Let me tell you something about my public schooling experience. They didn't tell us the whole reasoning on how Pythagorean's Theorem worked. They never went into details about the squares and their areas. We were just told "Yeah to find the side of the triangle do this." It's extremely unfortunate since most people don't learn well through just being told something works. Keep on giving people a real education, James! Great video as usual!
"I'm not smart because my school didn't force me to be smart" We all learned it that way, including Einstein as you saw in the video. Some of us were able to apply and extend that knowledge thanks to our own diligence, while others seem to make excuses for their own shortcomings.
@@B3Band You didn't quite get the point of my comment. A learning institution is an institution for learning, and if students aren't learning then it needs to change, which is also why a lot of people don't value standardized testing even though it is a common guage for univerisities and colleges. Someone who doesn't read fast (like me) may do poorly in the reading section not based on my understanding or comprehension, but based on my speed.
@@MaximusXavier You can thank teacher's unions that brought us all down to their stultified level - government monopoly on education . My best teachers in public school - all 13 years of it- were substitutes whom lacked tenure - not indoctrinated (yet) into "I don't give-a-shitism" - I am not hear to educate you just to count you "present" - want to learn? - go to the library (Rousseau Method). Thanks CPS (Chicago Public Schools) you made the USSR look good. A bad system is bad for all- the losses to humanity remains un-countable.
My father was bad at math - since the teachers didn't want to explain obvious things. But Pythagorean theorem spoke to him, and he found out by himself that it gives the area of 2 smaller squares built on the shorter edges of the triangle sum up to the are of a square that's built on the longest edge. He did the same with triangles - and yes it wasn't explained to him, because "he was bad at math" so why bother. Unfortunately it is his last "Achievement" in the area of Mathematics. EDIT: Typo - funny thing it was in the word "area" ;)
Funny thing is kids at my school say it is useless to teach those. But im very glad that they teach it, it actually help me rewrite the formula when i forget it.
@@BRORIGIN he didn't invent geometry but he came up with a proof for the Pythagoras theorem also he used differential geometry in his general theory of relativity
Teacher: what does the Pythagorean theorem say? Me: The wobble of the hypotenuse in a right angle triangle is equal to the sum of the wobbles of the other two sides
James! I’m 37 and how have been aware of pythagoras’ theorem, I actually thought I knew it quite well for a non-mathematician (I’m a medic) but I’ve never understood it until the first minute of your video! Thank you! I knew i’m a visual learner but never thought of it like that. Thanks very much - you do great work in educating us!
Same here - replacing my previous favourite which takes 8 copies of the original triangle and squares with sides equal to the three sides of the triangle and uses them to tile two larger squares with side equal to the sum of the other two sides. One square gets the aa and bb squares in opposite corners leaving two ab rectangles that can then be covered by four triangles; the other gets the remaining triangles in the corners to leave a tilted cc hole in the middle. Easier seen in a diagram than described in a youtube comment...
But will the students just have to take it on faith that the areas of the triangles vary in proportion to the squares of their bases? I don't recall learning that in high school geometry, certainly not before we covered the Pythagorean theorem.
@@stevethecatcouch6532 Sure, but since all of this triangles are similar it's very simple to demonstrate this property. And to be honest, we use this idea in high school when we study solids. At least we use that here in Brazil :)
@@stevethecatcouch6532 finding an explicit formula for the areas of each in terms of the angles they have in common should be pretty trivial, and a squared term should be present in that formula
@Rafael I wish I had an enthusiastic teacher like you, sir :/ My lecturers(not only Math), in school or college, reserve themselves only to the prescribed syllabus. But ironically struggle to complete that too, sometimes leaving many chapters n concepts for the students to self learn :/ I hope one of ur students don't post a comment like this on yt 😅
This actually reminds me. For a moving object that famous formula is (energy)^2=(momentum*c)^2+(mass*c^2)^2. Pythagoras' theorem! Right angled triangles in transforming space-time are pretty useful for working out special relativity.
When I saw the thumbnail, I thought „oh, he must be in some different time zone and it’s still April fools day there...” I’m glad I watched it regardless
Don't know if you know already, but he does a lot of work with Brady Haran's channel Numberphile (ua-cam.com/users/numberphile) where you'll also find lots of other talented and similarly charismatic mathematicians discussing a massive range of mathematical topics. Also, if you're into space and physics, Brady has other channels, too! For Physics (particularly astrophysics etc.) it's Sixty Symbols (ua-cam.com/users/sixtysymbols). For astronomical objects (particularly the Messier catalogue), it's DeepSkyVideos (ua-cam.com/users/deepskyvideos) They are all really fantastic, and I've learned a lot in the years I've been watching them. They're also entirely responsible for reigniting my passion for learning, so I hope they do for you too!
So elengant and obvious That sums up Einstein's greatest discovery Relativity The idea behind it is so simple and elegant and results in such a beautiful theory of reality Einstein truly had s completely different view of things
If I'm not wrong, then Pythagoras' theorem has the most no of proofs. Check out the Ted ed's video on this. There, a visual proof for this using tessellation is also given
I genuinely, uncontrollably, audibly exclaimed: "Holy shit" as soon as you drew the semi-circles on the triangle. Thanks for the great video, and please excuse my French.
By adding the information at the end, that a, b and c are any given numbers out of the digit sum system 1 to 9 and once squared they can never go back to that system, because there are no numbers with the digit sum of 3 or 6 in this system and playing this video backwards, you could understand Fermats last Theorem. I love this elementary nonesense, you are really great in what you are doing.
Little Albert's proof is great. I can see how introducing more triangles when we usually talk about squares might needlessly complicate things for school children who simply want to memorize the rule and repeat it during exams. But this allows you to visible see that the c-area is the same as the a-area plus the b-area.
This has been by far my favorite proof. Once you take care of distributivity to allow shapes other than squares, I don’t know if I should call it blunt or sharp, but it’s just _there._ It’s amazing how that can elude us until someone points at it and it’s suddenly too obvious.
In school textbooks we learnt the proof by similarity. You draw that line, and all the 3 triangles are similar to each other. Euate the side lengths, solve and you get a^2 + b^2 = c^2 in a couple of steps.
You first have to prove that the height of a right triangle always cuts it into 2 similar triangles that are also similar to the original triangle. Great proofs!
One sentence proof: For each of the smaller triangles one of their angles is a right angle and another one coincides with original triangle's angle, thus they are similar to original triangle by two angles.
I would say probably not. Unless someone can point me to an actual thing showing it. It would work for the volume of cuboids if they all have the same thickness. But that's just the same as the 2D case. It just works for 2D shapes. Doesn't work for 1D either, because a+b is not c (Unless the triangle is also a line).
@@lyrimetacurl0 In 1 dimension a triangle is always a line. There is no plane in 1 dimension. Also I believe that it was a joke because the case for cubes and higher was/is a famous problem called fermats last theorem which was proven to be correct on a paper with a crazy number of pages. So there is no solution to the equation A^n +B^n =C^n for n is an integer greater than 2
If the enlargement (scale) factor for two similar plane shapes is x, then the area will have a scale factor of x^2 and the volume will have a scale factor of x^3. So, if you can determine that all corresponding sides in your 3D Solid all have the same scale factor, then yes, this would also work for any solid - not just a cuboid/rectangular prism. One more thing, note that the converse of this isn't true.
To believe Einstein's proof, you have to accept that areas of similar triangles are proportional to the squares on the hypotenuse. But we only know that through the formula for area of a triangle. It's actually simpler than that. You just need to know the sides are proportional. Divide C into D and E, so D is a leg of triangle with hypotenuse B in the diagram, and E is a leg of the triangle with hypotenuse A. Then by similar truangles, we write: B/C = D/B => B^2 = CD and A/C = E/A => A^2 = CE Add the two results to get: A^2 + B^2 = CE + CD = C(E + D) = CC = C^2 Q.E.D.
This is really very elegant. You sidestep any need to write expressions for the areas of the triangles, and all you need to know is the three triangles are similar so corresponding sides have same ratios.
@@willjohnston2959 Hey thanks, man. Wish I could take credit for it. I saw it, ,maybe 40 years ago, in an old text laying on a math professor's desk. Title was something like "100 proofs of the Pythagorean Theorem," but don't quote me. I just picked up the book and looked for the shortest proof. Remembered the idea behind it. Woulda saved me lotsa time on my final high school geometry exam. ;)
The square of a triangle with the base X and the height H is XH/2. So, the squares of similar triangles (all linear dimentions of which are scaled against each other with a coefficient K) are scaled with a squared coefficient K^2. In our case the squared scaling coefficiens for of our three triangles against a triangle with, say, hypotenuse 1 are a^2, b^2 and c^2.
Just a note, if you take any triangle and create two similar triangles within it (from smaller sides). Only a right angle triangle leaves no overlap or gap. Hence ( as you wrote it) A+B=C.. awesome vid
Works for literally any shape as any shape's area is proportional to a square of that shape's side length. If you scale up all sides by 2, the area goes up by 4, so no matter what you use, if you attach it with the same side to sides of the triangle, the area will change as a square and it will work.
That's very elegant. In school I was very annoyed that I never got to see a proof of this theorem. In hindsight the standard proof would have been a bit much for us at that time. But this is simply beautiful. Very nice.
@@jonasdaverio9369 I collected 18 proofs for the theorem when I was in high school. The funniest one was done by one of the US presidents. It was published during his campaign to prove he was smart. What he did was double everything at the beginning of the proof and divide everything by 2 at the end. Typical politician, promise big, deliver little. Oops! I got that backwards. I did that more than 50 years ago. He went the other way. He divided the standard Pythagorean proof in half and doubled it at the end. A comment below names James Garfield as the president and a search finds this: www.maa.org/press/periodicals/convergence/mathematical-treasure-james-a-garfields-proof-of-the-pythagorean-theorem
@@jonasdaverio9369 Standard proof is Euclid's proof. I was taught this. Jeames Garfield's proof: io9.gizmodo.com/james-garfield-was-the-only-u-s-president-to-prove-a-m-1037750658
@@jonasdaverio9369 There is no "standard proof". Which I thought was your point, I was supporting it. There are many proofs. The Internet is a wonderful thing; I see others have collected many proofs, too.
Brilliant !! NB seems to me that the Einstein demo @06:45 jumps over one step : - Agree with C = A+B, (eq1) - For each triangle, there are X1, X2 and X3 such that C = X1C2, A = X2a2, B = X3B2 - A is similar to C, as they share two sides, so X1=X2 - B is similar to C, as they chare two sides, so X1=X3 - so, l'ts write X=X1=X2=X3 to eq1 QED.
Very nice. When you look at each equation for each shape. You can divide out the "factor" lets call it x from both sides of the equation x*a^2 + x*b^2 = x*c^2 (divide both sides by x) and still be left with a^2+b^2=c^2 Peace and cheers!
Hello. I've just listened to your interview on Numberphile where you told that you read the comments on this channel. I thought it was a good occasion to thank you for all the stuff you do
I know that according to Wikipedia there are over 300 proofs for this theorem("...the book Pythagorean Proposition, by Elisha Scott Loomis, contains 367 proofs."), I would like to see a more general proof such as one involving the law of cosines or the wedge product of the Hestenes Geometric Algebra. I like this video but I wonder about other viewpoints such as using vectors to create the initial triangle (vectorA + vectorB = vectorC)(suggested by the video with area A + area B = area C), definining the exponent of 2 in terms of triangles (3 edge lengths of a large triangle contain 3 to the exponent 2 (3^2) triangles(=9)(=1+3+5) and the Chinese gougou theorem illustrations.
I like Einstein's Proof. That was basically how I approached it when I was younger. The conventional direction wasn't intuitive to me. But then I never was very good at math, so it is of no surprise.
Einstein strikes again! My new favourite proof. One can actually work out an are formula for any regular n-gon based on the number of sides and the "radius" (centre to vertex). I don't remember it, and it's not exactly pretty... but it does provide a nice hint that if the shapes are scaleable, then Pythagoras holds.
OK. Take-away here is that, of course, any Euclidean area can be represented as a square or parts of squares, and those squares scale linearly with the scaling of the shape, so the area of the shape scales with the size just as though it was a square.
The proof at the end seems incomplete, as one would have to prove that the areas of the triangles are proportional to a^2, b^2 and c^2. Its not difficult to do this, but the most intuitive way to do it would be using the Pythagorean theorem which is circular, so it would make sense to show a way to prove this without Pythagoras. One way to see it intuitively would be to scale any of the triangles down until the hypotenuse is length 1. rotate the triangle until the hypotenuse is horizontal, then scale the triangle horizontally and vertically by a factor x, obviously the area of the triangle has been scaled by a factor of x^2 and since the hypotenuse was horizontal it was scaled by a factor of just x. The big X in the final equation would be the area of the triangle scaled down to hypotenuse length 1. On an abstract note, looking at the final picture we know that A, B, and C are all similar triangles. So if we scale triangle A by some scalar x such that the new area can be morphed into a square with side length a, by similarity applying that same scale factor to triangles B and C would allow us to morph those new areas into squares with side lengths b and c. Since A+B=C it must be true that Ax+Bx=Cx => a^2+b^2=c^2.
Interesting q. It's easy to see that if the wobble is a regular polygon, then there is a one-to-one correspondence between the polygon's shape n it's area. So, if we put some time n find the general formula for such a regular polygon, then u can use this result to find what shape it is if the area is given. But, if the wobble is an irregular shape, then I think it's highly unlikely to pin-point the shape. Cuz, in this case, the area function of the 'shape' variable isn't one-one anymore. Hope this helps :)
"It's elementary, Watson!" Any 3 similar 2d figures, scaled to the 3 sides of the right triangle, will obey the "modified" Pythagorean Theorem, because areas of plane figures always scale by the square of any corresponding linear dimension. Kudos to Al, for using that in such an elegant way! Fred
Why scaling the side length by a certain factor results in the area scaling by the same factor squared is true isn't immediately obvious to me. One way would be to approximate the shape with small squares, showing it for the approximation and letting the small square size go to zero. Is there a less heavy prove for this?
You can use similar triangles to show that, if all sides are longer by a factor of k, the altitude is also longer by a factor of k, so the area is larger by a factor of k^2.
A more straight forward proof is to consider c = c1 + c2, where c1 and c2 are one of the leg of the two smaller triangles. As the triangles are similar, the ratio of the sides are the same: a/c2 = c/a --> a^2 = c*c2 b/c1 = c/b --> b^2 = c*c1 thus, a^2 + b^2 = c*(c2+c1) = c*c = c^2
Latee on, Einstein went on to prove a separate special case of a generalization of the Pythagorean theorem: namely, that in natural units such that c = 1, E^2 = p^2 + m^2.
I´ve tried with the parabolas y=x(x-b);y=x(x-c) and y= x(x-a) and and it did not work.I have obtained a**3=b**3+c**3 and not a**2=b**2+c**2 so it works only with regular polygons (remebering that a circle is a regular polygon with infinite sides)
How do fractals not scale the same way as other figures? If your fractal encloses an area then the proof in the video will work. If it encloses zero area (some fractals do that, like the Vicsek fractal) then the theorem is trivial (zero times any number remains zero).
Just discovered your personal channel via numberphile where you're one of my favorite presenters! Is there a theory/observation related to the angle of the plane formed by 'folding' each of the attached polygon of sides A B and C 90 degrees? In other words if you were to make a 5th plane that connects the heights of the three walls made by folding the polygons along edges a b c, at what angle relative to the original triangle's plane would it be? Is this a function? Has this been studied? Thanks!!
I am really dim. I don't understand this. I don't understand why Xa2+Xb2=Xc2 is a scaling of A+B=C. I can't make that statement make sense. What am I missing? What is the logical conjuring trick?
Whenever I feel like people around me are idiots and I'm just a misunderstood genius, I'll come back to this video and think back to what I did at 11. If I was proving anything back then, it sure wasn't maths.
While the beginning was unnecessary in my case since I know how scaling areas works (though others might not, which must be why you included it), this is the clearest presentation of that proof I have seen so far.
Yup David's reply is good. Maybe I should have said, imagine the similar triangle with a hypotenuse of length 1 and area X, then that triangle is scaled by a^2, b^ and c^2.
@@davidgould9431 But that proof is circular. We know that the areas of the earlier shapes vary according to the squares of linear measurements, but used the Pythagorean theorem to prove that. If we now use that fact to prove the Pythagorean theorem, we have used the Pythagorean theorem to prove the Pythagorean theorem. You need to prove that the areas vary with the square of the linear measurement by some means independent of the Pythagorean theorem. There's a fairly easy proof of that for triangles, but it needs to be made.
Wow. That line is clever... like really clever... I always used the proof where you draw a square inside a square rotated such that the corners of the smaller square touch the sides of the outer square. You see that the outer square is now carved into 5 pieces. 4 identical right triangles and 1 square. If you label the sides of the triangles A, B and C you see that the side of the big square is A plus B. Now we take the area of the big square. We can do this in two ways. Either as the lenght of its sides squared (A+B)^2 = A^2+2AB+B^2 Or as the sum of the area of the smaller elements. The square is C^2 and the triangles are each ½AB. 4 triangles and a square that's C^2+2AB Then put the two equal and remove the 2AB from either side and A^2+B^2 = C^2 It looks terrible in words but is pretty elegant visually. So might be more of a Matt Parker proof
You know, what would happen if you used a fractal, like Sierpiński's Triangle? I'd assume this idea would no longer hold, then, because the area (well, measure) doesn't just scale up by the square of the side length. It scales up by the lb(3)-th power (where lb is the binary log, approximately 1.58th power) of the side length. So we would get that if the large Sierpiński's Triangle has measure X, then the one hanging off the a side would have measure X*(a/c)^1.58 and X*(b/c)^1.58. And they don't add up to X, I think? But I wonder if there's any sort of analogous statement to the Pythagorean Theorem for fractal shapes. It almost feels like a non-integer version of Fermat's Last Theorem.
5:32 the blue, white and red triangle combined form an interesting quadrilateral. it's like two right kites glued together. i wonder if there's a name for that. 🤔
It’s a trapezoid. The top-right and bottom-left sides must be parallel, since they’re both perpendicular to the side opposite C. I have a truly marvelous proof that any right trapezoid can be divided into two right kites, but it’s too large for this comment. Edit: my conjecture is false, and a counterexample is trivial. It’s left here as an exercise to the reader.
sofias. orange, I edited my comment. It’s actually just a special type of right trapezoid, can you find what other trait is necessary for a trapezoid to be divided into two kites?
@@felipevasconcelos6736 uum. i'm pretty sure i replied to you comment and it seemed to have disappeared. in fact i think i replied twice but maybe one reply got lost in a browser restart or something but i'm pretty sure i sent it the second time. i'm not sure what's happening.. :/
I didn't get Einstein proof part. How did he know how to scale by a^2, b^2, and c^2 rather than just a, b, and c? I got it by 1) A + B = C, thus by multiplying both sides by 2 2) 2A + 2B = 2C, which happen to be squares with a width of a, b, and c respectively, thus 3) a^2 + b^2 = c^2.
While watching I developed the idea of finding a shape in such away that the smaller ones are congruent to the bigger one. I didn't know this was the point of the proof.
Correct me if I'm wrong, but if you flipped the triangles at the end there (i.e., so the angle in C which is next to triangle A is instead put so it's next to triangle B) wouldn't they form rectangles? I don't know how that'd help solve anything, but it's cool?
My curiosity is the small number of dislikes for this video. Who watches a fascinating proof like this (with great lead up) and thinks it deserves a dislike?
Perhaps a disgruntled student who was disappointed about a low grade or some other perceived slight. The nom de plumes and anonymity sometimes lead to harsher criticisms in cyberspace.
I guess I always just saw the literal shapes of squares as being arbitrary visualizations of area. Of course any shape's area formulation will be a simple factor proportional to the dimension you're using, and you just ignore the factor n in na^2+nb^2=nc^2.
This is the simplest proof of the Pythagorean theorem that I've ever seen. I love how simple and elegant it is! Very much in the style of Albert Einstein as well - 'If spacetime affects matter (special relatively), then does matter also affect spacetime (general relatively)?'
Let me tell you something about my public schooling experience. They didn't tell us the whole reasoning on how Pythagorean's Theorem worked. They never went into details about the squares and their areas. We were just told "Yeah to find the side of the triangle do this." It's extremely unfortunate since most people don't learn well through just being told something works. Keep on giving people a real education, James! Great video as usual!
"I'm not smart because my school didn't force me to be smart"
We all learned it that way, including Einstein as you saw in the video. Some of us were able to apply and extend that knowledge thanks to our own diligence, while others seem to make excuses for their own shortcomings.
@@B3Band You didn't quite get the point of my comment. A learning institution is an institution for learning, and if students aren't learning then it needs to change, which is also why a lot of people don't value standardized testing even though it is a common guage for univerisities and colleges. Someone who doesn't read fast (like me) may do poorly in the reading section not based on my understanding or comprehension, but based on my speed.
@@MaximusXavier You can thank teacher's unions that brought us all down to their stultified level - government monopoly on education . My best teachers in public school - all 13 years of it- were substitutes whom lacked tenure - not indoctrinated (yet) into "I don't give-a-shitism" - I am not hear to educate you just to count you "present" - want to learn? - go to the library (Rousseau Method). Thanks CPS (Chicago Public Schools) you made the USSR look good.
A bad system is bad for all- the losses to humanity remains un-countable.
My father was bad at math - since the teachers didn't want to explain obvious things. But Pythagorean theorem spoke to him, and he found out by himself that it gives the area of 2 smaller squares built on the shorter edges of the triangle sum up to the are of a square that's built on the longest edge. He did the same with triangles - and yes it wasn't explained to him, because "he was bad at math" so why bother. Unfortunately it is his last "Achievement" in the area of Mathematics.
EDIT: Typo - funny thing it was in the word "area" ;)
Funny thing is kids at my school say it is useless to teach those. But im very glad that they teach it, it actually help me rewrite the formula when i forget it.
Wobbles are my new favorite mathematical shape.
My calculus teacher would use barrels and blobs in place of letters to help students not get confused by how variables are used.
This Einstein kid sounds pretty smart. I see a relatively bright future ahead of him.
You know he invented geometry, right?..
@@BRORIGIN I really hope you're kidding.
@@BRORIGIN he didn't invent geometry but he came up with a proof for the Pythagoras theorem also he used differential geometry in his general theory of relativity
@@wolframalpha8634 no, newton did that
It's quite deterministic as well, because the God doesn't play dice, you know...
Teacher: what does the Pythagorean theorem say?
Me: The wobble of the hypotenuse in a right angle triangle is equal to the sum of the wobbles of the other two sides
James! I’m 37 and how have been aware of pythagoras’ theorem, I actually thought I knew it quite well for a non-mathematician (I’m a medic) but I’ve never understood it until the first minute of your video! Thank you! I knew i’m a visual learner but never thought of it like that. Thanks very much - you do great work in educating us!
That has just become my new favorite proof. I’ll show it to my students
Same here - replacing my previous favourite which takes 8 copies of the original triangle and squares with sides equal to the three sides of the triangle and uses them to tile two larger squares with side equal to the sum of the other two sides. One square gets the aa and bb squares in opposite corners leaving two ab rectangles that can then be covered by four triangles; the other gets the remaining triangles in the corners to leave a tilted cc hole in the middle. Easier seen in a diagram than described in a youtube comment...
But will the students just have to take it on faith that the areas of the triangles vary in proportion to the squares of their bases? I don't recall learning that in high school geometry, certainly not before we covered the Pythagorean theorem.
@@stevethecatcouch6532
Sure, but since all of this triangles are similar it's very simple to demonstrate this property.
And to be honest, we use this idea in high school when we study solids. At least we use that here in Brazil :)
@@stevethecatcouch6532 finding an explicit formula for the areas of each in terms of the angles they have in common should be pretty trivial, and a squared term should be present in that formula
@Rafael I wish I had an enthusiastic teacher like you, sir :/
My lecturers(not only Math), in school or college, reserve themselves only to the prescribed syllabus. But ironically struggle to complete that too, sometimes leaving many chapters n concepts for the students to self learn :/
I hope one of ur students don't post a comment like this on yt 😅
Einstein went on to prove E=m (a squared +b squared) ;-)
This c = arbitrary letter
The c you're looking for = speed of light in a vacuum
Sorry to be a buzz kill :(
@@yaj126 Pretty sure that was the joke :P
This actually reminds me. For a moving object that famous formula is (energy)^2=(momentum*c)^2+(mass*c^2)^2. Pythagoras' theorem! Right angled triangles in transforming space-time are pretty useful for working out special relativity.
@@summertilling the second famous equation should be Ec=pc² xD
_i have a E=mc²_
_i have a Ec=pc²_
*boom*
_E²= (pc)² + (mc²)²_
Nah bro the c in e=mc^2 is for the speed of light
E=m(a²+b²)
hahaha
thats actually not that off...
Well as a start there are some right angled triangles in the actual proof of special relativity
That Albert boy must have been some genius, because finding a one-line proof after 2000 years is something really clever
When I saw the thumbnail, I thought „oh, he must be in some different time zone and it’s still April fools day there...” I’m glad I watched it regardless
Please post your video more often
They are really good
Don't know if you know already, but he does a lot of work with Brady Haran's channel Numberphile (ua-cam.com/users/numberphile) where you'll also find lots of other talented and similarly charismatic mathematicians discussing a massive range of mathematical topics.
Also, if you're into space and physics, Brady has other channels, too! For Physics (particularly astrophysics etc.) it's Sixty Symbols (ua-cam.com/users/sixtysymbols). For astronomical objects (particularly the Messier catalogue), it's DeepSkyVideos (ua-cam.com/users/deepskyvideos)
They are all really fantastic, and I've learned a lot in the years I've been watching them. They're also entirely responsible for reigniting my passion for learning, so I hope they do for you too!
@@benoitb.3679 Thanks. I will surely check them out
I love Einstein's proof - so simple, so elegant. Why didn't I think of it, it seems so natural and obvious!
Genius is just being the first one to point out the obvious as they say
So elengant and obvious
That sums up Einstein's greatest discovery
Relativity
The idea behind it is so simple and elegant and results in such a beautiful theory of reality
Einstein truly had s completely different view of things
Did u really understand the tool used here...i didnt...
Plz xplain me too
My thoughts, exactly 😅.
good stuff mate. that's cool that all these years later, the pythagorean theorem still has something to teach us.
If I'm not wrong, then Pythagoras' theorem has the most no of proofs. Check out the Ted ed's video on this. There, a visual proof for this using tessellation is also given
I genuinely, uncontrollably, audibly exclaimed: "Holy shit" as soon as you drew the semi-circles on the triangle.
Thanks for the great video, and please excuse my French.
You're right: your French is _awful._ It all sounds just like plain English!
Great video James, I really appreciate your work
Thank you
What an amazingly simple proof!
I was not aware of this. Thanks for showing us!
By adding the information at the end, that a, b and c are any given numbers out of the digit sum system 1 to 9 and once squared they can never go back to that system, because there are no numbers with the digit sum of 3 or 6 in this system and playing this video backwards, you could understand Fermats last Theorem.
I love this elementary nonesense, you are really great in what you are doing.
Little Albert's proof is great. I can see how introducing more triangles when we usually talk about squares might needlessly complicate things for school children who simply want to memorize the rule and repeat it during exams. But this allows you to visible see that the c-area is the same as the a-area plus the b-area.
This has been by far my favorite proof. Once you take care of distributivity to allow shapes other than squares, I don’t know if I should call it blunt or sharp, but it’s just _there._ It’s amazing how that can elude us until someone points at it and it’s suddenly too obvious.
In school textbooks we learnt the proof by similarity. You draw that line, and all the 3 triangles are similar to each other. Euate the side lengths, solve and you get a^2 + b^2 = c^2 in a couple of steps.
You first have to prove that the height of a right triangle always cuts it into 2 similar triangles that are also similar to the original triangle.
Great proofs!
One sentence proof: For each of the smaller triangles one of their angles is a right angle and another one coincides with original triangle's angle, thus they are similar to original triangle by two angles.
@@balsoft01 Right, thank you!
@@pairot01 You also have to prove that the areas of similar triangles are proportionate to the squares of corresponding sides.
@@stevethecatcouch6532 That's what the first part of the video proved, but in a vaccuum yesyou'd have to prove it.
@@stevethecatcouch6532 We've got Thales for that 😂
It must be easy to prove the same for cubes right?
Right?
Yes. The proof is truly marvelous unfortunately this comment section is too small to contain it
I would say probably not. Unless someone can point me to an actual thing showing it.
It would work for the volume of cuboids if they all have the same thickness. But that's just the same as the 2D case. It just works for 2D shapes. Doesn't work for 1D either, because a+b is not c (Unless the triangle is also a line).
@@lyrimetacurl0 In 1 dimension a triangle is always a line. There is no plane in 1 dimension. Also I believe that it was a joke because the case for cubes and higher was/is a famous problem called fermats last theorem which was proven to be correct on a paper with a crazy number of pages. So there is no solution to the equation A^n +B^n =C^n for n is an integer greater than 2
If the enlargement (scale) factor for two similar plane shapes is x, then the area will have a scale factor of x^2 and the volume will have a scale factor of x^3.
So, if you can determine that all corresponding sides in your 3D Solid all have the same scale factor, then yes, this would also work for any solid - not just a cuboid/rectangular prism.
One more thing, note that the converse of this isn't true.
I literally proved this for all regular polygons and similar triangles.
2 months ago :)) I’m so happy I did it before this video came up haha
Einstein : "Pythagoras theorem is not absolutely hard, it is relatively simple."
To believe Einstein's proof, you have to accept that areas of similar triangles are proportional to the squares on the hypotenuse. But we only know that through the formula for area of a triangle. It's actually simpler than that. You just need to know the sides are proportional.
Divide C into D and E, so D is a leg of triangle with hypotenuse B in the diagram, and E is a leg of the triangle with hypotenuse A. Then by similar truangles, we write:
B/C = D/B => B^2 = CD and
A/C = E/A => A^2 = CE
Add the two results to get:
A^2 + B^2 = CE + CD = C(E + D) = CC = C^2
Q.E.D.
This is really very elegant. You sidestep any need to write expressions for the areas of the triangles, and all you need to know is the three triangles are similar so corresponding sides have same ratios.
@@willjohnston2959 Hey thanks, man. Wish I could take credit for it. I saw it, ,maybe 40 years ago, in an old text laying on a math professor's desk. Title was something like "100 proofs of the Pythagorean Theorem," but don't quote me. I just picked up the book and looked for the shortest proof. Remembered the idea behind it. Woulda saved me lotsa time on my final high school geometry exam. ;)
The square of a triangle with the base X and the height H is XH/2. So, the squares of similar triangles (all linear dimentions of which are scaled against each other with a coefficient K) are scaled with a squared coefficient K^2. In our case the squared scaling coefficiens for of our three triangles against a triangle with, say, hypotenuse 1 are a^2, b^2 and c^2.
Just a note, if you take any triangle and create two similar triangles within it (from smaller sides). Only a right angle triangle leaves no overlap or gap. Hence ( as you wrote it) A+B=C.. awesome vid
Works for literally any shape as any shape's area is proportional to a square of that shape's side length. If you scale up all sides by 2, the area goes up by 4, so no matter what you use, if you attach it with the same side to sides of the triangle, the area will change as a square and it will work.
That's very elegant. In school I was very annoyed that I never got to see a proof of this theorem. In hindsight the standard proof would have been a bit much for us at that time. But this is simply beautiful. Very nice.
What do mean by "the standard proof"?
@@jonasdaverio9369 I collected 18 proofs for the theorem when I was in high school. The funniest one was done by one of the US presidents. It was published during his campaign to prove he was smart. What he did was double everything at the beginning of the proof and divide everything by 2 at the end. Typical politician, promise big, deliver little.
Oops! I got that backwards. I did that more than 50 years ago. He went the other way. He divided the standard Pythagorean proof in half and doubled it at the end. A comment below names James Garfield as the president and a search finds this:
www.maa.org/press/periodicals/convergence/mathematical-treasure-james-a-garfields-proof-of-the-pythagorean-theorem
@@aeromodeller1 I don't quite see the link with what I wrote
@@jonasdaverio9369 Standard proof is Euclid's proof. I was taught this. Jeames Garfield's proof: io9.gizmodo.com/james-garfield-was-the-only-u-s-president-to-prove-a-m-1037750658
@@jonasdaverio9369 There is no "standard proof". Which I thought was your point, I was supporting it. There are many proofs. The Internet is a wonderful thing; I see others have collected many proofs, too.
Brilliant !!
NB seems to me that the Einstein demo @06:45 jumps over one step :
- Agree with C = A+B, (eq1)
- For each triangle, there are X1, X2 and X3 such that C = X1C2, A = X2a2, B = X3B2
- A is similar to C, as they share two sides, so X1=X2
- B is similar to C, as they chare two sides, so X1=X3
- so, l'ts write X=X1=X2=X3 to eq1 QED.
Very nice. When you look at each equation for each shape. You can divide out the "factor" lets call it x from both sides of the equation
x*a^2 + x*b^2 = x*c^2 (divide both sides by x)
and still be left with
a^2+b^2=c^2
Peace and cheers!
Very elegant proof :]
that was a really neat proof and easy to follow. thanks for the video.
Wobbles immediately made it clear for me
amazing video, finally made me truly understand why the pythagorean theorem held for all shapes
I'm a simple man. I see James Grime, I click and like and am never disappointed
Hello. I've just listened to your interview on Numberphile where you told that you read the comments on this channel. I thought it was a good occasion to thank you for all the stuff you do
Indeed I do, and thank you.
Does Fermat last theorem work for all shapes?
I know that according to Wikipedia there are over 300 proofs for this theorem("...the book Pythagorean Proposition, by Elisha Scott Loomis, contains 367 proofs."), I would like to see a more general proof such as one involving the law of cosines or the wedge product of the Hestenes Geometric Algebra. I like this video but I wonder about other viewpoints such as using vectors to create the initial triangle (vectorA + vectorB = vectorC)(suggested by the video with area A + area B = area C), definining the exponent of 2 in terms of triangles (3 edge lengths of a large triangle contain 3 to the exponent 2 (3^2) triangles(=9)(=1+3+5) and the Chinese gougou theorem illustrations.
Just checked the date this was released, glad that confirms this is a 100% legitimate video.
I like Einstein's Proof. That was basically how I approached it when I was younger. The conventional direction wasn't intuitive to me. But then I never was very good at math, so it is of no surprise.
But what if the shapes are a fractal which has a different dimension than 2? Are those included in those 'any' shapes?
any 2d shape is K times a square, it's proportional to it, so this should hold even for fractals
@@JorgetePanete Nope, the area of fractals doesn't obey that rule, although they lie in 2d plane, their Hausdorff dimension can be less than 2.
@@chankhavu oh
I love your videos. Keep up the great work.
1:53 PI-thagorean theorem.
Half-taugorean theorem
Einstein strikes again! My new favourite proof. One can actually work out an are formula for any regular n-gon based on the number of sides and the "radius" (centre to vertex). I don't remember it, and it's not exactly pretty... but it does provide a nice hint that if the shapes are scaleable, then Pythagoras holds.
That's elegant and geometrically satisfying
OK. Take-away here is that, of course, any Euclidean area can be represented as a square or parts of squares, and those squares scale linearly with the scaling of the shape, so the area of the shape scales with the size just as though it was a square.
The proof at the end seems incomplete, as one would have to prove that the areas of the triangles are proportional to a^2, b^2 and c^2. Its not difficult to do this, but the most intuitive way to do it would be using the Pythagorean theorem which is circular, so it would make sense to show a way to prove this without Pythagoras. One way to see it intuitively would be to scale any of the triangles down until the hypotenuse is length 1. rotate the triangle until the hypotenuse is horizontal, then scale the triangle horizontally and vertically by a factor x, obviously the area of the triangle has been scaled by a factor of x^2 and since the hypotenuse was horizontal it was scaled by a factor of just x. The big X in the final equation would be the area of the triangle scaled down to hypotenuse length 1.
On an abstract note, looking at the final picture we know that A, B, and C are all similar triangles. So if we scale triangle A by some scalar x such that the new area can be morphed into a square with side length a, by similarity applying that same scale factor to triangles B and C would allow us to morph those new areas into squares with side lengths b and c. Since A+B=C it must be true that Ax+Bx=Cx => a^2+b^2=c^2.
Lovely video James - terrific . Thank you.
Interesting video, however....
white text on a light yellow background is not very visible.
Great video. Thanks for sharing!
Given the coefficients of a, b and c, find what shape are they representing.
Like 1 is a square, pi/8 is a semicircle...
Interesting q. It's easy to see that if the wobble is a regular polygon, then there is a one-to-one correspondence between the polygon's shape n it's area. So, if we put some time n find the general formula for such a regular polygon, then u can use this result to find what shape it is if the area is given. But, if the wobble is an irregular shape, then I think it's highly unlikely to pin-point the shape. Cuz, in this case, the area function of the 'shape' variable isn't one-one anymore. Hope this helps :)
"It's elementary, Watson!"
Any 3 similar 2d figures, scaled to the 3 sides of the right triangle, will obey the "modified" Pythagorean Theorem, because areas of plane figures always scale by the square of any corresponding linear dimension.
Kudos to Al, for using that in such an elegant way!
Fred
It can be extended to 3 D ( volumes) with same thickness solid on the sides of right ∆
Dr.James! love ur work!
Thank you!
Why scaling the side length by a certain factor results in the area scaling by the same factor squared is true isn't immediately obvious to me. One way would be to approximate the shape with small squares, showing it for the approximation and letting the small square size go to zero. Is there a less heavy prove for this?
You can use similar triangles to show that, if all sides are longer by a factor of k, the altitude is also longer by a factor of k, so the area is larger by a factor of k^2.
A more straight forward proof is to consider c = c1 + c2, where c1 and c2 are one of the leg of the two smaller triangles. As the triangles are similar, the ratio of the sides are the same:
a/c2 = c/a --> a^2 = c*c2
b/c1 = c/b --> b^2 = c*c1
thus, a^2 + b^2 = c*(c2+c1) = c*c = c^2
Latee on, Einstein went on to prove a separate special case of a generalization of the Pythagorean theorem: namely, that in natural units such that c = 1, E^2 = p^2 + m^2.
I´ve tried with the parabolas y=x(x-b);y=x(x-c) and y= x(x-a) and and it did not work.I have obtained a**3=b**3+c**3 and not a**2=b**2+c**2 so it works only with regular polygons (remebering that a circle is a regular polygon with infinite sides)
Many years ago, someone decided that areas would be measured in squares, and then we think that it is the only way to do it.
How does this work out for Fractals, which don't scale by a factor of c^2?
How do fractals not scale the same way as other figures? If your fractal encloses an area then the proof in the video will work. If it encloses zero area (some fractals do that, like the Vicsek fractal) then the theorem is trivial (zero times any number remains zero).
@@renerpho en.wikipedia.org/wiki/Fractal_dimension
I don't know how many times I minimised the screen to leave like and realized that I have already liked the video
is it 1
So, what you're syaig is, there's a fixed ratio for every polygon and a square with a side length of the base of that polygon.
Cool.
Love you James sir, plz come to India once we will love it.
Looking forward to it.
I really like the new blackboard app that lets your writing show up onscreen, Dr. Grime. No more chalk dust on your hands.
I'm just writing on paper and scanning it in. I wouldn't want people getting the idea that I know what I'm doing.
@@singingbanana That makes more sense. I know you like working with pencil and paper--or brown paper and Sharpie, depending on the channel.
Does that mean E = m (a^2 + b^2) ?
Yes, when a is the projected speed of light in the north-south direction and b is the speed of light in the projected east-west direction.
That was a very neat video!
Thanks for this great video
Just discovered your personal channel via numberphile where you're one of my favorite presenters! Is there a theory/observation related to the angle of the plane formed by 'folding' each of the attached polygon of sides A B and C 90 degrees? In other words if you were to make a 5th plane that connects the heights of the three walls made by folding the polygons along edges a b c, at what angle relative to the original triangle's plane would it be? Is this a function? Has this been studied? Thanks!!
This is actually more intuitive than the original version.
Jesus Christ the most elegant proof I've ever seen.
where did you get "singingbanana" from?
First proof for the theorem that I really do like. Einstein was smart.
I am really dim. I don't understand this. I don't understand why Xa2+Xb2=Xc2 is a scaling of A+B=C. I can't make that statement make sense. What am I missing? What is the logical conjuring trick?
Whenever I feel like people around me are idiots and I'm just a misunderstood genius, I'll come back to this video and think back to what I did at 11. If I was proving anything back then, it sure wasn't maths.
I consider my childhood to be one massive self-referential in-joke directed at my current self
Thank you Mileva Marić.
Eres genial amigo un abrazo desde San Jerónimo antioquia Colombia gracias
If I had been thought this proof at school I would be a better person now
While the beginning was unnecessary in my case since I know how scaling areas works (though others might not, which must be why you included it), this is the clearest presentation of that proof I have seen so far.
i dont understand how from A+B=C we go to Xa^2+Xb^2=Xc^2
@@davidgould9431 the problematic part is why is (area of right triangle)=(constant)*(hypotenuse^2)
Yup David's reply is good. Maybe I should have said, imagine the similar triangle with a hypotenuse of length 1 and area X, then that triangle is scaled by a^2, b^ and c^2.
@@davidgould9431 But that proof is circular. We know that the areas of the earlier shapes vary according to the squares of linear measurements, but used the Pythagorean theorem to prove that. If we now use that fact to prove the Pythagorean theorem, we have used the Pythagorean theorem to prove the Pythagorean theorem. You need to prove that the areas vary with the square of the linear measurement by some means independent of the Pythagorean theorem. There's a fairly easy proof of that for triangles, but it needs to be made.
Wow. That line is clever... like really clever...
I always used the proof where you draw a square inside a square rotated such that the corners of the smaller square touch the sides of the outer square.
You see that the outer square is now carved into 5 pieces. 4 identical right triangles and 1 square.
If you label the sides of the triangles A, B and C you see that the side of the big square is A plus B.
Now we take the area of the big square. We can do this in two ways. Either as the lenght of its sides squared (A+B)^2 = A^2+2AB+B^2
Or as the sum of the area of the smaller elements. The square is C^2 and the triangles are each ½AB. 4 triangles and a square that's C^2+2AB
Then put the two equal and remove the 2AB from either side and A^2+B^2 = C^2
It looks terrible in words but is pretty elegant visually. So might be more of a Matt Parker proof
Thank you. this was great!
You know, what would happen if you used a fractal, like Sierpiński's Triangle? I'd assume this idea would no longer hold, then, because the area (well, measure) doesn't just scale up by the square of the side length. It scales up by the lb(3)-th power (where lb is the binary log, approximately 1.58th power) of the side length. So we would get that if the large Sierpiński's Triangle has measure X, then the one hanging off the a side would have measure X*(a/c)^1.58 and X*(b/c)^1.58. And they don't add up to X, I think?
But I wonder if there's any sort of analogous statement to the Pythagorean Theorem for fractal shapes. It almost feels like a non-integer version of Fermat's Last Theorem.
What if the perimeter of the wobble is a fractal? Does that change things?
It's a fun and elegant proof - although substituting A=Xa^2 etc. needs to be worked out in greater detail.
If a similar triangle with side = 1 has area = X, then a triangle (similar) triangle with side = a has area = Xa^2.
Yes, that's true in Euclidean geometry, but it has to be proven for the proof to be complete.
The direct approach is obvious, but also tedious.
@@Hecatonicosachoron You are right, but one can assume that this fact is "well-known". :D
Do Check Mathologer's Video too on Pythagoras Theorem
Ah. I hadn't seen that (I have now).
@@singingbanana Thanks for reply.
5:32 the blue, white and red triangle combined form an interesting quadrilateral. it's like two right kites glued together. i wonder if there's a name for that. 🤔
It’s a trapezoid. The top-right and bottom-left sides must be parallel, since they’re both perpendicular to the side opposite C. I have a truly marvelous proof that any right trapezoid can be divided into two right kites, but it’s too large for this comment.
Edit: my conjecture is false, and a counterexample is trivial. It’s left here as an exercise to the reader.
@@felipevasconcelos6736 Without moving pieces around?
@@felipevasconcelos6736 "any right trapezoid can be divided into two right kites" that's pretty cool, thanks!
sofias. orange, I edited my comment. It’s actually just a special type of right trapezoid, can you find what other trait is necessary for a trapezoid to be divided into two kites?
@@felipevasconcelos6736 uum. i'm pretty sure i replied to you comment and it seemed to have disappeared. in fact i think i replied twice but maybe one reply got lost in a browser restart or something but i'm pretty sure i sent it the second time. i'm not sure what's happening.. :/
Using it with circles gave me an exciting feeling of being fundamental to the Field Equation.
That was elegant indeed
Einstein's overlapping triangles is rather a clever insight for a child of 11.
I didn't get Einstein proof part. How did he know how to scale by a^2, b^2, and c^2 rather than just a, b, and c? I got it by
1) A + B = C, thus by multiplying both sides by 2
2) 2A + 2B = 2C, which happen to be squares with a width of a, b, and c respectively, thus
3) a^2 + b^2 = c^2.
While watching I developed the idea of finding a shape in such away that the smaller ones are congruent to the bigger one.
I didn't know this was the point of the proof.
smaller ones are similar (not congruent) to the bigger one.
Correct me if I'm wrong, but if you flipped the triangles at the end there (i.e., so the angle in C which is next to triangle A is instead put so it's next to triangle B) wouldn't they form rectangles?
I don't know how that'd help solve anything, but it's cool?
because they are flipped over the sides of the triangle
or in other words the sides of the triangle are the axis of symmetry
Leave it to Einstein to make the most geometrically simple, yet incredibly effeective proof a none other then pithagoras theorem
I wonder what's the standard proof taught in schools ... because Einstein's proof is basically what I was taught in school.
We were taught Euclid's proof! Sigh. en.wikipedia.org/wiki/Pythagorean_theorem#Euclid's_proof
very nice
neat! especiallly liked the wobbles!
My curiosity is the small number of dislikes for this video. Who watches a fascinating proof like this (with great lead up) and thinks it deserves a dislike?
Perhaps a disgruntled student who was disappointed about a low grade or some other perceived slight. The nom de plumes and anonymity sometimes lead to harsher criticisms in cyberspace.
wait a minute
why is your username singingbanana
I guess I always just saw the literal shapes of squares as being arbitrary visualizations of area. Of course any shape's area formulation will be a simple factor proportional to the dimension you're using, and you just ignore the factor n in na^2+nb^2=nc^2.
i wonder what happens if instead squares, pentagons, semicircles.. you put right triangles
nvm lol
This is the simplest proof of the Pythagorean theorem that I've ever seen. I love how simple and elegant it is! Very much in the style of Albert Einstein as well - 'If spacetime affects matter (special relatively), then does matter also affect spacetime (general relatively)?'
"Spacetime tells matter how to move; matter tells spacetime how to curve." John Archibald Wheeler's summary of Einstein's GR.