Imagine teaching fractions in elementary school and a kid says "I'm not doing addition wrong, I'm computing the median" and then explains everything from this video.
@@alonelyphoenix8942 “By the time Gauss was 7 years old, his schoolmasters admitted that there was nothing more they could teach the boy.” ~David Burton, Elementary Number Theory
i kid you not, THE DAY after i watched this i had a math competition, and one of the problems was to find the fraction will the smallest denominator between 2 fraction, i tried trial and error and then remembered this video and immediately got the answer
i was the apprenticeship instructor for the roofing program in calgary for 13 yrs... my most fulfilling memory was being able to teach fractions and the metric system to 40 yr old roofers with learning disabilities, addiction problems and 'incomplete' education scenarios... seeing the look on someone's face when they actually get it and feel good about themselves...i was so blessed to help...
I'm in high school. Clicked with great curiosity to watch and expand my knowledge but it slowly kept getting more and more complex until my brain couldn't understand
That's ok! It's amazing that you have curiosity about math. The video can get a little advanced at places, it's not a video meant for everyone. Don't worry too much about it, just try your best and take things at your own pace, maybe come to it in the future when you have stronger foundations. Some of my other videos might be a little simpler and easier, if you want to give them a try, like this one: th-cam.com/video/3B-D3w292TI/w-d-xo.html
I am in highschool too, few months ago I saw it understood not very much, today rewatching understood more then that time(not everything obviously) so maybe in my next re watch I will understand more
Finally someone talked about this ! Last year I rediscovered most of this, in a attempt to find an algorithm that converts computer floating point numbers into a ratio, without suffering from the precision loss of floating point arithmetic. I couldn't find anything about this on the internet, until a friend of mine did. I'm happy more people learn about this simple but very interesting maths concept !
@@siddanthvenkatesh2744 I don't know about project Euler, but yes you avoid any operations on floating numbers to avoid losing precision, and just use comparisons in order to know if your ratio is close.
That was the first thing I thought of when thinking of a fun application for these concepts. A friend of mine actually fiddled around with similar ways of representing floats after I told them that the only way to never lose any precision was to always preserve the full chain of operations that occur when some variable becomes imprecise, then recalculate values on the fly with the desired precision when retrieving them in the future. I believe they did end up representing floats as ratios of VLIs, but I'm not sure how far they got in terms of being able to achieve any desired precision.
Actually floating point numbers are alreaady exact rational numbers; imprecision comes in only when you try to calculate a number that cannot be exactly represented as floating point value. However when converting to a fraction, that loss alredy happened, and if you convert to anything but the exact value of that float, you are going to adda second imprecision on top of it. That second imprecision *may* cancel out the first, but unless you take into account knowledge on how you arrived at it, generally it will not. Indeed, every floating point number (except for infinities and NaN, of course) can be written exactly as m*2^n where m and n are integers where m can be derived from the sign and mantissa bits, and n can be derived from the exponent bits (with some special handling for denormalized numbers; those however are also of the form above, just the formulas for m and n are slightly different). Of course in that representation you lose the signed zero (unless you store m in a signed-magnitude or 1-complement integer representation, which also has signed zero), but then, mathematically 0 = -0 anyway..
@@__christopher__ Floating point numbers aren't exact rational numbers because of infinite (binary) decimals that can't be represented. And in my algorithm for the conversion, you define the converted number as the one that if you convert back to a float, you get the original, so in a way there are no loss of precision during conversion. Also you can represent infinities, nan and -0 in a well-designed library with 1/0, -1/0, 0/0, 0/-1.
If ad-bc=1, you can use Pick's Theorem to prove a bunch of properties of the mediant. The parallelogram is a lattice polygon with area 1, and we know 4 points on its border. Since the area is the number of interior points plus half the border points minus 1, there must not be any points in the interior and no more points on the border, so there cannot be any rational numbers that would fall in this region or on its border.
It is just nice when u start from very simple things and sort of play around with it to discover interesting observations. It feels like going in the reverse direction when u are reading a theorem. Instead of having a very complicated unintuitive statement thrown at you and then having to use every single brain cell to figure out why that even works in the first place, this just feels very satisfying. It feels like the thought process flows naturally, without resistance.
Excellent video. I felt like every time you introduced a new concept, I had some questions pop into my head and thought "I'll have to google this after..." but then you answered the question in the video!
Wow, I saw the thumbnail and thought that the video would be something really simple, but then saw it was nearly 30 minutes long! Unexpectedly turned out to be an absolutely banger! Good job man!
Me too - i recall stern brocot series' "name" from some quantum or other foo-bar but only the name - its the first video somebody draw it for me/ made a picture of it :)
I planned on covering this topic as an interactive website for SoME1 for use as estimations. You came at it from a much better angle and covered more than I would have
This is a great video! I love this graph in the outro, it looks cool and really helps to rewind the video in the head. The theme is very cool too, I love when the video just starts investigating and playing with some concept just to see what happens. That's really my favorite part of math. Last time I caught that feeling when watching that video about hackenbush and surreal numbers. This was like my 2nd favorite math video of all time, and yours is really high up there! It's a shame that you couldn't fit everything in it, on 18:00 the audio quality changes for a moment meaning that this part was recorded after everything else, so you probably didn't have enough time. So please make part 2 to cover it!
Great video, I like it! It's interesting and yet incredibly calming. It draws you in and immerses yourself into the topic. I didn't think that a bit of classical piano could do that to a video. I can imagine myself calming down to it in the evening or something. A portion of Maths before bed. Your calm voice goes with it.
for those still confused about Simpsons paradox, it's because the pair of vectors which have lower slopes has most of its magnitude distributed into its higher slope vector, whereas the higher slope pair of vectors doesn't. basically it's the weighting. 5/5 is larger than 499/500 and 1/4 is larger than 1/5, but 6/9 is smaller than 500/505
Hi - i am sorry but i cant imagine your explanation - i think it can happen but not always so the "always iff explanation wont make always working examples" ? regarding the first part of video i simply (as a phycisist) thought about normalized vectors instead of circles or spheres :) and found out there are also some lower + lower pairs that wont make higher sum. in 2D continuus space there is the whole complex theorems space that can happen :)
This was really neat. I did see the graph visualisation coming, but seeing a visual example of Simpsons Paradox was also pretty cool. And the irrational approximations was also pretty cool
Here because the thumbnail interested (and confused) me. 0:19 The thumbnail actually supports this. It had 1/2+1/2=1 being wrong and 1/2+1/2=2/4 written in as being "right." 0:30 Yes, but the thumbnail had the addition operator, not the mediant operator.
This was very reminiscent of a first year linear algebra course where you just jump from one result to the next, it was actually a very nice style which I enjoyed. I imagine a lot of people will need to skip back and rewatch some proofs (I usually watch on x3 speed, but here I had to take it all the way down to 1.5), but I don't think this is a bad thing, it just meant it was dense with information. The results were certainly better motivated and appeared more naturally than in a university course, and I especially liked how you drew from many areas to show results. This is a perfect mix of technical writing, recreational mathematics, and use of the video medium. Well done!
@@1.4142 I've got a firefox extension that allows me to change the playback speed. Gives me much more resolution in choice of speed, lower and higher speeds, and keyboard shortcuts to change speed. Works on pretty much any video as well, not just on youtube.
1:17 This function has fascinated me for years. To me the practical application is when you divide something up and divvy it evenly amongst multiple people. Take a wandering band of fishermen living around ancient Sumeria for example. Let's say 13 guys catch 6 lb. of fish. They happen across some old allies they still love, and decide to have supper together. The other wandering band is 8 guys who caught 4 lb. of fish. How can they equally allot the fish to each fisherman, in the combined supper that night? 6/13 (+) 4/8 = 10/21 . Each person gets 10/21 lb of fish to eat. Now let's say you come along, and tell them, you need to reduce the fractions first. So, the equation becomes 6/13 (+) 1/2 = 7/15 lb. See why we need to do away with this rule? Now i'm much less familiar with vectors, than making sure i eat enough. But from what you're showing around 5:00 in, it looks like my example should work with vectors also. The demand to reduce the fractions isn't a good idea.
There are two ways to understand vectors: first, in terms of x and y components, and second, in terms of magnitude and direction. In your example, the components are pounds of fish and number of fishermen; the magnitude doesn't really have an intuitive physical meaning here, and the direction is the ratio of pounds of fish per fisherman. When you sum the pounds of fish and sum the number of fishermen separately, you are performing standard vector addition using components. When you compute the combined pounds-per-fisherman, you are computing the direction of the resultant vector. The mediant, however, is a different operation than vector addition. It operates only on the direction of the vectors, irrespective of their magnitude. In your example, magnitude is relevant, so it is unsurprising that this operation produces odd results.
Can we do something with this?! Mediants are how almost every US teacher grades their students. a/b and c/d being grades on b and c "point" test/assignments are merged (a+c)/(b+d). The mediants provide weighted averages of tests/assignments. Mediants are how almost every statistical study is done. You send out dozens of collectors to take small samplings and count all the positive events (sum of numerators) and divide by the total samples (sum of denominators) Two samples of 1 out of 2, sum to one sample of 2 out of 4. But a sample of 8/9 and a sample of 1/1 sum to 9/10.
Your Stern-Brocot (mediant/parallelogram) tree is interesting. Perhaps it has some utility in combinatoric questions, such as the open no-three-in-line problem. In any event construction of the tree is simpler than attempting to enumerate unique slopes on a grid from scratch.
I have a probabilistic argument that the denominator in Hurwitz's theorem should be q^2, which I find pretty neat. It goes like this: Suppose you have an arbitrary irrational number, x. How good can you expect the "best" rational approximation with denominator < q to be? Well, since fractions with denominator q form a lattice with spacing 1/q, the distance from x to the nearest such fraction, p/q, can't be more than 1/(2q). So it makes sense to take the number h = 2 |qx-p| (which is always between 0 and 1) as a measure of how "good" the rational number p/q is as an approximation of x. If x is "randomly" chosen, we can say that h is uniformly distributed between 0 and 1. If you test all denominators between 1 and q to find the one with minimum h, you essentially have q independent tries. The expected value of the minimum of q independent numbers chosen uniformly at random between 0 and 1 is 1/(q+1), so we should expect the "best" rational approximation to x with denominator less than q to differ from x by about 1/(2q(q+1)), or, asymptotically for large q, something of order 1/q^2, which meshes very nicely with Hurwitz's theorem. From this perspective, it becomes really interesting that there are some simple-ish approximations for π (for example, 355/113) that beat this bound by quite a lot. Most other irrational numbers you might come up with (say, e, or √2), don't have such exceptionally good approximations.
Cool ideas! I haven't looked into this space very deeply or rigorously, but I suspect the reason why you can beat the bound significantly with pi, but not so significantly with some other irrational numbers, as a lot to do with the continued fraction representation. If I had to conjecture, I think there are some ways to metricize how close these rational approximations can get, and there's probably a metric in which the golden ratio is the furthest away from its rational approximations.
@@zhulimath For sure with the pi thing, you can see it in terms of the continued fraction expansion, which is a really tidy way to get the best rational approximations (that I didn't know about when I was thinking about this the first time). But if you ask me, that just pushes the question back a step. If you start computing the continued fraction expansion of pi, you get 3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, .... That 292 corresponds 355/133 being such a weirdly good approximation, and it's surprising that it shows up so soon. Why does that happen? Unlike for quadratic irrational numbers or for e, for example, there's no clear pattern to the continued fraction expansion for pi. And maybe there's no satisfying answer, but it is interesting.
@@rossjennings4755 I think the lack of pattern is because pi is transcendental (cannot be formed from a non infinite equation) and if a continued fraction has a repeated pattern then the result can be written as the solution to an finite equation e.g. sqrt(2) has denominators 2,2,2,2,2,2... and so (2+2/x)=x has roots sqrt(x). e should be the same as it is transcendental but I think it just tends to have smaller denominators than pi (I have no idea why).
6:03 Honestly, this process is going on a list of things I've learnt about that make me feel more and more like x/0 should really be defined as infinity at least sometimes.
Overall good concept. Honestly, there's something flawed in the flow of your arguments, as far as presentation is concerned. I kept going blank because I didn't know "what this is leading to", even though every single concept (Farey sequences, Simpson's paradox, mediants) are all familiar to me. Consider mapping out the steps you're leading in advance, because this video's main points are dependent on catching VERBAL content, and not mathematical content.
The part with the slopes reminded me of "a natural construction for the real numbers" by "Norbert A'Campo", where a real number is defined as an equivalence class of objects called slopes, which are "almost linear" functions on the integers.
13:30 I got very confused here because I thought that this is the determinant of 2 vectors shown on the screen, which should be 2. You should've changed a b c d with actual numbers for a moment to make it a bit clearer.
Ahhh, that's a very good catch. I think I would have color-coded the ad-bc to be green and red, then substituted them into the matrix to make that clear. Sorry about the confusion. You're right here, and this is going to be a small technical error that will bug me until the end of time!
0:36 - 0:50 The reason behind this is that the mediant is technically a function which takes in pairs of pairs of integers, not pairs of rational numbers. This is a technical, but important distinction, because a rational number is not merely a pair of integers. To get more rigorous, the set of rational numbers Q is not the set of pairs of integers Z^2. However, they are closely related. Regardless, the point is, irrational numbers do not have this relationship to Z^2, so talking about the mediant in this context is meaningless. However, the mediant is not even well-defined on Q, only on Z^2. The relationship between Q and Z^2 is *not* preserved by the mediant, and this is what needs to be understood. For those readers who are interested in the abstract, technical details: the set Q is isomorphic (equivalent to, and in a very special kind of way; it can be effectively treated as equality) to the quotient (Z×Z\{0})/~, meaning that Z×Z\{0} (the Cartesian product of Z and Z\{0}, which os just Z without 0) is partitioned according to the equivalence relation ~. How is ~ defined? We say (a, b) ~ (c, d) if and only if a•d = b•c. From this definition, it is easy to prove that (a, b) ~ (c, d) if and only if (c, d) = k•(a, b) for some nonzero integer k. The last sentence above is an important detail. The mediant is well-defined for Z^2, but if we want it to be well-defined for Q, we need it to be compatible, unaffected, by the equivalemce relation ~. Essentially, this means that multiplying the involved pairs by nonzero integers should not affect their mediant. However, it very easily does. The mediant of (a, b) and (c, d) is the component-wise sum, equal to (a + c, b + d). However, this means that the mediant of k•(a, b) and λ•(c, d) is (k•a + λ•c, k•b + λ•d). In most cases, this is not equal to (a + c, b + d). As such, the mediant is not well-defined for Q. So, whenever they ask you to take the mediant of two rational numbers, and they insert the caveat that the rational numbers must be "written in standard form," what they are *actually* asking you to do is to take the mediant of pairs of integers, where both pairs satisfy gcd(x, y) = 1, and then take the rational number corresponding to said pair. The language and notation is misleading, when it comes down to it, but there are reasons behind why it is done like this, some of which this video addresses.
Zhu Li, you did the thing! (Sorry, couldn't resist that one.) I stumbled across Hurwitz's theorem a few years ago, but this is the first time I had ever seen a proof of it. I'll have to take some time to digest that to see if I really understood it -- some of those steps really flew by. But it's neat to see that you can do it in such an elegant, geometric way.
1:11 The non-simplest-form Mediant cannot be "any real number". It can be any _rational_ number _between_ the two fractions being added. Getting a result outside that range is impossible (unless you use a negative denominator for one of the fractions, but forbidding negative denominators is a far looser requirement). You acknowledge this at 7:43, but don't specifically call back to the different statement early in the video.
You are correct if the fractions are rational. As mentioned in the video, in order for the mediant of two numbers to be well-defined, they must be rational and in simplest form. If you do not accept that the fractions must be rational, then any real number can be the resulting mediant.
@@zhulimath Well, if the fractions are irrational but still with positive denominators, then the result can be any real number between the two fractions. Making them irrational won't break them out of the range; but yes, my "rational" correction was unjustified.
Great video but there is a thing that I don't understand. In 0:48, you say that the fraction must be in simplest form. But then you use 3/6 which isn't simplest form at 4:45.
Good question! We say that in order for mediants to be well-defined, they must be fully simplified. Here, "well-defined" means that there can only be one possible answer/result/outcome. However, we don't always need this well-defined property if we have enough context. The vector visualizations work independent of the mediant concept. Also, if we fix the numerator and denominator as they are, meaning that we do *not* let 1/2 = 2/4 for instance, then this is an alternate way to make mediants well-defined. (I don't mention this in the video due to time constraints.) In the case of Simpson's paradox, we can use the second approach to making mediants well-defined. We are visualizing the bucket probabilities as vectors, rather than fractional values, in a way where we want to preserve the exact numerators and denominators, because we have to combine the buckets later. We need to know the exact number of balls. The logic behind how the visualizations work is still the same in this case!
You can think of the mediant as translation of vectors. a/b (+) c/d which should be (a + c)/(b + d). given that the nominator is the x component and the denominator is the y component, you can compare a+c to c+a. ‘a’ vector translating by +c in the x axis, or ‘c’ vector translating by +a in the x axis
This video wasn't meant to be a out tf2, but I think I now better know how trimping with demoknight works. Granted it's mostly eyeballing how much you need to turn to gain an adequate amount of speed, but if you know how much is too much then I feel like estimating what the median of that is would help improve your trimping skills a ton!
This video is literally my mind mid-exam "So x divided by y = z." My other part of my consciousness: "hello vesauce here, what if we calculated it differently? What if we change it? And is this question have similarities to questions 5 in page 3?" 5 minutes fly by that moment 😂
Amazing video, but not for when I'm delaying going to sleep and only half conscious. Gonna mark the video to watch tomorrow and will give my opinion then, when I can understand anything.
I don't understand the argument starting at 15:13. What is to prevent, when honing in on the supposedly missing fraction, the mediant of the bounding fractions (to the left and the right) to miss the denominator of the missing fraction ? E.g., in your example, to jump straight to irreducible fractions with denominator >= 7 ? So that no other mediant from that point on could achieve 3/5 (given that the mediant will continue to produce fractions in lowest terms and the denominator of a mediant of two fractions is > either denominator.)
At 11:10, we explained and showed that when you take the mediant of two fractions, you must obtain the fraction in between with the lowest denominator (if ad-bc=1). If 3/5 is between your two fractions, and so is 4/7, your mediant cannot possibly be 4/7, because 3/5 has a smaller denominator.
In my mind, this means that when I mistakenly added fractions like that as a kid, I was actually doing some crazy big brain shit. Take that, primary school teachers
I know that your comment in particular is a bit of fun and not meant to be interpreted seriously, but it highlights what I think is an interesting idea: What makes something clever is often not the action or thing by itself, but the reasoning and ideas behind it.
> we should learn to embrace this exploration, and let our curiosity take us where we want. if we end up where we expected, great. if we end up somewhere completely different, that's also great. this is one of the most important parts of math, and missing it might make math seem boring when it's really the opposite (although i wouldn't say that all cases of being bored by math are caused by this). maybe not even math specifically, maybe it's way more general, but i don't know a lot about that. thank you very much for formulating it, i'm glad i watched this video. the reasons why i love math have been just vague feelings and intuition for a very long time, and it's really nice to understand them and maybe even be able to explain them to others. although to fully comprehend these reasons, i would probably need to make a list of them, and it would be a loooong list :D
0:42 . what does it mean for fraction to be of rational values and in simplest form. To me simplest form makes sense only for fraction with integers, so why say it has to not have irrationnal values?
You're correct in that there is no well-defined way to represent an irrational value as a fraction in simplest form. I am simply stating that in order for the mediant to be well-defined, not only do the values need to be rational, they **also** need to be in simplest form.
21:21 Everytime I see √5 I think "there must be a Golden Ratio hiding here...", but why? Then you showed the triangle and I understood why √5 is necessary, but I still don't know where φ is 🤔 edit: 23:38 lol, now the mystery is solved. Thank you for explaining
9:52 the proof isn't complete, gcd(a,b)=gcd(a,c)=gcd(d,c)=gcd(d,b)=1 doesn't imply a+c and b+d have no common divisor. For example 1+1 and 1+3 have 2 as a common divisor. Edit: I watched again and found out I was wrong 😅. Area([a+c, b+d], [c, d]) = 1 implis that a+c and b+d have no common divisor.
I accidentally did something like this, and got the right answer. Much later I needed to take the same code/math and update it to account for multiple fractions being added together, and it all fell apart. Took me a while to realize what the correct answer was, but it was a fun journey. I think it had something to do with accidentally solving for the reciprocal of the fraction, then forgetting that step occurred. Pretty simple with two fractions, much more complex and obvious with three.
Thanks a lot for this video! I've recently been learning about continued fractions and best rational approximations, and this idea of treating fractions as vectors really demistifies a lot of these concepts.
Imagine teaching fractions in elementary school and a kid says "I'm not doing addition wrong, I'm computing the median" and then explains everything from this video.
That's some Terry Tao shit
Bro's believes hes Gauss
@@alonelyphoenix8942
“By the time Gauss was 7 years old, his schoolmasters admitted that there was nothing more they could teach the boy.”
~David Burton, Elementary Number Theory
*Mediant. Median is a completely different concept & the terminology can get confusing quickly.
and then you get told off because these are not medians but mediants, and the task was to add fractions
i kid you not, THE DAY after i watched this i had a math competition, and one of the problems was to find the fraction will the smallest denominator between 2 fraction, i tried trial and error and then remembered this video and immediately got the answer
Baader Meinhof phenomenon
i was the apprenticeship instructor for the roofing program in calgary for 13 yrs...
my most fulfilling memory was being able to teach fractions and the metric system to 40 yr old roofers with learning disabilities, addiction problems and 'incomplete' education scenarios...
seeing the look on someone's face when they actually get it and feel good about themselves...i was so blessed to help...
I'm in high school. Clicked with great curiosity to watch and expand my knowledge but it slowly kept getting more and more complex until my brain couldn't understand
That's ok! It's amazing that you have curiosity about math. The video can get a little advanced at places, it's not a video meant for everyone. Don't worry too much about it, just try your best and take things at your own pace, maybe come to it in the future when you have stronger foundations.
Some of my other videos might be a little simpler and easier, if you want to give them a try, like this one: th-cam.com/video/3B-D3w292TI/w-d-xo.html
I am in highschool too, few months ago I saw it understood not very much, today rewatching understood more then that time(not everything obviously) so maybe in my next re watch I will understand more
Even if the other entries in SoME3 are incredible, this deserves at the very least an honourable mention.
Good news!
Finally someone talked about this ! Last year I rediscovered most of this, in a attempt to find an algorithm that converts computer floating point numbers into a ratio, without suffering from the precision loss of floating point arithmetic. I couldn't find anything about this on the internet, until a friend of mine did. I'm happy more people learn about this simple but very interesting maths concept !
Just wondering, did you do project Euler? If so then for that problem I think what I did was if the |ratio - floating point|
@@siddanthvenkatesh2744 I don't know about project Euler, but yes you avoid any operations on floating numbers to avoid losing precision, and just use comparisons in order to know if your ratio is close.
That was the first thing I thought of when thinking of a fun application for these concepts. A friend of mine actually fiddled around with similar ways of representing floats after I told them that the only way to never lose any precision was to always preserve the full chain of operations that occur when some variable becomes imprecise, then recalculate values on the fly with the desired precision when retrieving them in the future. I believe they did end up representing floats as ratios of VLIs, but I'm not sure how far they got in terms of being able to achieve any desired precision.
Actually floating point numbers are alreaady exact rational numbers; imprecision comes in only when you try to calculate a number that cannot be exactly represented as floating point value. However when converting to a fraction, that loss alredy happened, and if you convert to anything but the exact value of that float, you are going to adda second imprecision on top of it. That second imprecision *may* cancel out the first, but unless you take into account knowledge on how you arrived at it, generally it will not.
Indeed, every floating point number (except for infinities and NaN, of course) can be written exactly as m*2^n where m and n are integers where m can be derived from the sign and mantissa bits, and n can be derived from the exponent bits (with some special handling for denormalized numbers; those however are also of the form above, just the formulas for m and n are slightly different). Of course in that representation you lose the signed zero (unless you store m in a signed-magnitude or 1-complement integer representation, which also has signed zero), but then, mathematically 0 = -0 anyway..
@@__christopher__ Floating point numbers aren't exact rational numbers because of infinite (binary) decimals that can't be represented. And in my algorithm for the conversion, you define the converted number as the one that if you convert back to a float, you get the original, so in a way there are no loss of precision during conversion. Also you can represent infinities, nan and -0 in a well-designed library with 1/0, -1/0, 0/0, 0/-1.
If ad-bc=1, you can use Pick's Theorem to prove a bunch of properties of the mediant. The parallelogram is a lattice polygon with area 1, and we know 4 points on its border. Since the area is the number of interior points plus half the border points minus 1, there must not be any points in the interior and no more points on the border, so there cannot be any rational numbers that would fall in this region or on its border.
Strongly agree
Yes yes definitely agreed
It is just nice when u start from very simple things and sort of play around with it to discover interesting observations. It feels like going in the reverse direction when u are reading a theorem. Instead of having a very complicated unintuitive statement thrown at you and then having to use every single brain cell to figure out why that even works in the first place, this just feels very satisfying. It feels like the thought process flows naturally, without resistance.
I love it when different bits of math come together so sensibly and beautifully. Excellent video.
Damn criminally underrated video. Good work man. Keep em coming
It was lovely getting an intuition on where the mystifying square root of 5 comes from in Hurwitz’s theorem!
Excellent video. I felt like every time you introduced a new concept, I had some questions pop into my head and thought "I'll have to google this after..." but then you answered the question in the video!
I was not prepared for this wild ride
The piano is so relaxing, thank you!
Wow, I saw the thumbnail and thought that the video would be something really simple, but then saw it was nearly 30 minutes long! Unexpectedly turned out to be an absolutely banger! Good job man!
Me too - i recall stern brocot series' "name" from some quantum or other foo-bar but only the name - its the first video somebody draw it for me/ made a picture of it :)
I planned on covering this topic as an interactive website for SoME1 for use as estimations. You came at it from a much better angle and covered more than I would have
Frick yeah, new zhuli video just dropped!
School teachers: "You can't divide by 0"
Mediant: "Hold my beer"
The intro riff sounded vaguely familiar to me, so I checked the description and it’s Kapustin! Great taste in music and great video
It's like a random walk from flower to flower in a garden - then at the end you realise you know the shape of the whole garden.
I like how my thought upon seeing the thumbnail was, "So like vectors?"
Yep.
this channel is so underrated!
I can’t believe this channel isn’t bigger, you’re doing amazing!
This is a great video! I love this graph in the outro, it looks cool and really helps to rewind the video in the head.
The theme is very cool too, I love when the video just starts investigating and playing with some concept just to see what happens. That's really my favorite part of math. Last time I caught that feeling when watching that video about hackenbush and surreal numbers. This was like my 2nd favorite math video of all time, and yours is really high up there!
It's a shame that you couldn't fit everything in it, on 18:00 the audio quality changes for a moment meaning that this part was recorded after everything else, so you probably didn't have enough time. So please make part 2 to cover it!
Wow I was not expecting Kapustin when clicking on this video, thank you for that
Great video, I like it! It's interesting and yet incredibly calming. It draws you in and immerses yourself into the topic. I didn't think that a bit of classical piano could do that to a video. I can imagine myself calming down to it in the evening or something. A portion of Maths before bed. Your calm voice goes with it.
This is a banger!let me share it! Eye opener
I ran into these structures while solving some project Euler problem. This explanation was just so perfect. Thank you.
jajaja. I was wondering why you weren’t just using the determinant at the beginning, good that I stayed. Great video! Interesting as well.
Someone very wise said : '' Math is you, a paintbrush and an empty board with infinite possibilities. ''
جميل جداً، عمل متقن ومثير للاهتمام
شكراً على الجهود المبذولة في هذا الفيديو
for those still confused about Simpsons paradox, it's because the pair of vectors which have lower slopes has most of its magnitude distributed into its higher slope vector, whereas the higher slope pair of vectors doesn't.
basically it's the weighting.
5/5 is larger than 499/500 and 1/4 is larger than 1/5, but 6/9 is smaller than 500/505
Hi - i am sorry but i cant imagine your explanation - i think it can happen but not always so the "always iff explanation wont make always working examples" ?
regarding the first part of video i simply (as a phycisist) thought about normalized vectors instead of circles or spheres :) and found out there are also some lower + lower pairs that wont make higher sum. in 2D continuus space there is the whole complex theorems space that can happen :)
This was really neat. I did see the graph visualisation coming, but seeing a visual example of Simpsons Paradox was also pretty cool. And the irrational approximations was also pretty cool
I took a break from a math practice test to watch a half hour long video about math.
Truly a chad
Spectacular video and ending message. Keep exploring, gathering adjacrent math topics, and learn something new!
would love a ford circle video! love the way you break things down :3
Here because the thumbnail interested (and confused) me.
0:19 The thumbnail actually supports this. It had 1/2+1/2=1 being wrong and 1/2+1/2=2/4 written in as being "right."
0:30 Yes, but the thumbnail had the addition operator, not the mediant operator.
clickbait at its finest right?
This was very reminiscent of a first year linear algebra course where you just jump from one result to the next, it was actually a very nice style which I enjoyed. I imagine a lot of people will need to skip back and rewatch some proofs (I usually watch on x3 speed, but here I had to take it all the way down to 1.5), but I don't think this is a bad thing, it just meant it was dense with information. The results were certainly better motivated and appeared more naturally than in a university course, and I especially liked how you drew from many areas to show results. This is a perfect mix of technical writing, recreational mathematics, and use of the video medium. Well done!
You have 3x speed?!
@@1.4142 I've got a firefox extension that allows me to change the playback speed. Gives me much more resolution in choice of speed, lower and higher speeds, and keyboard shortcuts to change speed. Works on pretty much any video as well, not just on youtube.
Very good video. Criminally underrated.
1:17 This function has fascinated me for years. To me the practical application is when you divide something up and divvy it evenly amongst multiple people. Take a wandering band of fishermen living around ancient Sumeria for example. Let's say 13 guys catch 6 lb. of fish. They happen across some old allies they still love, and decide to have supper together. The other wandering band is 8 guys who caught 4 lb. of fish. How can they equally allot the fish to each fisherman, in the combined supper that night? 6/13 (+) 4/8 = 10/21 . Each person gets 10/21 lb of fish to eat.
Now let's say you come along, and tell them, you need to reduce the fractions first. So, the equation becomes 6/13 (+) 1/2 = 7/15 lb. See why we need to do away with this rule?
Now i'm much less familiar with vectors, than making sure i eat enough. But from what you're showing around 5:00 in, it looks like my example should work with vectors also. The demand to reduce the fractions isn't a good idea.
There are two ways to understand vectors: first, in terms of x and y components, and second, in terms of magnitude and direction. In your example, the components are pounds of fish and number of fishermen; the magnitude doesn't really have an intuitive physical meaning here, and the direction is the ratio of pounds of fish per fisherman.
When you sum the pounds of fish and sum the number of fishermen separately, you are performing standard vector addition using components. When you compute the combined pounds-per-fisherman, you are computing the direction of the resultant vector.
The mediant, however, is a different operation than vector addition. It operates only on the direction of the vectors, irrespective of their magnitude. In your example, magnitude is relevant, so it is unsurprising that this operation produces odd results.
Can we do something with this?! Mediants are how almost every US teacher grades their students. a/b and c/d being grades on b and c "point" test/assignments are merged (a+c)/(b+d). The mediants provide weighted averages of tests/assignments. Mediants are how almost every statistical study is done. You send out dozens of collectors to take small samplings and count all the positive events (sum of numerators) and divide by the total samples (sum of denominators) Two samples of 1 out of 2, sum to one sample of 2 out of 4. But a sample of 8/9 and a sample of 1/1 sum to 9/10.
Your Stern-Brocot (mediant/parallelogram) tree is interesting. Perhaps it has some utility in combinatoric questions, such as the open no-three-in-line problem. In any event construction of the tree is simpler than attempting to enumerate unique slopes on a grid from scratch.
This is the first video I’ve ever seen from this channel. I love the choice of music for the intro lol - the end of Kapustin‘s 3rd Concert Etude.
Besides the content of the video, the music of the video is amazing.
Great video, watched the whole thing, hope you get more views 😄
we have 3blue1brown at home
meanwhile at home:
no im just kidding, great video dude, your editing chops are amazing
I have a probabilistic argument that the denominator in Hurwitz's theorem should be q^2, which I find pretty neat. It goes like this: Suppose you have an arbitrary irrational number, x. How good can you expect the "best" rational approximation with denominator < q to be? Well, since fractions with denominator q form a lattice with spacing 1/q, the distance from x to the nearest such fraction, p/q, can't be more than 1/(2q). So it makes sense to take the number h = 2 |qx-p| (which is always between 0 and 1) as a measure of how "good" the rational number p/q is as an approximation of x. If x is "randomly" chosen, we can say that h is uniformly distributed between 0 and 1. If you test all denominators between 1 and q to find the one with minimum h, you essentially have q independent tries. The expected value of the minimum of q independent numbers chosen uniformly at random between 0 and 1 is 1/(q+1), so we should expect the "best" rational approximation to x with denominator less than q to differ from x by about 1/(2q(q+1)), or, asymptotically for large q, something of order 1/q^2, which meshes very nicely with Hurwitz's theorem. From this perspective, it becomes really interesting that there are some simple-ish approximations for π (for example, 355/113) that beat this bound by quite a lot. Most other irrational numbers you might come up with (say, e, or √2), don't have such exceptionally good approximations.
Cool ideas! I haven't looked into this space very deeply or rigorously, but I suspect the reason why you can beat the bound significantly with pi, but not so significantly with some other irrational numbers, as a lot to do with the continued fraction representation.
If I had to conjecture, I think there are some ways to metricize how close these rational approximations can get, and there's probably a metric in which the golden ratio is the furthest away from its rational approximations.
@@zhulimath For sure with the pi thing, you can see it in terms of the continued fraction expansion, which is a really tidy way to get the best rational approximations (that I didn't know about when I was thinking about this the first time). But if you ask me, that just pushes the question back a step. If you start computing the continued fraction expansion of pi, you get 3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, .... That 292 corresponds 355/133 being such a weirdly good approximation, and it's surprising that it shows up so soon. Why does that happen? Unlike for quadratic irrational numbers or for e, for example, there's no clear pattern to the continued fraction expansion for pi. And maybe there's no satisfying answer, but it is interesting.
I'm sure there are good reasons for it, but I'm not too well-versed on this (yet).
@@rossjennings4755 I think the lack of pattern is because pi is transcendental (cannot be formed from a non infinite equation) and if a continued fraction has a repeated pattern then the result can be written as the solution to an finite equation e.g. sqrt(2) has denominators 2,2,2,2,2,2... and so (2+2/x)=x has roots sqrt(x). e should be the same as it is transcendental but I think it just tends to have smaller denominators than pi (I have no idea why).
Great video, I'm subscribing!
Fun exercise: Find the product of 2*sin(πx) where x ranges over all fractions in the Farey sequence of order n, excluding the endpoints 0/1 and 1/1.
6:03
Honestly, this process is going on a list of things I've learnt about that make me feel more and more like x/0 should really be defined as infinity at least sometimes.
Emphasizing "sometimes" and given that "infinity" comes with as little baggage as possible, I completely agree!
Overall good concept. Honestly, there's something flawed in the flow of your arguments, as far as presentation is concerned. I kept going blank because I didn't know "what this is leading to", even though every single concept (Farey sequences, Simpson's paradox, mediants) are all familiar to me. Consider mapping out the steps you're leading in advance, because this video's main points are dependent on catching VERBAL content, and not mathematical content.
The part with the slopes reminded me of "a natural construction for the real numbers" by "Norbert A'Campo", where a real number is defined as an equivalence class of objects called slopes, which are "almost linear" functions on the integers.
This is a killer math video will be recommending to all my friends! 👍👍
13:30 I got very confused here because I thought that this is the determinant of 2 vectors shown on the screen, which should be 2.
You should've changed a b c d with actual numbers for a moment to make it a bit clearer.
Ahhh, that's a very good catch. I think I would have color-coded the ad-bc to be green and red, then substituted them into the matrix to make that clear.
Sorry about the confusion. You're right here, and this is going to be a small technical error that will bug me until the end of time!
This video gives me a brand new perspective of how to look at math. Thanks
I never thought something arising from this much of abstractness would attract me to this extent
0:36 - 0:50 The reason behind this is that the mediant is technically a function which takes in pairs of pairs of integers, not pairs of rational numbers. This is a technical, but important distinction, because a rational number is not merely a pair of integers. To get more rigorous, the set of rational numbers Q is not the set of pairs of integers Z^2. However, they are closely related. Regardless, the point is, irrational numbers do not have this relationship to Z^2, so talking about the mediant in this context is meaningless. However, the mediant is not even well-defined on Q, only on Z^2. The relationship between Q and Z^2 is *not* preserved by the mediant, and this is what needs to be understood.
For those readers who are interested in the abstract, technical details: the set Q is isomorphic (equivalent to, and in a very special kind of way; it can be effectively treated as equality) to the quotient (Z×Z\{0})/~, meaning that Z×Z\{0} (the Cartesian product of Z and Z\{0}, which os just Z without 0) is partitioned according to the equivalence relation ~. How is ~ defined? We say (a, b) ~ (c, d) if and only if a•d = b•c. From this definition, it is easy to prove that (a, b) ~ (c, d) if and only if (c, d) = k•(a, b) for some nonzero integer k.
The last sentence above is an important detail. The mediant is well-defined for Z^2, but if we want it to be well-defined for Q, we need it to be compatible, unaffected, by the equivalemce relation ~. Essentially, this means that multiplying the involved pairs by nonzero integers should not affect their mediant. However, it very easily does. The mediant of (a, b) and (c, d) is the component-wise sum, equal to (a + c, b + d). However, this means that the mediant of k•(a, b) and λ•(c, d) is (k•a + λ•c, k•b + λ•d). In most cases, this is not equal to (a + c, b + d). As such, the mediant is not well-defined for Q.
So, whenever they ask you to take the mediant of two rational numbers, and they insert the caveat that the rational numbers must be "written in standard form," what they are *actually* asking you to do is to take the mediant of pairs of integers, where both pairs satisfy gcd(x, y) = 1, and then take the rational number corresponding to said pair. The language and notation is misleading, when it comes down to it, but there are reasons behind why it is done like this, some of which this video addresses.
1:45 calculating arithmetic average of whole numbers?
5:06 this is easyer than what my teacher makes me do lol
Zhu Li, you did the thing! (Sorry, couldn't resist that one.)
I stumbled across Hurwitz's theorem a few years ago, but this is the first time I had ever seen a proof of it. I'll have to take some time to digest that to see if I really understood it -- some of those steps really flew by. But it's neat to see that you can do it in such an elegant, geometric way.
What a beautiful video. I should do more mathematics.
I'll make a mental note to rewatch this video. I think it's going to need multiple watches to absorb and understand everything.
Music makes this video perfect
1:11 The non-simplest-form Mediant cannot be "any real number". It can be any _rational_ number _between_ the two fractions being added. Getting a result outside that range is impossible (unless you use a negative denominator for one of the fractions, but forbidding negative denominators is a far looser requirement). You acknowledge this at 7:43, but don't specifically call back to the different statement early in the video.
You are correct if the fractions are rational.
As mentioned in the video, in order for the mediant of two numbers to be well-defined, they must be rational and in simplest form. If you do not accept that the fractions must be rational, then any real number can be the resulting mediant.
@@zhulimath Well, if the fractions are irrational but still with positive denominators, then the result can be any real number between the two fractions. Making them irrational won't break them out of the range; but yes, my "rational" correction was unjustified.
Amazing video. Came in there without any expectations, and it fulfilled my interests much more than I expected.
The video could be subtitled to include an answer: "You discover the beauty of math!"
4:10 I'm getting Chopin Nocturne vibes from the background music and kind of digging it.
Great video but there is a thing that I don't understand. In 0:48, you say that the fraction must be in simplest form. But then you use 3/6 which isn't simplest form at 4:45.
Good question! We say that in order for mediants to be well-defined, they must be fully simplified. Here, "well-defined" means that there can only be one possible answer/result/outcome.
However, we don't always need this well-defined property if we have enough context. The vector visualizations work independent of the mediant concept. Also, if we fix the numerator and denominator as they are, meaning that we do *not* let 1/2 = 2/4 for instance, then this is an alternate way to make mediants well-defined. (I don't mention this in the video due to time constraints.)
In the case of Simpson's paradox, we can use the second approach to making mediants well-defined. We are visualizing the bucket probabilities as vectors, rather than fractional values, in a way where we want to preserve the exact numerators and denominators, because we have to combine the buckets later. We need to know the exact number of balls. The logic behind how the visualizations work is still the same in this case!
@@zhulimath Thank you very much.
intro is fire
Is it SoME time already?? Heck yeah!
Hmm it seemed like junior high school math, but that vector approach was simply brilliant
26:45 - Is this flowchart available somewhere? If not, please, make it so.
No promises, but I will try to work on this sometime in the near future. Probably shouldn't be too hard, but like I said, no promises :)
Hello, I have made a poster for you!
drive.google.com/file/d/1tKWcIdzvF86TxqprY_i9Rhw56wByn2Jg/
@@zhulimath Thank you so much!
what software is used to animate these? 3b1b and this channel and many others all seem to have the same designs 13:05
I used Manim, the Python library 3b1b initially developed!
@zhulimath Oh! That explains a lot! Thank you!!
isn't simpson's paradox just…gerrymandering
You can think of the mediant as translation of vectors. a/b (+) c/d which should be (a + c)/(b + d). given that the nominator is the x component and the denominator is the y component, you can compare a+c to c+a. ‘a’ vector translating by +c in the x axis, or ‘c’ vector translating by +a in the x axis
This video wasn't meant to be a out tf2, but I think I now better know how trimping with demoknight works. Granted it's mostly eyeballing how much you need to turn to gain an adequate amount of speed, but if you know how much is too much then I feel like estimating what the median of that is would help improve your trimping skills a ton!
What is this comment lol
Heho vectors go brrrrrrrrr
Great video. Thank you
This video is literally my mind mid-exam
"So x divided by y = z."
My other part of my consciousness: "hello vesauce here, what if we calculated it differently? What if we change it? And is this question have similarities to questions 5 in page 3?" 5 minutes fly by that moment 😂
Amazing video, but not for when I'm delaying going to sleep and only half conscious.
Gonna mark the video to watch tomorrow and will give my opinion then, when I can understand anything.
What a nice video, I thought I understood math lol. Such a nice topic.
The title makes it look like it's just vectors, now I am starting to watch whether what I guessed was correct!
hi , can someone answer this question please
proove that for each n>1 there is a and b two integers such as 3/n =1/a +1/b
Truly delightful
Algorithm bump. I don't follow math well. Butt you did an excellent job. And the recap was really valuable. Wow
Deep. Elegant. AweSoME3.
I don't understand the argument starting at 15:13. What is to prevent, when honing in on the supposedly missing fraction, the mediant of the bounding fractions (to the left and the right) to miss the denominator of the missing fraction ? E.g., in your example, to jump straight to irreducible fractions with denominator >= 7 ? So that no other mediant from that point on could achieve 3/5 (given that the mediant will continue to produce fractions in lowest terms and the denominator of a mediant of two fractions is > either denominator.)
At 11:10, we explained and showed that when you take the mediant of two fractions, you must obtain the fraction in between with the lowest denominator (if ad-bc=1). If 3/5 is between your two fractions, and so is 4/7, your mediant cannot possibly be 4/7, because 3/5 has a smaller denominator.
@@zhulimath thanks, that clears it up!
In my mind, this means that when I mistakenly added fractions like that as a kid, I was actually doing some crazy big brain shit. Take that, primary school teachers
I know that your comment in particular is a bit of fun and not meant to be interpreted seriously, but it highlights what I think is an interesting idea: What makes something clever is often not the action or thing by itself, but the reasoning and ideas behind it.
> we should learn to embrace this exploration, and let our curiosity take us where we want. if we end up where we expected, great. if we end up somewhere completely different, that's also great.
this is one of the most important parts of math, and missing it might make math seem boring when it's really the opposite (although i wouldn't say that all cases of being bored by math are caused by this). maybe not even math specifically, maybe it's way more general, but i don't know a lot about that.
thank you very much for formulating it, i'm glad i watched this video. the reasons why i love math have been just vague feelings and intuition for a very long time, and it's really nice to understand them and maybe even be able to explain them to others. although to fully comprehend these reasons, i would probably need to make a list of them, and it would be a loooong list :D
0:42 . what does it mean for fraction to be of rational values and in simplest form. To me simplest form makes sense only for fraction with integers, so why say it has to not have irrationnal values?
You're correct in that there is no well-defined way to represent an irrational value as a fraction in simplest form. I am simply stating that in order for the mediant to be well-defined, not only do the values need to be rational, they **also** need to be in simplest form.
Simpson's paradox, in the example you gave, isn't unintuitive at all. "Paradox" is overkill there for sure.
Dang, this is so dense with information that i looked away for 2 seconds and suddenly i have no idea whats going on😂
Math class all over again
21:21 Everytime I see √5 I think "there must be a Golden Ratio hiding here...", but why? Then you showed the triangle and I understood why √5 is necessary, but I still don't know where φ is 🤔
edit: 23:38 lol, now the mystery is solved. Thank you for explaining
Amazing video. Subscribing.
Very cool video, thank you!
Very cool way of visualization
9:52 the proof isn't complete, gcd(a,b)=gcd(a,c)=gcd(d,c)=gcd(d,b)=1 doesn't imply a+c and b+d have no common divisor. For example 1+1 and 1+3 have 2 as a common divisor.
Edit: I watched again and found out I was wrong 😅. Area([a+c, b+d], [c, d]) = 1 implis that a+c and b+d have no common divisor.
I accidentally did something like this, and got the right answer. Much later I needed to take the same code/math and update it to account for multiple fractions being added together, and it all fell apart. Took me a while to realize what the correct answer was, but it was a fun journey. I think it had something to do with accidentally solving for the reciprocal of the fraction, then forgetting that step occurred. Pretty simple with two fractions, much more complex and obvious with three.
How the fuck do you have 15k subs, this is amazing!!!
I came for the mediants, I stayed for Chopin.
Thanks a lot for this video! I've recently been learning about continued fractions and best rational approximations, and this idea of treating fractions as vectors really demistifies a lot of these concepts.