Several viewers asked this question: how many points can you have, all an integer distance apart, of which no three points are on the same line? Think about it. The answer is below. . . . Answer: you can have any finite number of such points, all on one circle. Consider the (infinitely many) points on the unit circle with rational coordinates. Treating them as complex numbers, square them all. The resulting points are a rational distance apart. We can then apply the rational-to-integer trick as in the video. This construction can be thought of as an "inversion" around a circle of the stack of right triangles from the video.
But you'll have to take a finite subset before multiplying with the common denominator, right? (Also, you are using that half of the squares equal "the other half", e.g. {1, -1, i, -i} are all on the same line when squared, but yes, only two distinct points { 1, -1 } remain. Anyways, the squares are still on the unit circle so there can't be more than 2 on one line.) But IMO it's also nontrivial that the distances are rational, which isn't the case for any squares of "rational complex" numbers; e.g. (1/2+i/3)² - (1/5+i/7)² has an irrational norm. This also works thanks to the fact that they have unit norm.
@@M-F-H Yes, finite subset before multiplying. One way to check the distances are rational: if w and z are complex numbers on the unit circle, then |w^2 - z^2| is equal to |w/z - z/w|. Since w/z is also on the unit circle, that will be 2 |Im(w/z)|, which is rational.
I saw the train coming as soon as you added the first hyperbola. I bet Erdos sat bolt upright from a sound sleep when it struck him, and immediately awoke someone else to tell them. What a mind.
This way of working through the entire reasoning and a bit of history with it before reaching the conclusion is such a good way to teach theorems. Refreshing!
This is the first video from this channel that I was recommended by the youtube algorithm, and I immediately subscribed. I wish all math streamers state clearly their Erdős number as you did, so that I can test my hypothesis about correlation between that parameter and clarity of the presentation and quality of the content.
متعة من متع الحياة لا يتذوقها الا الذي أحب الرياضيات المسألة في غاية الجمال. بول اردش قامة في تواضعه وعلمه وعرض البرهان بهذا الشرح المستفيض يستوجب شكر صاحب الفيديو شكرا جزيلا
Actually stunning. When the video started I had no idea how to solve the problem, but by the end the solution seemed obvious and I couldn't remember not knowing it.
THIS is the way math should be taught! Thank-you for presenting this beautiful proof in an accessible, step-by-step manner that has a gentle, curious style that draws the listener in. Well done!!
I really appreciate how you can break things down into something someone without an advanced maths degree can understand. Nice work, Dr. Ravi! P.S. I remember learning about Pólya during my Math Education degree - that's a blast from the past!
An interesting proof for sure. My uncle, a mathematician who was a tenured professor in Hawaii that got me interested in mathematics as a teen, has an Erdos number of 3. Thank you for giving us this beautiful and memorable work.
Thanks to RUclips algorithm for recommending such a hidden gem of a channel. Amazing video! Never subbed & turned on the bell so fast! Keep making such videos
I was looking at the diagram of the hyperbolas when the proof hit me, wow, I haven’t felt the rush of joy from an elegant proof in far too long! Thank you
Recommended by the algorithm, fits my interests very well. The proof is elegant, Erdos is a legend, I will subscribe this channel. This is great, very well prepared. Thanks so much
But "DISH" could be misleading. The "ö" is actually closer to the sound in "bird" or "Sir". I'm german and we do have the "ö umlaut" as well. I wasn't sure how it may be pronounced in hungarian, but google translate essentially confirmed the pronounciation is pretty much the same (when switched to hungarian). Btw "Erdő" in hungarian means "Forest" while "Erdös" means "wooded" / "arboreous" / "forested". Though I don't speak hungarian :) I've been in hungary for 2 weeks years ago where we had a robotics and CNC course. Hungarian is a quite difficult language. Our teacher back then tried to learn the basics but has given up quickly :) But yes, he seems to pronounce it correctly. Though it's difficult to write it down. Luckily many dictionaries (like google translate) provide audio for many languages. So if you're not sure, google translate is actually a brilliant tool for that.
Thanks, David! Yes, I met Erdős a handful of times at conferences and such. One time, I took him to lunch in Manhattan when his minder (Joel Spencer) had to teach a class at NYU. That may have been a rare Erdős meal that didn't lead to a paper. 😀
@@BoppanaMath You (collectively) clearly didn't have enough coffee to transform it into a theorem! Thank you for the video - really nice format joining history and mathematics; the personal connections add the icing on the cake.
Very well presented. I would have tackled the problem using Pythagorean triangles. For example, take any two: 3,4,5 and 5,12,13. Work out the lowest common multiple (LCM) of the first number of each triple, namely the 3 from the first and the 5 from the second. Their LCM is 15. Scale the two triangles up to get two larger triangles, with the first side length 15, namely 15,20,25 and 15,36,39. We can now draw two right angled triangles with a common left side of length 15 and horizontal bases of lengths 20 and 36 respectively. (Their hypotenuses have lengths 25 and 39 respectively.) The example above just uses 2 Pythagorean triangles, but the process can be extended to cover 3 or more triangles or in fact any number of them. (Note, in the example used, there just happens to be a side of length 5 in both triangles, which could be used to simplify the example. I have ignored this fact because such a common side length is not always available.)
Thanks, Danny! I remember you from back in the day. If you're interested in such bounds, there has been recent progress on the diameter versus size of integer-distance point sets. Here is a recent paper that also mentions plenty of open problems: arxiv.org/abs/2401.10821.
That's pretty interesting way of proving something. The last thing I would have though while looking at this problem is "How does this relate to conic sections?"
Thanks for this very educational video! But even if there can't be infinitely many points (not on the same line) that are an integer distance apart, surely there must be an infinite number of starting triangles (finite sets) where all points are an integer distance apart? And that you can create such a triangle with as many points as you like, just as long as the points aren't infinitely many?
Yes, that's right. That's why the answer to the video question for 100 (or a million) points is yes, but the answer for an infinite number of points is no.
I met Erdos once, at the University. of Calgary. He showed me a beautilful problem, that I had actually been thinking about myself; but I can't remember now what it was. Anyways, my Erdos number is 4.
I can be shown also that there are infinitely many points on the plane at rational distances in which the condition is strengthen to "no three points on the same line." The solution consists of placing points on the unit circle at even multiples of the angle of a 3-4-5 right triangle.
Yes, I thought about discussing "no three on a line", but didn't want to go too long. Here's a related construction. Consider all the (infinitely many) points on the unit circle with rational coordinates. Viewing them as complex numbers, square them all. The resulting set of points has rational distances with no three on a line.
*YES!* _If you can name the number of points,_ then the answer to "Are there infinitely many points, not all on the same line, that are an integer distance apart?" is *YES.* The proof given at 7:26 is NOT an impediment, because it actually changed the fundamental nature of the original question by ASSUMING that all the triangles have to be Right Triangles. But that is an invalid assumption, as it was made only for convenience. Instead, ALL manner of triangles must be allowed. It matters not at all that they do not all stack up neatly. But wait... Is there an INTEGER number so large that it cannot be named? *NO!* A really large number of numbers are really inconvenient to name, but that does not make any and every one of them fundamentally impossible to name. Impractical, yes, but impossible? *NO!* Objecting that INFINITY breaks this system is wrong, because *INFINITY IS NOT A NUMBER.* Infinity is a CONCEPT. Also, Infinity is neither integer nor non-integer. @Boppana Math
Nice and elegant! Btw, what do you use for creating your slides? It's simple and gorgeous. The equations look nice as well. May I know your process? Thank you
Thanks! For creating the slides, I use Beamer, which is a slide package in LaTeX. For creating the diagrams, I use Asymptote, and then include the diagrams in my LaTeX file.
6:11 So does this mean that there are only a finite number of integer right triangles for any given vertical integer _a_ ? I.e., are there only a finite number of integer Pythagorean triples (a, b, c, where a^2 + b^2 = c^2) for any given integer a?
Yes. One way to see it is to rewrite a^2 + b^2 = c^2 as a^2 = c^2 - b^2, and then use difference of squares to get a^2 = (c - b)*(c + b). The integer a^2 has only a finite number of factors, and so there are only a finite number of choices for b and c.
@@BoppanaMath Actually, I showed that if a=(2^k)*(2n+1) where k>1 and (2n+1)=p_0^(d_0)p_1^(d_1).... I can build (d_0+1)(d_1+1).... integer right triangles. And I can't figure out why this multiplication has to be finite. What am I missing?
The proof uses only two sides of the triangle. Can we show something stronger if we also use the third side? E.g. a bound on the fraction of distances that are integral?
@@igorshinkar8046 In a related question, there has been some exciting recent work on the diameter of the point set versus the size of the point set: arxiv.org/abs/2401.10821. Many open problems remain.
You're saying it in an ambiguous way. You can "add" (clarify...) as many triangles as you wish -- it's always a finite number, but it can be as large as you want.
Good question. The answer is yes. You can place all the points on a circle. One example is here: en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Anning_theorem. A second example is to start with the stack of right triangles in the video, and then perform an "inversion" onto a circle; the distances will remain rational.
You can have an arbitrarily large finite number of such points. For one example, see this Wikipedia article: en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Anning_theorem. For a second example, you can take the construction in the video (the stack of right triangles) and perform an "inversion" onto a circle, which keeps the distances rational.
I am curious, wouldnt the fact that there is no positive finite integer that has a factor of every positive finite integer already prove the theorem because such a number would be needed for that set of points to exist?
If A is any integer 3 and above, but B is a prime integer value, C has no integer solution. The Minkowski Solution for ovates challenges the most general question here because of inconsistent application of the notion of Infinity. The Infinity which Multiplication is able to distinguish is much smaller than that which Addition can provide.
Is there a name for the set of intersections generated by a triangle? If not, I suggest an Erdös constellation. What other properties do these sets have?
Yes, I'm a little surprised that Erdös’s proof didn't make it into the Aigner-Ziegler book, especially since Erdös was an early advisor for the first edition of their book.
I dont understand why the solution for infinite points didnt look like an induction on the 100 point solution. It seems you could extend to 101, 102... Its your ending right triangle case, and it seems the number of intersection points grows with the size of the triangle. Does it not grow fast enough?
I'm not sure I understand your question. We can get integer distances for 101 points, 102 points, and so on. But we can't get integer distances for infinitely many points.
@BoppanaMath it's just not clear where a proof by induction would break down, if we can do 10,000 why not 10,001? You showed that the number of intersections grows, it's not clear why it can't grow to infinity. Sorry I'm sure it is a bad question and it's a failure of understanding on my part, appreciate the video a lot. Motivation was on point!
@@electrowizard2000 Thanks. Just because we can do any finite number of points doesn't mean we can do infinitely many. For example, there is a regular polygon with 3 points, 4 points, and so on, but there is no regular polygon with infinitely many points.
I have a doubt. If I understood the reasoning, this proves that, if you have n points, pairwise an integer distance apart, then the (n+1)-th point can only be in a finite number of places. But can't n become arbitrary large?
You can't always add an (n+1)-th point. But the proof isn't about an (n+1)-th point - it shows that there are only a finite number of possibilities for the other points, by picking *any* 3 randomly chosen points, A, B, C (or actually, any two pairs, and you should choose the two pairs which have the smallest distances that occur), for example d1 = AB and d2 = BC (or CD if you pick a different pair). Then there are at most 4(d1+1)(d2+1) possibilities [actually much less] for *all* of the other points can be. So, in principle there can be any (finite!) number of points, but you can't add an arbitrary number of additional points. The maximum number of points is restricted by the two smallest distances among the given points.
@@BoppanaMath Its a nineteenth century topic. Subsumed into intersection theory. Essentially given a curve/surface it gives a method of counting the intersections. This video reminded me of it
There isn't a largest such integer. There are non-collinear integer-distance sets with 1 point, 2 points, 3 points, and so on. But there isn't a non-collinear integer-distance infinite set.
@@BoppanaMath I don't understand that. When you say there are sets with 1 point, 2 points, 3 points, and so on, doesn't "and so on" mean that it goes on to infinity? If not, where does it stop? In my mind a sequence like 1, 2, 3 is one you can extend infinitely.
@Cartilog-z4f The only way to prove that you can construct all the points an integer distance apart expressly depends on starting by limiting yourself to a finite number of points. So any finite number no matter how large; but you can't "make the jump" to infinity.
I think by choosing two distinct points and a difference of the distances, for example, two distinct points A, B, along with a restriction that |PA-PB|=a, where 0
Each yellow hyperbola has foci A and B. Each blue hyperbola has foci A and C. So each yellow hyperbola is different from each blue hyperbola. For more information on the two foci of a hyperbola, see en.wikipedia.org/wiki/Hyperbola.
suppose n ∈ ℤ⁺, pₙ ∈ ℙ. then we can say: n mod φ(n) - pₙ = (n + pₙ) mod φ(n), where φ(n) is euler's totient function. i'd like to propose: -the equation holds only for (n = 1, 2, 3, 4, 6). P.S.: great videos so far, especially the one about the series problem!
I'm using a program called Asymptote to create a sequence of static images, and for some reason, Asymptote moves each diagram a bit with each new element. I'll find a way to avoid such movement in future videos.
The Anning-Erdős theorem also holds in 3 dimensions (and in every finite dimension). Here is the final sentence of their paper: "By a similar argument we can show that we cannot have infinitely many points in n-dimensional space not all on a line, with all the distances being integral." In the video description, I have included links to the Anning-Erdős paper and the Erdős paper.
You can have any finite number of such points, all on one circle. Consider the (infinitely many) points on the unit circle with rational coordinates. Viewing them as complex numbers, square them all. The resulting points are a rational distance apart. We can then apply the rational-to-integer trick as in the video. This construction can be thought of as an "inversion" around a circle of the stack of right triangles from the video.
For any baseline, when points are far enough away, the difference in length is one. Points further away (in the same direction) must have a difference of less than one, so we cannot increase the distance and keep both lengths integer. This applies in all directions.
@ Ok, I see the argument. You just kind of skipped a couple of details at the end. It appears that you are taking A,B,C to be S instead of just any three points in S. The argument lies in the fact that for any arbitrary A,B,C in S, there are only a finite number of positions able to house the rest of the points in S maintaining the assumptions. That finishing detail is not stated.
Still quite confused about the concept of infinity after studying a bit of calculus. Like even if for an arbitrarily large integer n, we can still find n points such that the hypothesis hold true, but it doesn’t mean it works for infinitely many points.
But you have to change your points with every new n! So, in other words, you cannot add new points to existing ones infinitely, eventually you will run out of points, and have to remove one or more to add new ones!
Do you remember what PA + PB = 1 would give? (An ellipse.) A hyperbola is related. In either case, the points A and B are the foci (focuses) of the hyperbola/ellipse. See en.wikipedia.org/wiki/Hyperbola.
Euler published more than 1500 papers. Euler's work is still being published to this day and he has been dead for over 300 years. So, I don't know where you are getting your information.
Ha ha. A mathematician I know (Danny Kleitman) has an Erdős-Bacon number of 3, since he had a cameo in the movie Good Will Hunting. My Erdős-Bacon number might be 6?
While the theorem is one of the most beautiful theorems, it's also one that should not be explained. A reader has to comprehend the proof to understand just how elegant the proof is.
you seem to have a hypothesis that there is no such thing as an infinite size integer. I can't accept this and it seems to be just defining infinite magnitude integers out of existence. If they do exist then you can make your original triangle have infinite length integer sides (we don't know which infinite length) which would contain infinitely many hyperbolas and lead to an infinite number of integer length triangles based on their intersections. Another way to look at this is to exploit the one to one mapping between rational numbers and integers to prove that for every point in your rational number construction (which contains an infinite number of points) there can be assigned a set of integer lengths to each distance via the mapping.
Intuitively, I would answer the question if an infinite number of points exists, not being on a single line with integer distances, as affirmative!Starting with a single right triangle of sides 3,4 and 5, and adding an infinite number of similar 3,4,5-right triangles joined at their sides with common lengths yields an infinite number of triangles and hence number of points in the 2D plane.
a proof that at times Math gets into the nowhere of total unusefulness 😅. saying otherwise.... where does that postulate convey to? .... nowhere in real life.
![Pythagoras Would Be Proud: High School Students' New Proof of the Pythagorean Theorem [TRIGONOMETRY]](/img/1.gif)
Several viewers asked this question: how many points can you have, all an integer distance apart, of which no three points are on the same line? Think about it. The answer is below.
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Answer: you can have any finite number of such points, all on one circle. Consider the (infinitely many) points on the unit circle with rational coordinates. Treating them as complex numbers, square them all. The resulting points are a rational distance apart. We can then apply the rational-to-integer trick as in the video. This construction can be thought of as an "inversion" around a circle of the stack of right triangles from the video.
But you'll have to take a finite subset before multiplying with the common denominator, right?
(Also, you are using that half of the squares equal "the other half", e.g. {1, -1, i, -i} are all on the same line when squared, but yes, only two distinct points { 1, -1 } remain. Anyways, the squares are still on the unit circle so there can't be more than 2 on one line.) But IMO it's also nontrivial that the distances are rational, which isn't the case for any squares of "rational complex" numbers; e.g. (1/2+i/3)² - (1/5+i/7)² has an irrational norm. This also works thanks to the fact that they have unit norm.
@@M-F-H Yes, finite subset before multiplying. One way to check the distances are rational: if w and z are complex numbers on the unit circle, then |w^2 - z^2| is equal to |w/z - z/w|. Since w/z is also on the unit circle, that will be 2 |Im(w/z)|, which is rational.
The proof was so elegant that the Theorem seemed so obvious!
I saw the train coming as soon as you added the first hyperbola. I bet Erdos sat bolt upright from a sound sleep when it struck him, and immediately awoke someone else to tell them. What a mind.
This way of working through the entire reasoning and a bit of history with it before reaching the conclusion is such a good way to teach theorems. Refreshing!
I agree!
You can’t just casually drop that you knew Erdős
He did it twice.
@@davidbroadfoot1864 I didn't know him but I wrote a paper (and more) with someone who did know Erdös and wrote a paper with him ... 😎
This is the first video from this channel that I was recommended by the youtube algorithm, and I immediately subscribed. I wish all math streamers state clearly their Erdős number as you did, so that I can test my hypothesis about correlation between that parameter and clarity of the presentation and quality of the content.
I also have Erdös number 2, maybe I should also make math videos...?🤔
Only you can answer that question.
@@hrvojedjurdjevic2123 That might be a necessary condition, but maybe not sufficient...?
متعة من متع الحياة لا يتذوقها الا الذي أحب الرياضيات
المسألة في غاية الجمال. بول اردش قامة في تواضعه وعلمه
وعرض البرهان بهذا الشرح المستفيض يستوجب شكر صاحب الفيديو شكرا جزيلا
Actually stunning. When the video started I had no idea how to solve the problem, but by the end the solution seemed obvious and I couldn't remember not knowing it.
Couldn't Remeber Not Knowing It
Inciteful!
it is obvious, that you love math and want to show others the beauty of ideas, solutions and clear thinking.
thank you very much for this.
THIS is the way math should be taught! Thank-you for presenting this beautiful proof in an accessible, step-by-step manner that has a gentle, curious style that draws the listener in. Well done!!
To be clear, when I say "curious", I mean that it makes the *listener* curious. :)
I really appreciate how you can break things down into something someone without an advanced maths degree can understand. Nice work, Dr. Ravi!
P.S. I remember learning about Pólya during my Math Education degree - that's a blast from the past!
Thanks, Kyle! Yes, Pólya wrote several books on problem-solving heuristics in math.
An interesting proof for sure. My uncle, a mathematician who was a tenured professor in Hawaii that got me interested in mathematics as a teen, has an Erdos number of 3. Thank you for giving us this beautiful and memorable work.
Thanks to RUclips algorithm for recommending such a hidden gem of a channel. Amazing video! Never subbed & turned on the bell so fast! Keep making such videos
I recall attending a small (10 or so students) proving-conjectures session with Erdos in the '60s. That man could see around mathematical corners.
I was looking at the diagram of the hyperbolas when the proof hit me, wow, I haven’t felt the rush of joy from an elegant proof in far too long! Thank you
Recommended by the algorithm, fits my interests very well. The proof is elegant, Erdos is a legend, I will subscribe this channel. This is great, very well prepared. Thanks so much
I often see math videos on RUclips, most of I forget after a while, but this one will stay in my head. Thanks!
You knew Erdős? Wow, such an honour!
So good to hear you pronouncing Erdős correctly as AIR-DISH. So many people say AIR-DOSS
But "DISH" could be misleading. The "ö" is actually closer to the sound in "bird" or "Sir". I'm german and we do have the "ö umlaut" as well. I wasn't sure how it may be pronounced in hungarian, but google translate essentially confirmed the pronounciation is pretty much the same (when switched to hungarian).
Btw "Erdő" in hungarian means "Forest" while "Erdös" means "wooded" / "arboreous" / "forested". Though I don't speak hungarian :) I've been in hungary for 2 weeks years ago where we had a robotics and CNC course. Hungarian is a quite difficult language. Our teacher back then tried to learn the basics but has given up quickly :)
But yes, he seems to pronounce it correctly. Though it's difficult to write it down. Luckily many dictionaries (like google translate) provide audio for many languages. So if you're not sure, google translate is actually a brilliant tool for that.
I love it when two parts of math meet so naturally.
Absolutely elegant presentation!
You really helped me to realize how to approach difficult problems.
Thanks a lot.
Your voice is very pleasant to listen to
Nice! This solution, if you think about it, is much easier than many math Olympiads problems.
Awesome video, as usual. I had no idea you had met Erdos!
Thanks, David! Yes, I met Erdős a handful of times at conferences and such. One time, I took him to lunch in Manhattan when his minder (Joel Spencer) had to teach a class at NYU. That may have been a rare Erdős meal that didn't lead to a paper. 😀
@@BoppanaMath You (collectively) clearly didn't have enough coffee to transform it into a theorem!
Thank you for the video - really nice format joining history and mathematics; the personal connections add the icing on the cake.
So beautiful and great as always.
Thank you.
Hoping for more videos.
Thanks for this lovely video. The hyperbolas were unexpected and beautiful. Its so cool that you met Erdos. My Erdos # is 4.
Very well presented. I would have tackled the problem using Pythagorean triangles. For example, take any two: 3,4,5 and 5,12,13. Work out the lowest common multiple (LCM) of the first number of each triple, namely the 3 from the first and the 5 from the second. Their LCM is 15. Scale the two triangles up to get two larger triangles, with the first side length 15, namely 15,20,25 and 15,36,39. We can now draw two right angled triangles with a common left side of length 15 and horizontal bases of lengths 20 and 36 respectively. (Their hypotenuses have lengths 25 and 39 respectively.)
The example above just uses 2 Pythagorean triangles, but the process can be extended to cover 3 or more triangles or in fact any number of them. (Note, in the example used, there just happens to be a side of length 5 in both triangles, which could be used to simplify the example. I have ignored this fact because such a common side length is not always available.)
Very nice construction!
These videos are truly amazing.
Nice lecture! The bound proven here seems too high. Let the three sides of the initial Pythagorian triangle be x, y, and z, where WLOG x
Thanks, Danny! I remember you from back in the day. If you're interested in such bounds, there has been recent progress on the diameter versus size of integer-distance point sets. Here is a recent paper that also mentions plenty of open problems: arxiv.org/abs/2401.10821.
now this is interesting... there's no infinite set of points that satisfy the condition but infinitely many finite sets... 🤔
I don't think I've subscribed faster. Thanks for sharing!
Exceptionally well delivered!
That's pretty interesting way of proving something. The last thing I would have though while looking at this problem is "How does this relate to conic sections?"
A stunning proof that I'm joyful to know now!
Thanks for this very educational video!
But even if there can't be infinitely many points (not on the same line) that are an integer distance apart, surely there must be an infinite number of starting triangles (finite sets) where all points are an integer distance apart? And that you can create such a triangle with as many points as you like, just as long as the points aren't infinitely many?
Yes, that's right. That's why the answer to the video question for 100 (or a million) points is yes, but the answer for an infinite number of points is no.
You explained it so well.
Lovely lecture. Thank you.
I met Erdos once, at the University. of Calgary. He showed me a beautilful problem, that I had actually been thinking about myself; but I can't remember now what it was. Anyways, my Erdos number is 4.
ABSOLUTELY WONDERFUL PRESENTATION.... I WENT IN A DIFFERENT WORLD... LIKE MEDITATION... THANKS GOD BLESS
I can be shown also that there are infinitely many points on the plane at rational distances in which the condition is strengthen to "no three points on the same line." The solution consists of placing points on the unit circle at even multiples of the angle of a 3-4-5 right triangle.
Yes, I thought about discussing "no three on a line", but didn't want to go too long. Here's a related construction. Consider all the (infinitely many) points on the unit circle with rational coordinates. Viewing them as complex numbers, square them all. The resulting set of points has rational distances with no three on a line.
*YES!* _If you can name the number of points,_ then the answer to "Are there infinitely many points, not all on the same line, that are an integer distance apart?" is *YES.* The proof given at 7:26 is NOT an impediment, because it actually changed the fundamental nature of the original question by ASSUMING that all the triangles have to be Right Triangles. But that is an invalid assumption, as it was made only for convenience. Instead, ALL manner of triangles must be allowed. It matters not at all that they do not all stack up neatly.
But wait... Is there an INTEGER number so large that it cannot be named? *NO!* A really large number of numbers are really inconvenient to name, but that does not make any and every one of them fundamentally impossible to name. Impractical, yes, but impossible? *NO!*
Objecting that INFINITY breaks this system is wrong, because *INFINITY IS NOT A NUMBER.* Infinity is a CONCEPT. Also, Infinity is neither integer nor non-integer. @Boppana Math
Nice and elegant! Btw, what do you use for creating your slides? It's simple and gorgeous. The equations look nice as well. May I know your process? Thank you
Thanks! For creating the slides, I use Beamer, which is a slide package in LaTeX. For creating the diagrams, I use Asymptote, and then include the diagrams in my LaTeX file.
6:11 So does this mean that there are only a finite number of integer right triangles for any given vertical integer _a_ ?
I.e., are there only a finite number of integer Pythagorean triples (a, b, c, where a^2 + b^2 = c^2) for any given integer a?
Yes. One way to see it is to rewrite a^2 + b^2 = c^2 as a^2 = c^2 - b^2, and then use difference of squares to get a^2 = (c - b)*(c + b). The integer a^2 has only a finite number of factors, and so there are only a finite number of choices for b and c.
@@BoppanaMath Actually, I showed that if a=(2^k)*(2n+1) where k>1 and (2n+1)=p_0^(d_0)p_1^(d_1).... I can build (d_0+1)(d_1+1).... integer right triangles. And I can't figure out why this multiplication has to be finite. What am I missing?
@@metulusoy In the prime factorization, all but a finite number of the d_i are zero. So the product of the d_i + 1 is a finite number.
The proof uses only two sides of the triangle. Can we show something stronger if we also use the third side? E.g. a bound on the fraction of distances that are integral?
Intriguing idea. I don't know.
@BoppanaMath maybe there is a tradeoff bound between how many integers and their values
@@igorshinkar8046 In a related question, there has been some exciting recent work on the diameter of the point set versus the size of the point set: arxiv.org/abs/2401.10821. Many open problems remain.
So it's saying that for any given triangle, there are just a finite set of integer-sided triangles that can be added to it?
You're saying it in an ambiguous way. You can "add" (clarify...) as many triangles as you wish -- it's always a finite number, but it can be as large as you want.
Can you have an arbitrary finite number of points, where no more than two are on the same line and all are pairwise integer distances apart?
Good question. The answer is yes. You can place all the points on a circle. One example is here: en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Anning_theorem. A second example is to start with the stack of right triangles in the video, and then perform an "inversion" onto a circle; the distances will remain rational.
Thanks a lot!
24^2=25^2-7^2; ( 7 +3 =10)
24^2=26^2-10^2; (10+ 8=18)
24^2=30^2-18^2; (18+14=32)
24^2=40^2-32^2; (32+13=45)
24^2=51^2-45^2; (45+25=70)
24^2=74^2-70^2
How many points in the plane can have only integer distances while no 3 of them lie in a line?
You can have an arbitrarily large finite number of such points. For one example, see this Wikipedia article: en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Anning_theorem. For a second example, you can take the construction in the video (the stack of right triangles) and perform an "inversion" onto a circle, which keeps the distances rational.
I am curious, wouldnt the fact that there is no positive finite integer that has a factor of every positive finite integer already prove the theorem because such a number would be needed for that set of points to exist?
All that would show is that our infinite construction for rationals couldn't be extended to integers. It wouldn't rule out other constructions.
@@BoppanaMath thank you for clarifying
If A is any integer 3 and above, but B is a prime integer value, C has no integer solution.
The Minkowski Solution for ovates challenges the most general question here because of inconsistent application of the notion of Infinity. The Infinity which Multiplication is able to distinguish is much smaller than that which Addition can provide.
Very nice!
Is there a name for the set of intersections generated by a triangle? If not, I suggest an Erdös constellation. What other properties do these sets have?
very well done.
For an infinite amount of points, you could just make the distances infinitely long right? Just curious, and great video!!!
Thanks. The distance between two points in the plane is always finite. We don't have a "point at infinity".
This could be added to "Proofs from THE BOOK" by Aigner and Ziegler, a tribute to Erdös’s genius.
Yes, I'm a little surprised that Erdös’s proof didn't make it into the Aigner-Ziegler book, especially since Erdös was an early advisor for the first edition of their book.
What’s the limit then finite limit suggest some ??
I dont understand why the solution for infinite points didnt look like an induction on the 100 point solution.
It seems you could extend to 101, 102... Its your ending right triangle case, and it seems the number of intersection points grows with the size of the triangle. Does it not grow fast enough?
I'm not sure I understand your question. We can get integer distances for 101 points, 102 points, and so on. But we can't get integer distances for infinitely many points.
@BoppanaMath it's just not clear where a proof by induction would break down, if we can do 10,000 why not 10,001? You showed that the number of intersections grows, it's not clear why it can't grow to infinity. Sorry I'm sure it is a bad question and it's a failure of understanding on my part, appreciate the video a lot. Motivation was on point!
@@electrowizard2000 Thanks. Just because we can do any finite number of points doesn't mean we can do infinitely many. For example, there is a regular polygon with 3 points, 4 points, and so on, but there is no regular polygon with infinitely many points.
I have a doubt. If I understood the reasoning, this proves that, if you have n points, pairwise an integer distance apart, then the (n+1)-th point can only be in a finite number of places. But can't n become arbitrary large?
You can't always add an (n+1)-th point. But the proof isn't about an (n+1)-th point - it shows that there are only a finite number of possibilities for the other points, by picking *any* 3 randomly chosen points, A, B, C (or actually, any two pairs, and you should choose the two pairs which have the smallest distances that occur), for example d1 = AB and d2 = BC (or CD if you pick a different pair). Then there are at most 4(d1+1)(d2+1) possibilities [actually much less] for *all* of the other points can be. So, in principle there can be any (finite!) number of points, but you can't add an arbitrary number of additional points. The maximum number of points is restricted by the two smallest distances among the given points.
any chance of a video on the Schubert calculus given this one on intersection theory?
I barely know anything about Schubert calculus, but I have included your suggestion in my list of candidate topics.
@@BoppanaMath Its a nineteenth century topic. Subsumed into intersection theory.
Essentially given a curve/surface it gives a method of counting the intersections.
This video reminded me of it
If there are not infinitely many integers that satisfy it, then there must be a largest integer that does. What is that integer?
There isn't a largest such integer. There are non-collinear integer-distance sets with 1 point, 2 points, 3 points, and so on. But there isn't a non-collinear integer-distance infinite set.
@@BoppanaMath I don't understand that. When you say there are sets with 1 point, 2 points, 3 points, and so on, doesn't "and so on" mean that it goes on to infinity? If not, where does it stop? In my mind a sequence like 1, 2, 3 is one you can extend infinitely.
@Cartilog-z4f The only way to prove that you can construct all the points an integer distance apart expressly depends on starting by limiting yourself to a finite number of points. So any finite number no matter how large; but you can't "make the jump" to infinity.
Why some hyperbolas from two sets can't be same? (Only *distinct* hyperbolas have at most 4 intersections)
I think by choosing two distinct points and a difference of the distances, for example, two distinct points A, B, along with a restriction that |PA-PB|=a, where 0
Each yellow hyperbola has foci A and B. Each blue hyperbola has foci A and C. So each yellow hyperbola is different from each blue hyperbola. For more information on the two foci of a hyperbola, see en.wikipedia.org/wiki/Hyperbola.
It seemed so trivial after seeing the proof
Do you consider all the points of a circle in the same line?
The points on one circle are not on the same line.
How many papers are erdos 0
Great!
"we found 3 and 4, now can you find 100?" and "mathematicians like to make things difficult" 😂
suppose n ∈ ℤ⁺, pₙ ∈ ℙ. then we can say:
n mod φ(n) - pₙ = (n + pₙ) mod φ(n),
where φ(n) is euler's totient function.
i'd like to propose:
-the equation holds only for (n = 1, 2, 3, 4, 6).
P.S.: great videos so far, especially the one about the series problem!
the side dish: construction of rational numbers solution is even better than the main dish: Erdos'a proof 😅
Is there some reason that, when you add elements to a diagram, the diagram bumps a little to the right? It's really bizarre and distracting.
I'm using a program called Asymptote to create a sequence of static images, and for some reason, Asymptote moves each diagram a bit with each new element. I'll find a way to avoid such movement in future videos.
This is one for The Book
What about to generalise this interesting theorem to tridimensionale space.
The Anning-Erdős theorem also holds in 3 dimensions (and in every finite dimension). Here is the final sentence of their paper: "By a similar argument we can show that we cannot have infinitely many points in n-dimensional space not all on a line, with all the distances being integral." In the video description, I have included links to the Anning-Erdős paper and the Erdős paper.
How many points can you have, all an integer distance apart, of which no *three* points are on the same line?
You can have any finite number of such points, all on one circle. Consider the (infinitely many) points on the unit circle with rational coordinates. Viewing them as complex numbers, square them all. The resulting points are a rational distance apart. We can then apply the rational-to-integer trick as in the video. This construction can be thought of as an "inversion" around a circle of the stack of right triangles from the video.
@@BoppanaMath Thank you, that makes sense!
For any baseline, when points are far enough away, the difference in length is one. Points further away (in the same direction) must have a difference of less than one, so we cannot increase the distance and keep both lengths integer. This applies in all directions.
Next video the Friendship Theorem I hope
I'll put it on my list of candidate topics.
When first hyperbola appeared I already knew where this is going
If I ever started a religion it would have a prophecy that a child would be born with an Erdős-Bacon number of zero.
The infinite case at 7:41 does work, you just need inf factorial as your multiple. Sure... it isn't practical, but this solution never was.
Didn't you assume a finite set of points in order to prove it was a finite set of points?
No. I never assumed there were a finite number of points.
@ Ok, I see the argument. You just kind of skipped a couple of details at the end. It appears that you are taking A,B,C to be S instead of just any three points in S. The argument lies in the fact that for any arbitrary A,B,C in S, there are only a finite number of positions able to house the rest of the points in S maintaining the assumptions. That finishing detail is not stated.
@@tmwtpbrent14 Maybe the written proof here will help: en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Anning_theorem
Still quite confused about the concept of infinity after studying a bit of calculus. Like even if for an arbitrarily large integer n, we can still find n points such that the hypothesis hold true, but it doesn’t mean it works for infinitely many points.
Yet this video is very nice after all🎉
For real?
Any examples?
But you have to change your points with every new n! So, in other words, you cannot add new points to existing ones infinitely, eventually you will run out of points, and have to remove one or more to add new ones!
great video :)
It sound like my geometry knowledge has failed me. I had a hard time thinking through why |PA-PB| = 1 should be a hyperbola.
Do you remember what PA + PB = 1 would give? (An ellipse.) A hyperbola is related. In either case, the points A and B are the foci (focuses) of the hyperbola/ellipse. See en.wikipedia.org/wiki/Hyperbola.
yeah this was awesome
Looking at the proof made me feel like an idiot
Euler published more than 1500 papers. Euler's work is still being published to this day and he has been dead for over 300 years. So, I don't know where you are getting your information.
According to the book "Euler Through Time", Euler wrote about 850 papers. The Euler Archive makes a similar claim: eulerarchive.maa.org.
*Perhaps you could arrange to interview Kevin Bacon on this channel*
Ha ha. A mathematician I know (Danny Kleitman) has an Erdős-Bacon number of 3, since he had a cameo in the movie Good Will Hunting. My Erdős-Bacon number might be 6?
Your content superposes with the subtitles.
You should not use the bottom of the screen.
Good point. I hadn't thought of that. How much of the screen should I avoid? The bottom 10 percent?
@@BoppanaMath Just click on the subtitles button and check it on the screen.
While the theorem is one of the most beautiful theorems, it's also one that should not be explained. A reader has to comprehend the proof to understand just how elegant the proof is.
What if all points coincide, then they are not on a single line, simple
nice
Paul Erdős is not married.
He was also the king of amphetamines. 🙂
you seem to have a hypothesis that there is no such thing as an infinite size integer. I can't accept this and it seems to be just defining infinite magnitude integers out of existence. If they do exist then you can make your original triangle have infinite length integer sides (we don't know which infinite length) which would contain infinitely many hyperbolas and lead to an infinite number of integer length triangles based on their intersections. Another way to look at this is to exploit the one to one mapping between rational numbers and integers to prove that for every point in your rational number construction (which contains an infinite number of points) there can be assigned a set of integer lengths to each distance via the mapping.
Intuitively, I would answer the question if an infinite number of points exists, not being on a single line with integer distances, as affirmative!Starting with a single right triangle of sides 3,4 and 5, and adding an infinite number of similar 3,4,5-right triangles joined at their sides with common lengths yields an infinite number of triangles and hence number of points in the 2D plane.
It is not clear
noice.
a proof that at times Math gets into the nowhere of total unusefulness 😅.
saying otherwise....
where does that postulate convey to? .... nowhere in real life.