I had a terrible maths teacher back in highschool, that whenever I solved a problem using a non-standard approach, she said that I was killing an ant with an RPG shot. I hated that witch. That being said, using diff eqs to prove Pythagoras is killing an ant with an RPG.

@lucasdeoliveira5316

7 ай бұрын

He used to have a whole series dedicated to killing ants with RPG shots. It was appropriately called Overkill. I miss that series.

@atzuras

7 ай бұрын

sorry that's more like blasting a dead ant with a nuke.

@andersdahlner1052

7 ай бұрын

Had a teacher who said I cheated when I did fractions in the head. Lost my interest in school when I was 10-11 years.

@gibbogle

7 ай бұрын

It's true that there are much nicer geometric proofs of Pythagoras that don't need any mathematics.

@Cyrus72

7 ай бұрын

If I were a teacher, I would give extra points for creative solutions.

I always thought it was weird when mathematicians called mathematical proofs "beautiful" but now I get it.

I like how instead of simplifying tricky things, he complicates the most generic and simplest of things 😂

@deependrasinghis_ronaldo

7 ай бұрын

sure, looks like emphasizing certain reasoning/discussion and downplaying little stuffs.. will help make even better videos. michael seems to treat every step with same significance and importance (at least in his voice tone and time given to it)... which probably isn't the best thing for a new math learner.

VERY Interesting proof. I've never actually seen an analytical proof of the Pythagorean theorem I was just always told draw 2 squares from the legs of the right triangle and observe their areas add to the area of a square drawn from the hypotenuse of the right triangle. So very cool to see this

I think the step DE -> y(x+dx) - y(x) needs a bit more love. For the proof to work you would have to show that DE = y(x+dx) - y(x) + O(dx^2).

@gabrielbarrantes6946

7 ай бұрын

Yeah, this argument is what a physics would do, but a mathematician would do better calculations...

My favorite i s the geometrical one where you have a big square that is cut twice perpendicular to the sides, so that you end up with four parts where two of them are squares (A and B) and two are identical rectangles.The rectangles are then cut diagonally to create four identical right angled triangles. If you put the right angles of those triangles in the corners of the original square, they will leave a square shaped hole 'C' in the center with the sides made up by the hypotenuses of the triangles. Since the four triangles plus the newly formed square hole fill up the same square as the one when they were combined with squares A and B, the total area of C has to be the same as A + B.

@flamencoprof

7 ай бұрын

Using an opened by letter-knife typical mail envelope which has sides of length 1 and 2, you can prove Pythagoras' Theorem simply by folding. It is akin to what you have said. You start with two 1 x 1 squares, and end up with one root 2 by root 2 square.

@jakubfrei3757

7 ай бұрын

yeah, its nice and simple proof, unlike this insanity

@ouwebrood497

7 ай бұрын

Still my favorite proof, because it's so simple an elegant.

@gholler1

6 ай бұрын

@@jakubfrei3757 I prefer the geometric proof too but this proof is illustrative, in an easy to understand geometric context, of how applied mathematics (physic, biology, finance...) can transform a real world problem into a differential equation to be solved by reasoning with small increments and limits

@El0melette

3 ай бұрын

I think that proof came from Euclid

My favorite proofs are based on similar triangles. Let the triangle be ABC with opposing side lengths a, b, c, with c the hypotenuse. Drop a perpendicular from the right angle vertex C to the hypotenuse AB. Label the point of intersection D. Then triangles ABC, ACD, and CBD are all similar. From here, we can construct two proofs. The first uses the fact that areas of similar triangles are proportional to the squares of corresponding sides. Call the area of ACD x and the area of CBD y. Then the area of ABC is x+y. The areas are in proportion x:y:x+y and the hypotenuses are in proportion a^2:b^2:c^2, which is the Pythagorean theorem. The second proof uses the property that corresponding sides of similar triangles are in equal ratio. From △ABC ~ △ACD, we have that AD/AC = AC/AB, or AC^2 = AB × AD. From △ABC ~ △CBD, we have that BD/BC = BC/AB, or BC^2 = AB × BD. Adding these two equations, we have that AC^2 + BC^2 = AB × (AD + BD). But since AD + BD = AB, we end up with AC^2 + BC^2 = AB^2.

Starting with a triangle ABC which has a right angle at B, construct the altitude from B to a point D on AC. Given that both new triangles ADB and CDB share an angle with ABC, in addition to a right angle, we have that all three triangles are similar. By similarity, |AD|/|AB|=|AB|/|AC| and |CD|/|BC|=|BC|/|AC|. Clearing denominators gives us |AD|*|AC|=|AB|² and |CD|*|AC|=|BC|² Adding the two results in (|AD|+|DC|)*|AC|=|AB|²+|BC|² which, since D is between A and C, gives the Pythagorean Theorem

There was a book written in the 40's and published by I believe the NCTM that had a lot of proofs of this theorem.

@leif1075

7 ай бұрын

A lot? Wjat is a lot ? I'm not sure I can see more than. Afeq waus to prove this?

@ojas3464

7 ай бұрын

Loomis gives over 300 Proofs of Pythagorean Theorem. Assuming it had been around for over 3K years, would average an additional proof every decade

@sethkingman2118

7 ай бұрын

@@leif1075dunno about the book but there are likely hundreds. Check the Wikipedia page for the theorem to get an idea of the scope.

My favorite proof uses dimensional analysis, that is “checking the units” (I am a physics and math student): We have a right triangle with sides a and b, and hypotenuse c, and an acute angle theta. We know that a right triangle is completely and uniquely specified by its hypotenuse c and its acute angle theta. Therefore, the area a right triangle is a function of the hypotenuse and the acute angle: A(c, theta). c has dimension of length, [L], theta is dimensionless, and area has dimension of length squared, [L^2]. Therefore, by dimensional analysis, the area has to be proportional to c^2, and it’s dependence on theta is arbitrary: A(c, theta) = f(theta) * c^2. The important thing is that it’s the same function f(theta) for every right triangle. Now from the vertex of the right angle, draw a perpendicular to the hypotenuse, creating three similar right triangles with the angle theta (two small triangles and the original triangle). The area of the original triangle is A(c, theta). For the two smaller triangles, note that their hypotenuses are a and b, and therefore their areas are A(a, theta) and A(b, theta). The sum of these areas equals the area of the original triangle, which gives us the equation: A(a, theta) + A(b, theta) = A(c, theta) f(theta) * a^2 + f(theta) * b^2 = f(theta) * c^2 a^2 + b^2 = c^2 which is the Pythagorean theorem. I really like this proof since it’s extremely simple, elegant and can be taught to any kid who knows that length is measured in meters and area is measured in meter squared. This proof essentially says that the fact that area is measured as squared length forces the Pythagorean theorem to be true.

@sergiogiudici6976

Ай бұрын

I am a Physicist too and I know we love dimensional analysis. It Is 90% of our everyday job.

@shaiavraham2910

Ай бұрын

@@sergiogiudici6976 I had a test in Wave Mechanics, and I knew nothing about the material. For the entire time of the test, I am just saying to myself: “My answer has units of energy, so I’ll just leave it like this, and pretend it works”. I was shocked to find out I got a 3 digit grade.

Sir your every segment is mind blowing..

Given triangle of sides a:b:c, where a

I like this channel - it reminds me of my two years of university math courses actually almost three (majored in CS). Also would like to say that there are some 80s vibes over his videos dunno why...

Very cool Michael. Math is awesome.

I liked how the proof manages to use only bits of calculus that don`t themselves require the pythagorean theorem. Very neat

Thank you professor!

Next: Let's prove 1+1=2 using the Riemann zeta function 😂

Parabéns, grande professor Michael! As suas aulas são um show. A sua contribuição, sem dúvida alguma, é de grande valia para quem gosta de matemática como eu dezenas de milhões mundo afora.

I think my favorite proof of the Pythagorean theorem is that it follows trivially from Ptolemy's theorem (in my opinion one of the coolest results from ancient mathematics).

Dude! The quality of your chalk! That's no cheap chalk! 8:41 , I've never used red chalk that made such vivid colour. Also Very cool application of Geometry and Limits. First time seeing it used like this! Could anyone point me in a direction of where an accessible entry hole is for this rabbit hole?

Love the chalk marks on Penn's shirt. So on brand!

My favorite proof of the Pythagorean Theorem is just repeated skewing of parallelograms, because it preserves area. What I'd be interested in seeing though is a proof of the hyperbolic equivalent of the pythagorean theorem in something like Minkowski space, where you have Δt²-Δx² = Δτ² because of the +--- metric.

So the Pythagorean theorem doesn't necessarily hold in this form in non-Euclidean geometries. This suggests that there is some subtle flaw in this proof. I think in general the angles of a triangle won't sum to pi, and perhaps even the tangent to "circles" don't form necessarily right angles.

@arnouth5260

7 ай бұрын

In non-euclidean geometries the angles of a triangle don’t sum to 180°, which is used in the proof.

@andrewkarsten5268

6 ай бұрын

Yes that is right, the sum of the angles of triangles is not pi in noneuclidean geometry, and actually how far away the sum is from pi depends on how large the triangle is and the curvature of the space it lives in.

the only one I could think of was: consider triangle ABC, where B has the right angle. set point P on AC so that the line PB is perpendicular to AC. due to same angles, triangles ABC, ABP, and BCP are similar. this mean that: AB/AP=AC/AB; BC/PC=AC/BC --> AB^2=AC*AP ; BC^2=AC*PC. adding the last 2 equations gives AB^2+BC^2=AC(AP,+PC) but AP+PC=AC and thus AB^2+BC^2=AC^2.

I'd be interested to see you explore the recent "trig-less" proof involving the limit of a series of similar triangles, that those two schoolgirls came up with, recently

I can't help thinking about how Pythagoras, Archimedes, and their contemporaries would react if they were presented with this. Would they accept or reject the proof?

@davidcroft95

7 ай бұрын

Probably reject, they had no idea of infinitesimals (look the Zeno's Paradoxes)

@sturlabingen2471

7 ай бұрын

Iirc Archimedes did use calculus like thinking to arrive at results but at the time they were not accepted as rigouros enough so he also created purely geometric proofs@@davidcroft95

@munchkinhut

7 ай бұрын

Hell no! (in that they would reject the proof) They would understand that the proof is based on the assumption that pythagorean theorem exists in the first place. (Classic Euclid) Archimedes would have enclosed the possibilities into a trap that only one can survive in. THERE CAN BE ONLY ONE!! Drawn it out on sand and get killed by some a cretin whose race wouldn't figure it out for another 1000 years. The Pythagoreans would all eat a salad and kill anyone that mentioned it.

@iabervon

7 ай бұрын

@@davidcroft95 I think Archimedes would have been okay with infinitesimals in the geometry (based on his method of deriving the volume of a sphere), but all of the symbol manipulation would have been outside of the mathematics whose validity they would have any idea how to judge. Their only notion of variables was the measures of parts of a diagram, so expressing and solving that differential equation would be a major conceptual leap, along with a large number of theorems they didn't know.

@gibbogle

7 ай бұрын

Since they had no knowledge of calculus they would have been baffled.

This is my favorite proof♥️

honestly, my favorite would probably be the classic scalar product approach. not only does it prove the pythagorean theorem for vectors, but also for ANY vector space with a positive definite scalar product, which i think is pretty nifty.

I had a few ancient proofs but for the students I teach.. this is one for the ages !

Try this: a²+b²=(a+b)²-2ab Consider: c²+2ab Factoring: c²(1+2(a/c)(b/c))=c²(1+2sinAcosA) The value 2ab/c² is the double angle trig ratio sin2A, (see ‡ at the bottom) Consider: (a+b)²/c² This is equivalent: (sinA+cosA)² Expanding out: sin²A+cos²A+2sinAcosA Subs for 2sinAcosA & due to “†” & “‡” (bottom), this is: 1+sin2A. From “1”, have: c²+2ab=c²(1+sin2A) From “2”, have: (a+b)²/c²=1+sin2A. Therefore: c²=(c²+2ab)/(1+sin2A)=(c²+2ab)/(a+b)²/c² Multiplying by (a+b)²/c² on both sides: (a+b)²=c²+2ab => (a+b)²-2ab=a²+b²=c² …QED 😊. (†)From the compound angle formulas(‡) for sin & cos, you get double angle formulas for those, and from cos2A you get: 1-2sin²A=2cos²A-1=cos2A => 2=2(cos²A+sin²A) Therefore: 1=(cos²A+sin²A) (‡)Proof in this video: “Angle sum identities for sine and cosine” by blackpenredpen

Place F on AE such that angle ACF is a right angle. (F is then close to D). Then the small triangle DCE is clearly similar to the original triangle for any angle, not just in the limit. That to me feels like a more rigourous set up, and we can do the same argument from that point, as in the limit F and D are the same.

@yannld9524

7 ай бұрын

You meant "FCE is clearly similar" ? This is wrong because the angle at F is not right

Wow! Fascinating ...

Using an opened by letter-knife typical mail envelope which has sides of length 1 and 2, you can prove Pythagoras' Theorem simply by folding.

This deserves a Field medal

Nice result, but there's a bit if hand waving there when you assume that the differences between the two triangles are infinitessimal when you approach the limit.

Taking DE -> y(x+dx) - y(x) in the limit that dx->0 is a bit strange to me, because in that limit we also have that y(x+dx) -> y(x), so this just seems like a complicated way of saying DE -> 0.

@erikb.celsing4496

7 ай бұрын

y(x+dx)-y(x) always approaches 0 no matter what kind of (nice) function y is. It’s is the quotient that does not

@chasebender7473

7 ай бұрын

That is only true when y is a continuous function, which was never proved

@perappelgren948

7 ай бұрын

Agree, formally Δx is already bound by the limit taking, so it’s no more available for usage in the RHS. But that is not Michael’s point here, what he is really saying is that higher order terms vanish.

Fantastic proof. Who originated it?

At 9:50 I think your reasoning was wrong. Using the same kind of argument you could say: as x approaches zero 2x appeoaches x, therefore lim(2x/x)=lim(x/x)

When I clicked on this video, I expected the more recent proof found by a couple of college students a couple of years ago that relies on infinite series and works only for non-isosceles right triangles: Let the length of the larger leg be a and of the smaller leg be b; then on the leg of length b, construct a similar right triangle in which the longer leg has length b, and extend the hypotenuses of both triangles until they intersect (this is why the legs cannot have the same length), and keep drawing line segments perpendicular to the previous one, marking off similar right triangles infinitely often. Then, looking at the next right triangle, its longer leg has length b=a*(b/a) and its shorter leg has length b²/a=b*(b/a); from this, the nth triangle after the first has legs of length a*(b/a)^n and b*(b/a)^n and area ½ab*(b/a)^(2n), from which the whole collection of right triangles has area ½ab/(1−(b/a)²)=½ba³/(a²−b²). If the hypotenuse of the original right triangle has length c, then the hypotenuse of the next one has length cb/a and the extension of that original hypotenuse, as a leg of the larger triangle, has length c/(1−(b/a)²)=ca²/(a²−b²), while the extension of that next hypotenuse has length abc/(a²−b²); the triangle formed by dropping a line segment from the vertex between the sides of lengths c and a perpendicular to the other "extended hypotenuse" is also similar to the original right triangle…and I'm honestly not sure about how the proof goes from there, but you're supposed to add the area of this to the area of the whole collection and equate this to the area of a larger right triangle that contains all of the others, and after a bunch of cancellations, the end result should be c²=a²+b², but I kept making some sort of mistake while finishing it up.

Excellent. 👍

As soon as you have "similar triangles exist" the Pythagorean theorem follows immediately.

The radius intersects de and not ad.

This how a physicist would prove the Pythagorean theorem.

I would have solved the DEQ by integrating y*dy=x*dx. Since the integration gives rise to a constant, the initial condition y(0)=a is incorporated by setting x=0.

@andrewkarsten5268

6 ай бұрын

That’s what he did but without introducing unnecessary constants of integration that need to be solved later. Using definite integrals when you can during DE solving is always better as a practice

13:58

No comment regarding the left-hand derivative of y(x)?

@d4slaimless

7 ай бұрын

What do you mean?

@neilbellinson5292

7 ай бұрын

@@d4slaimless A derivative, as a Newton quotient, is a limit as delta x goes to zero not just as delta x goes to zero from the right.

The limit definition of the derivative of y is the limit of the difference quotients, IF that limit exists. You should justify the limit’s existence before saying it is y’.

@divisix024

7 ай бұрын

He already found it to be x/y beforehand, which a fortiori shows the existence.

@tcmxiyw

7 ай бұрын

@@divisix024 I agree with you. Michael is an excellent expositor of mathematics, in no small part because he always verifies the hypotheses of a theorem before he applies it. I intended by my comment, which is as a minor critique at most, not to say that he hadn’t shown it, but that observing all the conditions of the definition were met before it is used illustrates how to correctly use definitions. I should have stated that thought more clearly.

c=a+b c^2 = (a+b)^2 = [a^2 + b^2] + [2ab] (binomial expansion) c^2 a^2+b^2 The "proof" in the video is only valid in the imagination. (Pythagoras was also confused).

Was that differential equation solution necessary, if I were him (which I'm obviously not) I would have written dy/dx instead of y' and then cross multiplication and taken integral of both sides, I am just wondering.,

@0Coeus

7 ай бұрын

Thats essentially what he did. He just wrote it down somewhat more rigorously. The method you described is still a method for solving a differential equation :)

@BenfanichAbderrahmane

7 ай бұрын

dy/dx is not a fraction 😃

@d4slaimless

7 ай бұрын

selimakar7201, what you suggest is called "abusing the notation". So Mr. Penn just solved it rigorously.

@ojas3464

7 ай бұрын

yy' = x can be rewritten in Leibnitz notation ½d(y²) = ½d(x²)

@ianfowler9340

7 ай бұрын

@@BenfanichAbderrahmane Ahh, but it does behave like a fraction (for multiplying and dividing only) because "diff of y = f ' (x) * diff of x" where diff of x is defined to be: delta x. And then, diff of y is defined to be: dy =f '(x)*delta x = f '(x)dx Therefore in the end: "dy by dx" = diff of y / diff of x That's why separation of variables works the way it does. BTW there is no requirement that diff of x be inf. small and the dame for diff. of y. The key is the their ratio will always be the slope of the tangent. It's the way we define dx and dy that allows dy divided by dx to dy by dx

ydy-xdx = 0 implies d(y^2-x^2) = 0, so y^2-x^2 = cst = a^2 since y)0)=a

How can you prove that y is C1?

@xinpingdonohoe3978

7 ай бұрын

And what is C1? Am I missing something?

@user-nm5ge9ht3c

7 ай бұрын

@@xinpingdonohoe3978 A function being C1 means that it is continuous, differentiable and its derivative is continuous

@xinpingdonohoe3978

7 ай бұрын

@@user-nm5ge9ht3c oh, that's what he was referring to.

✨✨ AWESOME ✨✨ Defn of Calculus..... Maths in Motion ...

12:40 Bro really wrote out a whole "u substitution" just for u = y. It's simple, cancel the dt. You have Integral y dy.

Why isn’t the limit of y(x+delta x)-y(x) just 0?

@gamesandthoughts2388

7 ай бұрын

If delta(x) -> 0, and y(x) is a continuos function then you are right, it is zero.

@Happy_Abe

7 ай бұрын

@@gamesandthoughts2388so length of DE should be approaching 0

@gibbogle

7 ай бұрын

Of course it is. We have in the limit 0/0, the value of that limit depending on how both 0s are approached. This is fundamental calculus.

@Happy_Abe

7 ай бұрын

@@gibbogle okay good, I thought Michael was saying something else, thanks

@Yougottacryforthis

7 ай бұрын

@@Happy_Abeit's only zero if y(x) is continous

I used to understand this shit in college.

Pedagogically the t substitution in the integral is problematic, no? In the integral the t is a dummy variable and could have been any variable name. I worry though that the video presentation is meant to make it look like some obvious choice was made given the equations presented thus far. The obvious choice then would be "y dy" instead if "t dt", based on the chain rule.

@andrewkarsten5268

6 ай бұрын

No

The argument about the tangent line creating a right angle seems unnecessary, unless I'm missing something. Triangle CDE is similar to triangle ABE because they share angle E, and they both contain a right angle. Then as dx-->0, triangle ABE becomes arbitrarily close to triangle ABC, so in the limit, that's also similar to CDE. But...I probably am missing something...

@gibbogle

7 ай бұрын

It's necessary to use the fact that only in the limit is CDE a right angle.

@jacemandt

7 ай бұрын

@@gibbogle angle CDE is constructed to be a right angle to begin with, though.

@gibbogle

7 ай бұрын

@@jacemandt Sorry, you are right, CDE is a right angle by construction. ACD approaches a right angle as dx ->0.

Is this argument circular? (pun very much intended) (d/dt) (sin(t)^2 + cos(t)^2) = 2sin(t)cos(t) - 2sin(t)cos(t) = 0 So for all t, sin(t)^2 + cos(t)^2 = sin(0)^2 + cos(0)^2 = 1 and therefore, Pythagoras.

The best proof of Pythagoras is the oldest proof. It uses constructed squares and is the first one given in Wikipedia. I believe the Chinese were using this at the time of Euclid.

@Michael-ch8hq

7 ай бұрын

I believe euclid was using it at the time of the Chinese

I was doing really well until it got to the integration part. I'll come back to it, just in case I meet Pythagoras in any afterlife :) :)

You seem to be saying that the two hypotenuses are parallel in the limit. For some reason that does not suit well with me...

@RandyKing314

7 ай бұрын

methinks the twain shall be one

A nine inch pizza and a twelve inch pizza have as much food as a fifteen inch pizza

Nice

Can't yiu solve without drawing the little right triangle at the top? A lit of ppl might not think of that

Why complicate the solution of the differential equation? Why not just separate variables and integrate y dy = x dx ?

@andrewkarsten5268

6 ай бұрын

It’s not complicating it, it’s treating it with the proper rigor. In general, when solving DE, it’s always better to use bounds of integration when possible.

@godfreypigott

6 ай бұрын

@@andrewkarsten5268 Perhaps you'd care to explain *WHY* separation of variables is not rigorous. What is lost in doing it that way? What are the potential pitfalls? With regard to your second sentence: (1) *WHY* is it "always better"? By what metric are you measuring "better"? I suspect your interpretation of "better" is "it looks slicker". (2) In any case, why do you believe separation of variables doesn't allow you to involve bounds?

@andrewkarsten5268

6 ай бұрын

@@godfreypigott the formal rigorous definition of the antiderivative is that F(x) is an antiderivative of f(x) if F(x)=\int_a^x f(t) dt for some constant a. Also, when solving a DE, just throwing up a ∫ symbol on each side, while visually pleasing and somewhat useful for problem solving, is not a rigorous approach for a proof. To actually do things rigorous enough for a proof, you are supposed to integrate both sides with respect to the same variable, be it definite or indefinite integrals. While your approach is the correct “symbol pushing” approach as my old professor would call it, and is extremely useful for getting your hands on a solution to a new problem, when writing a proper proof it is not rigorous enough. Sorry 🤷‍♂️

@godfreypigott

6 ай бұрын

@@andrewkarsten5268 You keep using the term 'rigorous' without explaining *WHY* SOV is not rigorous. That is, what is lost by using SOV. If you really want to be fully rigorous then surely you should be going back to integration by first principles. If not, where do you arbitrarily draw the line? The answer of course is that you draw the line at the point where overgeneralisation could lead to unforeseen complications. But even that issue can be addressed by properly considering special cases. So tell me, what complications could potentially arise by using SOV instead of the method presented here? Alternatively, what special cases does SOV not handle in relation to this problem? These should be the only criteria for labelling a method as non-rigorous, otherwise "rigour" is being used unthinkingly only as a means of asserting superiority without justification.

@andrewkarsten5268

6 ай бұрын

@@godfreypigott it seems you didn’t read my response. I used the formal definition of the antiderivative, which are definite integrals. I also explained when integrating, you need to integrate with respect to the same variable on both sides. SOV you just throw up a ∫ on both sides but are integrating with respect to different variable on each side. SOV is a good and useful symbol pushing approach to get your hands on a tangible solution, but it is not sufficient for a proof. Also, if you pay close enough attention, you’ll see that doing it the proper way with a u substitution gives the same overall solution. When working a problem to find a solution, SOV is fine. When writing a proper proof, you need to use what I have listed above. Read this comment completely before responding again.

I'm having a bit of a problem with this proof. It's kind of "high level," but the idea is what would have to change here if we weren't doing this on a plane? The answer is the triangles ABC and CDE: we would not be able to say that they are similar in the limit if we were, say, doing this on the surface of a sphere. I am not certain, but I'm getting the idea that because we are using the similarity argument, we are forcing the Pythagorean Theorem: ie. we are assuming the result that we are trying to prove. I may be wrong. I'll have to think about it some more.

@xinpingdonohoe3978

7 ай бұрын

If we weren't on a plane, these wouldn't be planar triangles, so what would we even be doing?

@minamagdy4126

7 ай бұрын

The Pythagorean theorem does rely on the parallel postulate, so talking about it applying to non-planar triangles makes no sense. The proof here doesn't overstep the bounds of Euclidean geometry, so it's fine.

@williamthomsomkelvin1765

7 ай бұрын

Pythogorean Theorem is a result valid only in 2-dimensional Euclidean Geometry

@TheEternalVortex42

7 ай бұрын

The Pythagorean Theorem is equivalent to being on a plane. Indeed that's an assumption of every proof of it lol

@zh84

7 ай бұрын

I saw from the start that the proof is only valid in Euclidean geometry: in non-Euclidean planes there are no such things as similar triangles (which is an alternative form of the parallel postulate.) However, the Pythagorean theorem is only valid in the Euclidean plane anyway, so this isn't a problem.

thaNK YOU REVIEW CAN NOT KEEP ALL FRESH

Do you just have a hundred TAs running frantically behind the camera shaking that that won’t draw a nice triangle?!?!?😂😂😂

We don't need the circle, CDE will always be similar to ABC

Ο Μάικλ Πεν είναι ο καλύτερος! This is exactly what Isaac Newton would had done! This is the conclusive proof that Mathematics is definitely better than sex!

@gibbogle

7 ай бұрын

I doubt that Newton ever had sex with another person, so he no doubt would have agreed.

👍

I like Garfield's proof.

My preferred proof is Perigal's dissection.

@charlesbrowne9590

7 ай бұрын

I googled ‘Perigal’ and, yes, that is a pretty proof. Thx.

@d4slaimless

7 ай бұрын

I think I like this one as well. Especially since it is the one from infinite family of proofs.

Pythagoras is actually invented by famous hindu mathematician "buddhayan", he gave Taylor series too but christian missionaries stole his works

The famous . . . presented by the infamous Mike Penn.

Interesting but I think there are way more simpler and intuitive ways to prove the theorem.

Opop

After y.y' = x, why not just observe that the derivative of y^2 is 2y.y', giving y^2 = x^2 + c?

@robertpearce8394

7 ай бұрын

He did state that he was using the rigorous definition of anti-derivative.

Too much of hand waiving argument for my taste. In the limit, the triangle in question becomes a point. A proper estimate (upper bound and lower bound) should have been the way.

I am an economic hostage.

The proof is based on calculus which is based on pythagorus theorem . Isn't it meaningless

I kind of feels to me that using a circle and tangents is a bit of cheating, but I'm not certain if it is.

@OuroborosVengeance

7 ай бұрын

It is not. Its bas>c geometry: you can always make a circle out of a point and a lenght/radius

@OuroborosVengeance

7 ай бұрын

And the similarity of the small triangle to the original one holds even without using limits

@jacemandt

7 ай бұрын

@@OuroborosVengeance Hmmmm, without limits, I think the small triangle is similar to ABE, not ABC

Similarity, alone, can be used to prove Pythagoras. So, your proof is an unnecessary exercise.

The provided proof is “circular reasoning,” as it uses trigonometry (which is a result of the Pythagorean theorem

@mathematicskid

7 ай бұрын

It does not use trigonometry. It uses calculus, which is not trigonometry.

@donach9

7 ай бұрын

Where exactly was the trigonometry? I see no sin or cos or any length squared until we use the power rule for integration

@mathematicskid

7 ай бұрын

@@donach9 It does not use trigonometry.

I’m sad to announce that the angle ABC is sadly a convex angle at 270 degrees. CBA on the other hand is a right angle.

@OuroborosVengeance

7 ай бұрын

Not true. You declare the orientation when you write the symbol for internal angle.

@Umbra451

7 ай бұрын

@@OuroborosVengeance yes, actually true.

You can't really call that a proof

@donach9

7 ай бұрын

Why not?

This is famous? He loses me as soon as he constructs the blue ' perturbation'. What's the point?