9 ms·

There are thousands of different proofs of the Pythagorean theorem, and some of them are really cool. The purely trigonometric proof that was found by some high

by WCSTombs 10mo ago

There are thousands of different proofs of the Pythagorean theorem, and some of them are really cool. The purely trigonometric proof that was found by some high school students recently is a great one. However, I think the greatest proof of all is this little gem that has been attributed to Einstein [1]. Take any right triangle. You can divide it into two non-overlapping right triangles that are both similar to the original triangle by dropping a perpendicular from the right angle to the hypotenuse. To see that the triangles are similar, you just compare interior angles. (It's better to leave that as an exercise than to describe it in words, but in any case, this is a very commonly known construction.) The areas of the two small triangles add up to the area of the big triangle, but the two small triangles have the two legs of the big triangle as their respective hypotenuses. Because area scales as the square of the similarity ratio (which I think is intuitively obvious), it follows that the squares of the legs' lengths must add up to the square of the hypotenuse's length, QED. It's really a perfect proof: it's simple, intuitive, as direct as possible, and it's pretty much impossible to forget. [1] https://paradise.caltech.edu/ist4/lectures/Einstein%E2%80%99s%20Boyhood%20Proof%20of%20the%20Pythagorean%20Theorem%20-%20The%20New%20Yorker.pdf https://paradise.caltech.edu/ist4/lectures/Einstein%E2%80%99...

bryanrasmussen 10mo ago
unfortunately doesn't work for me because of difficulty visualizing things, so I suppose there are probably a good number of people with the same problem. So I guess for one particular subset of the population it is difficult, impossible to understand, and because it cannot be understood it will not be remembered. Not complaining just noting the amusing thing that different explanations may have all sorts of problems with it. Although if there was a video of it I guess I would understand it then. Not sure if everyone with visualization issues would though.
- WCSTombs 10mo ago
  To be fair, I'm constrained by plain text on Hacker News. The argument I wrote down requires a diagram to be fully understood, so I described it in words expecting the reader to draw it themselves, or at least mentally visualize it (for those used to doing it). To be clear though, as far as I know every proof of the Pythagorean Theorem requires some sort of diagram, and the one I gave requires literally the least amount of drawing out of all the proofs (which is a bold claim, but call it a conjecture). That's why I felt comfortable writing out the proof just in words.
zeroonetwothree 10mo ago
This proof assumes that the area a triangle is some function k c^2 of the hypotenuse c where k is constant for similar triangles. This doesn’t seem super obvious to me, and it’s a bit more than just assuming area scales with the square of hypotenuse length, it indeed needs to be a constant fraction. To me that truth isn’t necessarily any less fundamental than the Pythagorean theorem itself. But to each their own. BTW Terrence Tao has a write up of this proof as well: https://terrytao.wordpress.com/2007/09/14/pythagoras-theorem/ https://terrytao.wordpress.com/2007/09/14/pythagoras-theorem...
- fiso64 10mo ago
  I don't get his "modern" proof. Specifically the step where he says "it's easy to see geometrically that these matrices differ by a rotation" seems to be doing a lot of heavy lifting. The first matrix transforms e1 to (a,-b), the second scales e1 to (c,0). If you can see that you obtain one of these vectors by rotating the other, then you've shown that their lengths are equal (i.e. a²+b²=c²), which is what we want to show in the first place.
  - degamad 10mo ago
    You're assuming that we know that the length of vector (a, -b) is a²+b². We don't know that. We start by assuming that the position vector (a, -b) has length c. This implies that we can rotate that vector until it becomes the position vector (c, 0). As you note, we can create the two vectors above from (1, 0) using linear transformation matrices [(a, b), (-b, a)] and [(c, 0), (0, c)] So we could create the position vector (c, 0) by starting at (1, 0), applying the linear transformation [(a, b), (-b, a)], then applying a rotation to bring it back to the e1 axis. Thus for some rotation matrix R, R × [(a, b), (-b, a)] = [(c, 0), (0, c)] The determinant of a rotation matrix is 1, so the determinant of the left side is 1×(a²+b²), while the determinant of the right side is c², which is how we end up with a²+b²=c². Now the only thing which I'm not sure of is whether there's a way to show that the determinant of a rotation matrix is 1 without assuming the Pythagorean identity already.
    - 10mo ago
zeroonetwothree 10mo ago
I think proof #6 on this page is easier to follow and uses the same similar triangles. But then it’s just some basic algebra without assuming anything about areas of similar triangles :) https://www.cut-the-knot.org/pythagoras/index.shtml#6 https://www.cut-the-knot.org/pythagoras/index.shtml#6
- WCSTombs 10mo ago
  That one's also very neat!
dataflow 10mo ago
> it follows that... "Now just draw the rest of the owl." Does it not feel like you skipped something here? The areas add up and area scales quadratically, therefore... Pythagorean Theorem? It definitely is not clear how this follows, even after the questionable assumption that it's obvious area scales quadratically.
- codethief 10mo ago
  He didn't skip anything but he left the "obvious" details (for a mathematician) to the reader: Let C be the area of the big triangle, A and B be the areas of the two small triangles. By construction we know that C = A + B. Moreover, a, b, c are the hypotenuses of the triangles A, B and C. The area scaling quadratically with the similarity ratio means that A = (a/c)² C, and B = (b/c)² C. Now, plug this into A + B = C, cancel C, rearrange.
  - dataflow 10mo ago
    The math is obvious enough, I agree. But the description of the approach feels like it's lacking something - specifically, something along the lines of "now write down the scaling equations and simplify the area summation." I feel like it's not at all clear they're switching to an algebraic argument there.
    - WCSTombs 10mo ago
      The thing is that in my head there is no algebraic argument: we go from (1) similarity ratios being A:B:C and (2) the first two areas adding up to the third area, straight to the conclusion of A^2 + B^2 = C^2. I think your point about a step being missing here is valid, but when I search my intuition, it's still not coming up as algebraic. I suspect this is the same for others like me who are inclined to think geometrically, but I'd like to hear their opinions. Here's an attempt at filling in the geometric intuition with something more concrete. You know how it's common to visualize the theorem with squares on the three sides of the triangle and saying that the two small squares add up to the big one? And then everyone stares at it and says "huh?" because that fact is far from obvious from that diagram. Here's the thing though, we're free to choose different area units if we want. So just choose units where our triangle itself with a given hypotenuse H has area H^2 units. Then we can give the argument above without any extra factors and cancellations. To fully justify the "choose any units," you do need to check that it's logically consistent, which you could say is more missing steps, but I think this idea is far more fundamental than the Pythagorean Theorem. Our use of squares to define the fundamental units of area really is a completely arbitrary choice. We call them "square units," which already biases us to think of area in a specific way, but there's absolutely no reason we can't use any other shape. Of course squares are convenient because you can stack them up neatly and count them, but that doesn't seem to be helpful at all in this context, so it's natural to choose something else.
card_zero 10mo ago
Einstein's proof relies on the fact that the theorem works with any shape, not just squares, such as pentagons: https://commons.wikimedia.org/wiki/File:Pythagoras_by_pentagons.svg https://commons.wikimedia.org/wiki/File:Pythagoras_by_pentag... Or any arbitrary vector graphics, like Einstein's face. So in the proof, the shape on the hypotenuse is the same as the original triangle, and on the other two sides there are two smaller versions of it, which when joined have the same area (and shape) as the big one. Fair enough. However, none of the hundreds or thousands of proofs explain it. They all prove it, like by saying "this goes here, that goes there, this is the same as that, therefore logically you're stupid," but it still seems like weird magic to me. Some explanation is missing.
- chronial 10mo ago
  Draw a square around Einstein's face. Call the side length of the square a and the area of the square A. We have A=a^2. Einstein takes up some portion p < 1 of that area, so Einstein has area E = pA. Now we scale the whole thing by factor f. So the new square has side lengths fa, and thus area A' = (fa)^2 = f^2×a^2 = f^2×A. Since the relative portion the face takes up doesn't change with scaling, the face now has size pA' = p×f^2×A = f^2 × pA = f^2 E. Does that help or was that not the part you were missing?
  - card_zero 10mo ago
    No, that part is fine: I'm happy with the fact that it works with arbitrary shapes. What bothers me is that the area on the hypotenuse is equal to the sum of the areas on the other two sides, when the triangle has a right angle. This somewhat like saying that I'm troubled by the fact that 1+1=2, I know. But that's a potentially distracting sidetrack, let's not get into that one.
- lupire 10mo ago
  What definition of area are you using in the first place, for non-swuare objects? Most people find area intuitive and informal, but if you describe area formally, it should be easy to use your definition to account for scaling.
JanisErdmanis 10mo ago
Indeed, a wonderful proof. It does, though, make one implicit assumption that if one stretches the fabric by the same amount, all holes in it stretch by the same amount. In particular, it assumes that triangle stretching is size-independent. Perhaps there are fabrics where that is not true...
- lupire 10mo ago
  You mean non-Euclidean fabric? https://www.johndcook.com/blog/2022/09/08/trig-hyperbolic-geometry/ https://www.johndcook.com/blog/2022/09/08/trig-hyperbolic-ge... https://math.hmc.edu/funfacts/spherical-pythagorean-theorem/ https://math.hmc.edu/funfacts/spherical-pythagorean-theorem/ https://en.wikipedia.org/wiki/Minkowski_distance https://en.wikipedia.org/wiki/Minkowski_distance
  - JanisErdmanis 10mo ago
    That is one possibility. Probably the only one, though.
- layer8 10mo ago
  The Pythagorean theorem is only true in Euclidian space, where that “stretching” assumption is true. So you are right about there being assumptions, and indeed they are imposing limitations to the applicability of the theorem.
lupire 10mo ago
> The purely trigonometric proof that was found by some high school students recently is a great one. It was geometric, using trigonometric vocabulary.
- SJC_Hacker 10mo ago
  Trig relies on some geometric assumptions. The definition of sin and cos is going to require a right triangle, for instance,
SJC_Hacker 10mo ago
> The purely trigonometric proof that was found by some high school students recently is a great one. I failed to understand what was so cool about that proof. It relied on concepts such as Cartesian coordinate systems, and the measure of an angle (not just a pure geometric concept), and even concepts like convergence of infinite sums, which weren't purely geometric. Geometry had been formalized in the 20th century and had moved past informal proofs
globalnode 10mo ago
That is interesting and made me think. Only after following some of the other subcomments did I manage to understand it. Personally, replacing the word similarity ratio with scale factor made all the difference. At first I thought it was a circular argument, relying on pythag to prove pythag but that scale factor is the key actually, and the fact that side lengths scale linearly but the area scales quadratically. It feels like a similar trick we see when adding logarithms gives us multiplication.
- WCSTombs 10mo ago
  Yeah, the similarity ratio/scale factor and its connection to area is the key part of the proof. Sorry, that could have been clearer.
  - globalnode 10mo ago
    All good! thanks for the maths though, just when I think I understand it... I dont! Its one of those love hate relationships, I love it, it hates me.
layer8 10mo ago
Also, iterating that triangular subdivision gives rise to the Pythagorean tree fractal: https://en.wikipedia.org/wiki/Pythagoras_tree_(fractal) https://en.wikipedia.org/wiki/Pythagoras_tree_(fractal)

bryanrasmussen 10mo ago

unfortunately doesn't work for me because of difficulty visualizing things, so I suppose there are probably a good number of people with the same problem. So I guess for one particular subset of the population it is difficult, impossible to understand, and because it cannot be understood it will not be remembered. Not complaining just noting the amusing thing that different explanations may have all sorts of problems with it. Although if there was a video of it I guess I would understand it then. Not sure if everyone with visualization issues would though.

WCSTombs 10mo ago
To be fair, I'm constrained by plain text on Hacker News. The argument I wrote down requires a diagram to be fully understood, so I described it in words expecting the reader to draw it themselves, or at least mentally visualize it (for those used to doing it). To be clear though, as far as I know every proof of the Pythagorean Theorem requires some sort of diagram, and the one I gave requires literally the least amount of drawing out of all the proofs (which is a bold claim, but call it a conjecture). That's why I felt comfortable writing out the proof just in words.

zeroonetwothree 10mo ago

This proof assumes that the area a triangle is some function k c^2 of the hypotenuse c where k is constant for similar triangles. This doesn’t seem super obvious to me, and it’s a bit more than just assuming area scales with the square of hypotenuse length, it indeed needs to be a constant fraction. To me that truth isn’t necessarily any less fundamental than the Pythagorean theorem itself. But to each their own. BTW Terrence Tao has a write up of this proof as well: https://terrytao.wordpress.com/2007/09/14/pythagoras-theorem/ https://terrytao.wordpress.com/2007/09/14/pythagoras-theorem...

fiso64 10mo ago
I don't get his "modern" proof. Specifically the step where he says "it's easy to see geometrically that these matrices differ by a rotation" seems to be doing a lot of heavy lifting. The first matrix transforms e1 to (a,-b), the second scales e1 to (c,0). If you can see that you obtain one of these vectors by rotating the other, then you've shown that their lengths are equal (i.e. a²+b²=c²), which is what we want to show in the first place.
- degamad 10mo ago
  You're assuming that we know that the length of vector (a, -b) is a²+b². We don't know that. We start by assuming that the position vector (a, -b) has length c. This implies that we can rotate that vector until it becomes the position vector (c, 0). As you note, we can create the two vectors above from (1, 0) using linear transformation matrices [(a, b), (-b, a)] and [(c, 0), (0, c)] So we could create the position vector (c, 0) by starting at (1, 0), applying the linear transformation [(a, b), (-b, a)], then applying a rotation to bring it back to the e1 axis. Thus for some rotation matrix R, R × [(a, b), (-b, a)] = [(c, 0), (0, c)] The determinant of a rotation matrix is 1, so the determinant of the left side is 1×(a²+b²), while the determinant of the right side is c², which is how we end up with a²+b²=c². Now the only thing which I'm not sure of is whether there's a way to show that the determinant of a rotation matrix is 1 without assuming the Pythagorean identity already.
  - 10mo ago

zeroonetwothree 10mo ago

I think proof #6 on this page is easier to follow and uses the same similar triangles. But then it’s just some basic algebra without assuming anything about areas of similar triangles :) https://www.cut-the-knot.org/pythagoras/index.shtml#6 https://www.cut-the-knot.org/pythagoras/index.shtml#6

WCSTombs 10mo ago
That one's also very neat!

dataflow 10mo ago

> it follows that... "Now just draw the rest of the owl." Does it not feel like you skipped something here? The areas add up and area scales quadratically, therefore... Pythagorean Theorem? It definitely is not clear how this follows, even after the questionable assumption that it's obvious area scales quadratically.

codethief 10mo ago
He didn't skip anything but he left the "obvious" details (for a mathematician) to the reader: Let C be the area of the big triangle, A and B be the areas of the two small triangles. By construction we know that C = A + B. Moreover, a, b, c are the hypotenuses of the triangles A, B and C. The area scaling quadratically with the similarity ratio means that A = (a/c)² C, and B = (b/c)² C. Now, plug this into A + B = C, cancel C, rearrange.
- dataflow 10mo ago
  The math is obvious enough, I agree. But the description of the approach feels like it's lacking something - specifically, something along the lines of "now write down the scaling equations and simplify the area summation." I feel like it's not at all clear they're switching to an algebraic argument there.
  - WCSTombs 10mo ago
    The thing is that in my head there is no algebraic argument: we go from (1) similarity ratios being A:B:C and (2) the first two areas adding up to the third area, straight to the conclusion of A^2 + B^2 = C^2. I think your point about a step being missing here is valid, but when I search my intuition, it's still not coming up as algebraic. I suspect this is the same for others like me who are inclined to think geometrically, but I'd like to hear their opinions. Here's an attempt at filling in the geometric intuition with something more concrete. You know how it's common to visualize the theorem with squares on the three sides of the triangle and saying that the two small squares add up to the big one? And then everyone stares at it and says "huh?" because that fact is far from obvious from that diagram. Here's the thing though, we're free to choose different area units if we want. So just choose units where our triangle itself with a given hypotenuse H has area H^2 units. Then we can give the argument above without any extra factors and cancellations. To fully justify the "choose any units," you do need to check that it's logically consistent, which you could say is more missing steps, but I think this idea is far more fundamental than the Pythagorean Theorem. Our use of squares to define the fundamental units of area really is a completely arbitrary choice. We call them "square units," which already biases us to think of area in a specific way, but there's absolutely no reason we can't use any other shape. Of course squares are convenient because you can stack them up neatly and count them, but that doesn't seem to be helpful at all in this context, so it's natural to choose something else.

card_zero 10mo ago

Einstein's proof relies on the fact that the theorem works with any shape, not just squares, such as pentagons: https://commons.wikimedia.org/wiki/File:Pythagoras_by_pentagons.svg https://commons.wikimedia.org/wiki/File:Pythagoras_by_pentag... Or any arbitrary vector graphics, like Einstein's face. So in the proof, the shape on the hypotenuse is the same as the original triangle, and on the other two sides there are two smaller versions of it, which when joined have the same area (and shape) as the big one. Fair enough. However, none of the hundreds or thousands of proofs explain it. They all prove it, like by saying "this goes here, that goes there, this is the same as that, therefore logically you're stupid," but it still seems like weird magic to me. Some explanation is missing.

chronial 10mo ago
Draw a square around Einstein's face. Call the side length of the square a and the area of the square A. We have A=a^2. Einstein takes up some portion p < 1 of that area, so Einstein has area E = pA. Now we scale the whole thing by factor f. So the new square has side lengths fa, and thus area A' = (fa)^2 = f^2×a^2 = f^2×A. Since the relative portion the face takes up doesn't change with scaling, the face now has size pA' = p×f^2×A = f^2 × pA = f^2 E. Does that help or was that not the part you were missing?
- card_zero 10mo ago
  No, that part is fine: I'm happy with the fact that it works with arbitrary shapes. What bothers me is that the area on the hypotenuse is equal to the sum of the areas on the other two sides, when the triangle has a right angle. This somewhat like saying that I'm troubled by the fact that 1+1=2, I know. But that's a potentially distracting sidetrack, let's not get into that one.
lupire 10mo ago
What definition of area are you using in the first place, for non-swuare objects? Most people find area intuitive and informal, but if you describe area formally, it should be easy to use your definition to account for scaling.

JanisErdmanis 10mo ago

Indeed, a wonderful proof. It does, though, make one implicit assumption that if one stretches the fabric by the same amount, all holes in it stretch by the same amount. In particular, it assumes that triangle stretching is size-independent. Perhaps there are fabrics where that is not true...

lupire 10mo ago
You mean non-Euclidean fabric? https://www.johndcook.com/blog/2022/09/08/trig-hyperbolic-geometry/ https://www.johndcook.com/blog/2022/09/08/trig-hyperbolic-ge... https://math.hmc.edu/funfacts/spherical-pythagorean-theorem/ https://math.hmc.edu/funfacts/spherical-pythagorean-theorem/ https://en.wikipedia.org/wiki/Minkowski_distance https://en.wikipedia.org/wiki/Minkowski_distance
- JanisErdmanis 10mo ago
  That is one possibility. Probably the only one, though.
layer8 10mo ago
The Pythagorean theorem is only true in Euclidian space, where that “stretching” assumption is true. So you are right about there being assumptions, and indeed they are imposing limitations to the applicability of the theorem.

lupire 10mo ago

> The purely trigonometric proof that was found by some high school students recently is a great one. It was geometric, using trigonometric vocabulary.

SJC_Hacker 10mo ago
Trig relies on some geometric assumptions. The definition of sin and cos is going to require a right triangle, for instance,

SJC_Hacker 10mo ago

> The purely trigonometric proof that was found by some high school students recently is a great one. I failed to understand what was so cool about that proof. It relied on concepts such as Cartesian coordinate systems, and the measure of an angle (not just a pure geometric concept), and even concepts like convergence of infinite sums, which weren't purely geometric. Geometry had been formalized in the 20th century and had moved past informal proofs

globalnode 10mo ago

That is interesting and made me think. Only after following some of the other subcomments did I manage to understand it. Personally, replacing the word similarity ratio with scale factor made all the difference. At first I thought it was a circular argument, relying on pythag to prove pythag but that scale factor is the key actually, and the fact that side lengths scale linearly but the area scales quadratically. It feels like a similar trick we see when adding logarithms gives us multiplication.

WCSTombs 10mo ago
Yeah, the similarity ratio/scale factor and its connection to area is the key part of the proof. Sorry, that could have been clearer.
- globalnode 10mo ago
  All good! thanks for the maths though, just when I think I understand it... I dont! Its one of those love hate relationships, I love it, it hates me.

layer8 10mo ago

Also, iterating that triangular subdivision gives rise to the Pythagorean tree fractal: https://en.wikipedia.org/wiki/Pythagoras_tree_(fractal) https://en.wikipedia.org/wiki/Pythagoras_tree_(fractal)