыл бұрын

this sort of thing is precisely why I hate doing euclidean proofs

*17:51** “The point in all of this is that while visual intuition is great and visual proofs often give you a nice way of elucidating what’s going on with otherwise opaque rigor, visual arguments and snazzy diagrams will never obviate the need for critical thinking. In Math you cannot escape the need to look out for hidden assumptions and edge cases.”* *-3Blue1Brown*

The last one seems to be a sleight of hand based on the "this drawing is not 100% accurate, but you get the idea" trick. The key element is forcing point P to fall inside the triangle, by making BD longer than DC (while stating that they're congruent) and by calling AP the BAC angle bisector (although it's not). If we draw the two bisector lines accurately, we'll find that they intersect below segment BC (outside the ABC triangle), dismantling the entire "proof". Also, the "SSA" and "SAA" sound a bit weird; but I suppose you used these shortcuts to save time, since it was quite easy to prove that those triangles were congruent the right way (if you were willing to make the video a little longer).

Another way to approach the surface area/circumference is to understand the definition of the limit. In the case of the circumference, this is taking the 'right hand limit'. You will then need to find a another sequence from the inside of the circle whose perimeter is also 8 (which you won't). This is precisely why stuff like squeeze theorem is quite difficult to use in reality, contrary to what introductory calculus tells you.

7:54 I think I got it. I decided to attempt following along with the construction in desmos geometry tool, and quickly found that point P always lies outside the triangle, except in the case of an isosceles triangle, where the lines are equal and there is no single way to define P. I'll have to watch the rest of the proof to know *exactly* where this throws a wrench in things, though

@Xeridanus

Жыл бұрын

That's how I got there, about 19 secs later than you :P. I used proof by absurdum. If AB and AC are the same, P cannot by defined so the whole thing falls apart. Notably, you can't state that BP and CP are the same length. Or FP and EP for that matter.

I'm quite proud of myself for actually figuring out that isoceles triangle proof. And I had a few things to add about it: - If the triangle truly was isoceles, then the angle bisector of A and the perpendicular bisector of B and C are actually THE SAME LINE, meaning their intersection point (P) is undefined. So if anything, the existence of point P is enough to prove that the triangle is not isoceles (or rather, that those two particular sides are not equal to each other). - AF really does equal AE, and FB really does equal EC. However, since either E or F must be outside the triangle, that means that one of the sides must equal AE+EC and the other must equal AE-EC. So essentially what we've done with this "proof" is, we found a new way to construct a line equalling the average of AB and AC, plus another line equalling half the difference between AB and AC. (Of course, if that was your original goal, there are far easier ways to do this - but I think it's interesting nonetheless.)