Math in the Simpsons: Homer's theorem
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- Опубликовано: 3 сен 2015
- After putting on some glasses he found in a toilet Homer feels very smart and declares: "The sum of the square root of any two sides of an iscosceles triangle is equal to the square root of the remaining side." Well, sounds like Pythagoras theorem but it's not. The Mathologer sets out to track down this mystery theorem to its lair and dissects the hell out of it.
Enjoy :)
I was more worried about him putting on those toilet glasses without washing them first.
I like to think they have him go to the sink to put them on as a tease. "Oh good, he's at least going to rinse them." Then you remember this is the man who once was craving beer so much he licked the dirt under the bleachers at a baseball stadium.
Cartoon germs don't cause infections unless the plot calls for it.
Ross Plavsic wow, you look really similar to him
I love your videos.
I won't even touch it.
Woahhhh so the simpsons was just referencing the wizard of oz. that’s a deep joke
shottysteve and the wizard of oz was just the result of the writers
Wow only 2 likes on a verified comment
make a new video already
make a new video already
make a new video already
Scarecrow doesn't get a brain, he just get a diploma.I think that the reason.
XD
He said he got a brain :) 2:55
@@just_is He has a brain, he had one all along, but he didn't get a new brain :-)
Scarecrow is just like every other person with a college diploma :)
even back then they knew how useless college was.
@@aidenaune7008 college in America is a class gate to limit upward mobility.
Oh man, I got that Homer's line was an homage to Wizard of Oz, and I could get that Homer got the Pythagorean theorem wrong, but I never noticed that the original line in Wizard of Oz was wrong!
But Scarecrow is a Doctor of Thinkology!
I know this is a year old...
One of main themes in Oz is that magic can't solve your problems. The wizard actually doesn't do anything in the world (allegory for false promises of politicians) and the work is left to the outsider Dorothy.
Scarecrow thinks he's been fixed, but he was already "fixed," he just didn't know it.
@fangere
That's beautiful.
Thank you.
🥰
@@MattMcIrvin so scarecrow works in the liberal arts?
I don't remember noticing that either! The lines go by so quickly it's hard to notice exactly which words they are using or have time to think about it!
It should be called the placebo theorem as all the instances we see it are the individuals thinking they're smarter.
Well one of them was practicing lines for the scarecrow, so technically it's right there.
Aaron Reamer
That's not what a placebo is.
@@ImDemonAlchemist The definition of a placebo is "A medicine or procedure prescribed for the psychological benefit to the patient rather than for any physiological effect."
You could say that homer receiving the glasses or the scarecrow receiving his "brain" making them think they are smarter when in fact they are not as a placebo effect.
This guy is a perfect example of the Dunning Kruger effect...
@@awulfy9052 Exactly. Its the Dunning Kruger effect, not a placebo effect.
Are you just pointing to a white wall and memorizing what to say?
He's holding a remote so I imagine that when he looks towards the camera, he's looking at a screen with a sort of slideshow on it
There's a projector that projects the slides onto the wall. The clean slides are superimposed in post.
@@itzmistz no, it's not. it's a green screen, he has a monitor to the side where he looks at a teleprompter or notes or a slideshow, and the edit is placed over later. no projector
@@PhazedAU You wouldn't be able to see shadow on the 'green screen'. Also look at 1:37, the text is clearly on his hand from the projector
To be honest, it could be a combination of both. I do see a bit of green
This line is also referenced in an episode of Hey Arnold, where Arnold’s grandpa goes back to elementary school to get his grade school diploma. Funny thing is Dan Castellaneta (who voices Homer) also voiced Arnold’s grandpa, whom recites this line to the principal in order to secure his diploma.
Minkowski metric in spacetime satisfies a reverse triangle inequality
Can I bear your children?
@@calvinsawyer1961 Yeah no prob LOL
@@obi6822 I'm a dude so I'd have to father ur children actually which would defeat the purpose
@@calvinsawyer1961 I assumed so. I am a dude too btw hahaha
@@obi6822 if I was a woman I'd bear your children. How bout that?
This triangles could exist in a cilinder
Vicio ONE MORE TIME!!!! Better the inside of a sphere
These* :v
I suppose if you take a cylinder at least 4 but less than 6 units in circumference and wrap the big side around and join it with the two shorter sides. I'd hesitate to call it a triangle though. It would be more like a letter C with the gap joined by a little v at right angles.
What I thought
@@misael8200 you're not going to talk about the cylinder?
I share this video with my students. Veeery goooood!!
Estava pensando exatamente isso. Quando eu estava no ensino fundamental/médio não conseguia visualizar as equações dessa forma, era tudo muito abstrato, depois desse vídeo consegui compreender algumas coisas da época da escola.
Spanish spanish Spanish spanish, whatever the dude above me said.
@@ADrunkCrayfish It´s portuguese, dude.
@@ADrunkCrayfish that ain’t Spanish
@@irioncampello6055 Well, I certainly em read it in Spanish, and am portugueses speaking! 😂
Hold on: If b=0, then we have a line. Then, solve for a using the first equation, and you get (a)^(1/2) = - a^(1/2), so a=0. Thus, you are left with a point. That's the joke! They have a point! :)
This is THE answer.
Sorry, I don't speak Egyptian. Can you translate?
LOOOL
Hold on, I don't think it'll work like that. We got b=0 for the second equation, so we can't use that in the first equation. These are not a set of equations, rather a matter of either/or . Also yeah ik this is meant as a joke lol :-)
Maths are interesting to begin with but immediately becomes ten times more enjoyable when explained by someone with a German accent.
He is not German bitchface
@@user-sj2vg8hb5q Austrian? Swiss?
Right here Right now yes, he is German. If you don’t think so, just google him “Burkard Polster”
@@user-sj2vg8hb5q wth he is are u sutpid
David^2 - S^2 = Cohen^2 gives us a hint: the "D" from David stands for Donut, the S stands for Sign and the C stands for Colossal donut. When homer points at the colossal donut, we can see all of these 3 points (donut, sign, colosal donut) in one frame. If we connect these 3 points we get a triangle where a is the height of the sign including the colosal donut. You can also measure the angle of homers arms (alpha): 10, and the credits give us the number 24m as the length of b. We can now calculate the length of the hypothenose c: 24/cos(10) which is 24.3. Now we can calculate a: sin(10)*24.8 which is about ~4m. This means that the man holding the colosal donut plus the colosal donut is 4 meters high. They are about the same size so we can divide by 2 to get the size of the colosal donut: 2 meters!!!
SinOfficial this is like the kind of comment i sometimes make but this is way better! Good job at figuring that out!!
I accept this as fact
Why can’t I be smart like this. DOH!!
if you had used tau instead of pi in your calculation, you wouldn't have had to divide by 2 at the end.
@@prezadent1 homygod
Actually, Homer's mistake was.....
…he didn't wash the glasses before putting them on his face.
Copied :(
Pink eye time
They were already washed, just not with water from a preferred source
remember, he's got a crayon stuck in his brain.
The simple fact that this guy so sincerely loves both math and the Simpsons makes me like him immensely.
8:13 one of the co-producer's name is "David² + S² = Cohen²"
well spotted!
Nice!
Haha I made it harder than it was and I thought it was the right triangle made to scale the small donut to the colossal donut XD Good job!
it was shown at 8:39 anyway
That actually says '2+' on each power, which is actually probably better read as a contradiction of Fermat's last theorem, and if I remember my Simpsons correctly, is not the last such contradiction in the episode :) check the equations in the background when Homer is in the 3rd dimension...
What about a triangle on the surface of... a doughnut?
+HerraTohtori Well, with more complex surfaces you first have to make up your mind what exactly you mean by a triangle. I've left a few comments earlier on in which I talk about this. Maybe have a look :)
+Mathologer A closed curve made of 3 geodesics. Yes it can be done on a torus.
I was thinking that would make sense considering Homer's favorite junk food! As for the math to prove it, I'll leave that to folks with more time and math training than me. If true, maybe Wizard of Oz screenplay writers (or Baum himself, if those exact words are found in the book) had donuts on the mind and/or knew something about tori.
it is a torus
They already did it!
Season 10, Episode 22 "They saved Lisa's brain"
Stephen Hawking: "Your theory of a donut-shaped universe is intriguing, Homer. I may have to steal it"
(Dun dun duuuunn)
Awesome job providing the clips, ALL of them, including the Scarecrow.
He even tested the Scarecrow Theorem in non-Euclidean geometry! I didn’t see that coming.
Any Lorenz geometry model usually works without triangle inequalities. Not sure now, but maybe homer's theorem holds true for minkowski space? Where triangle inequality is reversed?
Good idea :)
Human Effigy no Minkowski's, but it works on a sphere in Minkowski space
i didn't get a word of this, but Mathologer replied means this wasn't bs
so liked the comment
I just want to say that I really enjoy this channel. It's difficult to find interesting videos about cool bits of mathematics, and so far I have found 2 channels that deliver this: numberphile and mathologer. Keep doing what you're doing!
I totaly agree!!!
You're missing 3blue1brown, especially if you're already somewhat advanced in your maths education. But even if you're not, there's plenty of cool stuff on that channel too, definetely check it out.
(i know the comment is old af, but if you haven't seen it since then, GO FUCKING DO IT :p)
@@FelipeV3444 actually i was going to comment this... Lol😂
A man in the lightmode void talks about the mistakes Homer Simpson makes while looking at an omnipresent context and visual providing object that reacts to both his words and the content it showed previously.
i'd say it's a well planned powerpoint presentation
On a sphere, it is kind of possible to have a+b
In spherical geometry opposite points on the sphere are considered equivalent: this is because it changes the 5th postulate to say that parallel lines do not exist: lines can only be maximum circles (circles made by a place cutting the center of the sphere). All lines are perpendicular and cross at ONE point: so they consider the opposite points as one single point. So some of those points on your bigger side are part of the original triangle and the others are excedent like a side prolonged even ending on the same points. The distance in Riemannian geometry is given by the SMALLER maximum circle because a metric can not be a multivalued function and the metric by definition must obey the triangular inequality (or the hell will go loose and several contradictions arise because the metric should capture the intuitive notion of distance as being additive, and being symmetrical (in a loose sense that I am too lazy to explain). So your construction is not a triangle is a triangle with line segments added (my explanation is a little convoluted because I am lazy, maybe later I explain better).
@@agranero6 no they aren't
It works on a sphere when you ignore the any sides part. You can create a triangle were two of the sides equal a quarter of the circumference of the sphere and the other one spans around the equator. The angels between the equator line and the other two are always 90°, therefor the triangle is iscosceles. The third side can now vary from 0 to the circumference of the sphere. So if you subtract the other two sides (which equal half of the circumference) you still got the possibility to have half of the circumference left. Since in this example a equals b 2*(square root of a * square root b) equals 4*a. Since a equals a quarter of the circumference you get the solution when c spans the whole circumference. It doesn't look like a triangle but technically it is a triangle on a sphere I guess.
There are examples in which an instance of that formula, sqrt(s)=sqrt(x)+sqrt(y) may be found.
The triangle inequality is reversed in Minkowski space, so that's a candidate.
Secondly, it might be possible to find instances of that on some surfaces, such as a variant of the pseudosphere or some other surface of revolution of some cusp-containing curve.
Finally, something similar to it can be found in particular Lp spaces. For example, a space with a norm |s|^p = |x|^p + |y|^p will have something akin to the required formula for, say, p=1/2
What I find very intriguing about the last option is that circles, when drawn on a euclidean plane, will look like Lamé curves (with the power parameter being 1/2).
In short it can be done in spaces with a quasi-norm.
Definitely the best answer so far :) (Minkowski space has been suggested before)
I don't understand but this sound really brilliant. What kind of math this is?
Goldmos1
It's geometry and vector spaces.
Late to the party, I know. But my point is that sqrt(a) has TWO values.
sqrt(b) + sqrt(a) = sqrt(a) might not work, but sqrt(b) - sqrt(a) = sqrt(a) could easily be true, as well as -sqrt(b) + sqrt(a) = sqrt(a)
@@Goldmos1 topology I believe, I'm taking my first course in it now (MATH 525). I'm in my final year as an undergraduate and the stuff in the post seemed like stuff I could probably start to grasp. And I'm in North America, you can learn any mathenatics you want. You just have to be passionate and eyeballs deep in student loans! (The later may be optional if your really gifted or driven, but scholarships are few and self study is difficult)
On a sphere you can get it so a+b
If I remember correctly if you know the length of two sides of any triangle (a and b) the third side (c) has to be:
a-b < c
OMG Crystal math lmao
First comment on a post from 2016 in 2020
Second
@@protat0 no one cares
Every adhd medication is similar to meth
That part of Wizard of Oz always (well at least after middle school) pissed me off .
+DrRawley I think the point of it was as an in-joke: the Wizard never gave nothing to the Tin Man that he didn't already have, and all.
Qermaq I know :( That part pissed me off too. It's all a lie.
+DrRawley But WE know it is. That makes us richer. :)
The wizard was a dick.
+DrRawley Seen Wicked?
you can make a triangle where a+b
Ive never seen your channel but i found this very intriguing! Keep up the good work! 👍😁
8:12 found pythagorus in the credits
David^2 S^2 = Cohen^2
Subscribed! relating math to the Simpsons/futurama is a great motivator to get me through my homework! Thank you
it's been a long time since I used any high level of math. mostly basic stuff, Pythagorean theorem always comes in handy, and geometry in general. I do grow increasingly fascinated with Eratosthenes. This guy was simply amazing. Kind of sad, put in all those endless hours of head splitting work, worry, study, panic, study more, obsess, and in the end I have to periodically give myself math test so I don't forget all of it. everything today is charts, computers, and more charts. I remember i started my job and could figure everything with mobil calculator, pencil, and paper. Co-workers were jealous I believe and said why figure it out like that it's in the tables. One professor I had said - I feel sorry for you if technology ever crashes. At The time I didn't care The exams were so damn long and hard that without a calculator I would have had a nervous breakdown trying to crunch it all before I ran out of time. Now I understand though. The most important stuff you will need in life is college algebra and geometry maybe some trig but probably not. However when you have that knowledge it feels good. In a job interview I got asked a math problem and immediately pointed out the flaw in the question and offered a math solution to solve it. The other mathlete in the room laughed and of course no job for me. However, it felt damn good.
The theorem could work if the triangle was placed in a spherical cube where it’s centroid is at the vertex of the spherical cube plane.
Love these. My favourite bit of Simpsons math was when Homer had to count himself to be sure he was just one man.
Frame by frame from 8:14, you quickly get 3 and 4 dots on the donuts, 5 teeths in Homer’s mouth... That’s the first pythagorean triple!
+The Einhaender I’m afraid that’s it... 8:20 He said "it’s a tough one" and "there is a *hint* in the credit". Then at 8:38, they give the credit hint. I can’t believe the solution is this obvious. If it was all, they would say, the *solution* is in the credit, or something a bit more allusive I guess... David S Cohen is the math guy he must have done something clever in the sequence, not just adding a few squares in the credit... ;-)
My comment was totally desperate, I know it can’t be about the dots on the donuts. I searched for triangles that could have some obvious ratios, I couldn’t find any right triangle! Or maybe some circle with a crossed diameter, no chance... I’m afraid I’ll just be disappointed in the end. In ... Anyways, the show is great.
+boumbh The funny thing is that DAVID^24+S^24=COHEN^24 is not possible according to Fermat's last theorem
+Aishwarye Chauhan Actually, it just says that if it is true, then DAVID, S and COHEN cannot all be positive integers.
Fennec Besixdouze Oh, there is a corollary or something, right? I was thinking on Fermat's original proposition, and I forgot about generalizations.
leonardo21101996 exactly. I missed the whole been integer part haha
if you take an hyperbolic plan build on a cone with an angle enough close to 2π, I think you can make a triangle that will satisfy the condition of a+b
I like that you’re talking to the camera guy, its fun having you two bounce math off each other instead of just one guy talking into the void
This is such a great channel.
I don't know about math, but in physics, if you use a spacetime graph, the hypotenuse is the shortest side
+Jasper Tan thats a minkowski space (split-complex plane), but i don't think it works there either, not for all triangles at least.....
+John Galmann It does as long as all the lines are timeline - and that gives rise to the so-called twin "paradox" (not a paradox at all, of course, just the result of the triangle inequality in a Minkowski space).
So when Homer said that, he was obviously referring to lines in Minowski spacetime.
Home Simpson secret genius confirmed.
Jasper Tan citation needed... That silly formula for the Minkowski metric doesn't make much mathematical sense, especially considering that the distance between two distinct simultaneous events is an imaginary number(?!)... Even assuming that is the case, imaginary numbers aren't comparable, so the hypotenuse is neither shorter nor longer. :-\
@@irrelevant_noob Of course you can order imaginary numbers; you can't order complex numbers.
a+b
I think Futurama has more Math in it then the Simpsons one of its creators holds a PhD in Math & Physics.
That dude is super chill and the math looked like legit math so I guess this added value to my day
If the triangle is inside of the sphere, the two shortest lines can split from the longest line right before it makes the full radios. it would be a weird shape. but it would have three corners and it would give the two short sides a opportunity to be infinitely shorter then the longest line. Also works for the outside of the triangle ofcourse :)
not radios, But diameter.
Realized the same, but I though it was a translation error. Didn't know there was a whole video about.
RUclips always surprise me.
How did you manage to get me so hooked on watching this
I don't even pay attention in class XD
David²+S²=Cohen²
"pah, the way people act around here, you'd think the roads were paved with gold"
"they are"
Omg that poor scare crow xD
He makes us all look bad.
It can be done on the surface of a cone where the largest side of a triangle is towards the base of the cone compared to the other sides. Also after boiling down variables, I came to a = b + sqrt(2ab) to satisfy sqrt(c)=sqrt(a)+sqrt(b) where a is not equal to b and all three are real non zero numbers. It was a few napkins long of some algebra but it's possible there is a small mistake in there.
"Homer knows isosceles triangles? It's ridiculous." Hahaha. :)
His laugh is adorable. Love it!))
At 7:08, just replace line segment c with the remaining portion of the great circle that c lies on (ie the grey line extending outward from c in the diagram)
I enjoy your videos, even though I rarely understand what you’re talking about. I’m hoping to learn something.
The Wizard of Oz's scarecrow got Homer Simpson's brain!
... or the scarecrow is Homer Simpson's REAL dad!
@@SeanJTharpe He's nothing but hay and cloth. I doubt he's got genitals.
Simon Singh has a great book on the Mathematics in the Simpson's. Many of the writers held STEM degrees.
It can't work with any triangle since every triangle respects the triangle inequality - no matter in which space we embed it. One would have to give up the requirement that the points are connected by geodesics. As long as c is a geodesic a+b
Seen a bunch of your content but seeing you giggle like that when saying "wronger" made me subscribe
If I had learned math this way in school, I think I would less suck at it today. Still I am learning things here.
I think I have an easier proof for the isocele triangles that 2*sqrt(a) =/= sqrt(b).
You can construct an other isoceles triangle with the equal sides which are still a and the remaining size which would be b' =/= b.
Assuming the theorem mentionned is true, you have that sqrt(a) = sqrt( b )/2 = sqrt( b' )/2 which is a contradiction.
+nico.og Cool :)
With the 2 Simpsons clips at the end the Pythagorus theorem is at 8:13 David + S = Cohen (All squared) and 8:29 on the math book.
I love this channel.
7:00 The answer is *never* on any Riemannian manifold ... if "length" is defined as *geodesic distance* ... because the geodesic is the *shortest distance* between two points, which forces the triangle inequality. Now, on a *pseudo-Riemannian* manifold (even flat, like Minkowski space), that's another story.
This leads naturally to a question for you: do the flight distances of New York, Miami, Chicago and Houston fit in *any* Euclidean geometry, if they are treated as straight lines? If not, then what's the minimum curvature they must have before they do? What about other sets of 4 cities on the Earth, like London, Tokyo, New York and Johannesburg? Which geometries will 4 cities fit on, as a function of how much curvature their flight paths are endowed with? (Yes, some cases require a 2+1 dimensional Minkowski Geometry).
What about 5 or 6 cities? And since the Earth is *not* a sphere, what happens if you try to fit 6 cities, as a function of the curvature you give all the flight paths, assuming they're all given the same curvature? How much information can be said about the dimensions of the Earth - as well as the cities' *latitudes and relative longitudes* - on the assumption that the 6 cities fit on a ellipsoid? Try it with { New York, Miami, Chicago, Houston, Los Angeles, Seattle}, as well as {London, New York, Tokyo, Johannesburg, Melbourne, Rio de Janeiro}.
Sorry - geodesics are NOT necessarily the shortest routes between any two points. Geodesics are only LOCALLY the shortest routes between two oints.
You can't violate the triangle inequality, a+b>c, with some weird surface because no matter what surface and metric you're using, by definition the metric has to satisfy the triangle inequality. The only way is if you choose the sides of the triangle to be something other than geodesics (shortest paths), in which case you don't really have a triangle, just some 3 vertex shape.
I didn't say straight line, I said geodesic, which exist in any space, not just the plane.
One instance where it works? Lets say the circumference of the sphere is 2... Lets have the side c be equal to maybe 1.5 and then the other two sides can just connect from the end points of the side c... That would make a+b
Well done finding the Scarecrow origin of this!
But A+B < C does work on a sphere. You just have to go the long way around the sphere. So the Mercator projection would look like ____________/\______________Edit: I'm sure you've gotten this a thousand times. I tried to find a similar comment, but if it's not in Top Comments....
Not sure if I'm being an idiot and I would like more insight on this but wouldn't that Mercator projection make a hemisphere with a triangle missing instead of a triangle since the inside angles would exceed 180 degrees
When your literature teacher interprets a passage in a book
The video no one really ever needed but it's always good to educate the masses.
a+b≤c should work on a sphere. if c goes ¾ times around the sphere (equator) with a and b intersecting at the north pole, then a and b are ¼ times the circumference each.
Isn’t the gag that the Scarecrow got a diploma, not an actual brain?
Someone's putting quite a lot of faith in the writers
2:59 Is it normal that i see some similarities with the mathloger ?
He got the brain, went to Australia and became matematician.
There's the time homer solves fermat's last theorem. But they used an edge case where the answer is incorrect in decimals that a regular calculator doesn't show
3:04 ".. He got a bad deal..."
LOLOLOLOLOLOL!
In one of the episodes in which Homer tried to become an inventor there is a reference to Ferma's last theorem :).
+Jelle (NL) Ah, yes, that's a nice one. There are actually two occurrences of "counterexamples" to Fermat's last theorem in the Simpsons. The one you mention is the second one. The first one pops up in Homer^3 (Homer cubed) where Homer stumbles into a 3d world. Very neat stuff. There is also one mention of Fermat's little theorem in the Futurama Simpsons crossover episode.
+Mathologer keep the counterexamples in quotation marks. As per the numberphile video those are only close to a solution, not exact. (even the parity of the sum is wrong).
P.S. just to make sure no one thanks Ferma's last theorem is debunked.
There is no metric space where this equality could happen. In particular it is not true for any space with metric (Riemannian manifold: en.wikipedia.org/wiki/Riemannian_manifold), the sphere included. In such a strange world we would have a distance function which is does not satisfy the triangular inequality ...
Actually, on a sphere (or any contained surface) you could make a triangle with two obtuse angles. At this point a+b>c no longer necessarily holds.
Surely a + b < c would work on a spherical surface if c is greater than half a circumference. Eg. on a sphere of radius r, if c is 2r*pi-r/1000 then any combination of a, b would work if a+b > r/1000. An interesting special case is if c is exactly half an equator ie. c=r*pi and we have a and b meet at the north pole. There a + b = c exactly, and the three angles are 90,90,180 degrees respectively.
It is quite wrong ... but ... it can get even wronger 🤣🤣🤣
The Pythagorean Theorem but it's the opposite day
At around 7 minutes, trying to make the Mutilated Pythagorean Theorem (MPT) work on a spherical triangle - the triangle you show won't satisfy it, but there are spherical triangles that do. If you put the apex at a pole, and _c_ along the equator, then _a_ and _b_ are ¼-great-circle arcs ( _a_ = _b_ = ½πR), and _c_ can be any length in the open interval, 0 < _c_ < 2πR.
E.g., if R = 2/π, then _a_ = _b_ = 1, 0 < _c_ < 4. If you could make _c_ the entire equator, you'd have _(a,b,c)_ = (1,1,4), which satisfies the MPT; that is, for sides taken in the order _a,b,c_ ; √a + √b = √c.
If you make _a_ = _b_ a little shorter than 1 and at slightly different "longitudes", then they can be adjusted so that the great circle joining them the long way, will be _c_ = _4a_ = _4b_ , and the MPT will hold.
[Interesting to note: the MPT is homogeneous of degree ½, so it scales by any constant factor without changing.]
I stumbled into this on my recommendations. I don't know what the hell this channel is but by the thrice damned I'm going to subscribe. The algorithms brought me here for a reason probably I think.
The homer theorem would work in hyperbolic space in some cases i think
+Carol Vitez Yeah that's what I was speculating.
+Carol Vitez wouldn't it work on a torus?
+dalmation black
yes it would I think :)
Your theory of a Donut shaped universe intrigues me.
@techtrashing: Play some old Super Nintendo RPGs that feature a world map. The world map will usually loop from "west" to "east" and "south" to "north", thus forming a donut shaped world.
3:05 He got a brain just not a very good one.
In Lobochevsky (hyperbolic): a + b < c
Nikolai Ivanovich Lobochevsky?
@@carultch yes. in the hyperbolic geometry
so close... at 7 minutes the trigangle on the sphere: mark with c the rest of the circular arc,, not the small part, and suddenly: a+b
It could work with complex numbers where i*i=-1, then:
a*i + b*i +2sqrt(a*b*i*i) = a*i + b*i - 2sqrt(a*b) = c*i
might lead to some solutions.
p.s. oh... its 5 years old - saw 5th of september and didnt noticed the year :D
You didn't account for Non-Euclidean Geometry!
Ia Cthulhu Fhtagn!
he did tho, spherical geometry ain't euclidian ya cook!
????????????
en.wikipedia.org/wiki/Spherical_geometry#Relation_to_Euclid%27s_postulates
The video suggested drawing a triangle on a sphere but a+b>c is still satisfied. If you draw a triangle on an ellipsoid you can get lines where a+b>c is now false. Think of a football cut in half from one pointed end to the other. the a and b lines come down from the center on the top to the center on the sides. The c line starts at the bottom of a, around the ellipse towards the small rounded end and back to the bottom of b. The real question is whether a triangle drawn on something other than a flat plane can still be considered a triangle.
This video led me to thinking about "triangles" with side lengths including complex numbers....and, well, I ain't educated enough to really make much sense of it.
A right-angle triangle with sides root negative 1 and +2....would have a hypotenuse = root 3, which is shorter than one of the other sides....violating a + b > c.....but, does that inequality still actually matter?
Could the homer theorem work beyond real numbers?
While it may not be a solution, it might be an interesting topic for another video. Pythagorus theorem doesn't seem to work in the complex plane. A "simple" triangle (0), (0 + i), (1 + 0) has sides length 1, -i and (i - 1).....but the squares of 1 & -i add up to zero, not -2i.
I thought any metric that is constructed must still obey the triangle inequality. Even if it a + b = c.
Definitely, if you are in a "metric space" the triangle inequality has to be satisfied. But there are weird "reasonable" spaces that are not metric spaces. There have been a few suggestions in earlier comments :)
That's such a math teacher reaction to a bit such as 'crystal math'.
watch my maths videos to learn something.
Love your giggle. Math is interesting and fun 😍
A world where a+b can be less than c can be gotten by taking that sphere diagram of yours, and having c go the LONG way around the circle instead of the short way. Boom, a+b
I know this is old but I'm going to take a crack at these Pythagorean clips.
In the first clip, David S. Cohen's name is written as "David^2+S.^2 = Cohen^2", quite clever ;)
Of course, the second time around, A^2+B^2 = C^2 is just on the "MATH BOOK"
I know this is a dumb question...I guess I'm just not nerdy enough, but I don't get the joke. how is "David squared plus S squared = Cohen squared" clever? Is there some hidden meaning? Is there some sort of language wordplay thing going on there? I understand the pythagorean theorem, I understand the reference, but I don't get the joke.
+FeMaiden Maybe it's clever because no one ever looks at the credits so it was at least harder to find than the other example.
oh yeah, I looked back and I see the joke...it was just wordplay like on the halloween episodes they do that with the credits like "James Hell Brooks" instead of "James L Brooks"
I just thought maybe it was some sort of like...higher mathematics joke like a reference to a famous equation or something.
I did the calculations thinking that David^2+S.^2 = Cohen^2 would correlate numerically, if, for example, each letter associated with a number value (A=1, B=2, C=3)... but I didn't find anything. Someone can check my math, but I got DAVID (4+1+22+9+14)=40^2= 1400 Plus S (19)=19^2=361, so together 1961 equals COHEN (3+15+8+5+14)=45^2=2025. So, all together, 1961=2025 which obviously doesn't add up.
Tim Westchester 1400+361=1761, not 1961.
in the wizard of oz part, he really got a brain, the brain let him think logically, regardless of his answer being correct or not.
Hm, that statement doesn't make too much sense. I mean aren't you contradicting yourself? Logic would imply correctness, no?if he's not correct then that's Logic not working?
Logic does not imply correctness.
Im Dixie
Stating random incorrect facts from nowhere is the opposite of logic, no?
No, that is not the opposite of logic. hth
he did not seem to think in the slightest. he was smarter than any animal before (he could speak!), and the wizard changed nothing.
You CAN make a "triangle" on a sphear where a + b < c if c > r * pi
If we modify it just a little bit, so it says: c = sqrt(a) + sqrt(b) then some triangles do satisfy this, like a = 1, b= 2, c = 1+sqrt(2), but this is no longer Homer's triangle. Though, who knows. maybe the Scarecrow-Simpson triangle needs complex dimensions. I haven't tried that. :D