Welcome under Another Roof, where there is always another proof. Videos about mathematics, mathematical logic, and the history of mathematics. Did I mention I like mathematics?
I tried searching for the specific sets of cards where all 900 target numbers are reachable. It turns out that every single one has at least one large number in it, meaning the rat pack is at a disadvantage. Luckily, the 3-large and 4-large are also absent from the list, so they don't fare very well either.
I also tried searching for solutions involving fractions, because although they're not allowed in Countdown, other calculation games of this sort like 24 typically do allow them. Apparently with the cards 1, 1, 2, 2, 3, 100 you can get 122 using fractions. Still trying to figure out how...
My limited understanding of AI suggests that we can set a step limit and employ a backtracking mechanism. It's possible that Alpha Geometry can explore multiple solution paths concurrently and remain within the desired complexity bounds.
The more it is trained on multiple solution paths including beautiful creative human thinking the more refined the output will be. The fact that it was even able to generate this cumbersome circuitous route is still inspiring to me because I see its potential. And I’m speaking as a pure mathematician focused on algebraic number theory.
"first we take y = f(x) and rename it into F(x, y) = 0. We haven't really done anything just changed the notation". This statement is wrong. You made algebraic geometers happy by doing that, which is a goal in itself.
I first saw Cube back in 1998 or '99 when it premiered on Sci-Fi Channel. It had always been my impression that it was a made-for-TV or straight-to-video movie, bc I don't remember it ever having been in theaters prior to being aired on television. A lot of low-budget, independent filmmakers of the '70s, '80s and '90s (especially those working in the horror and sci-fi genres) preferred this model for getting their product into the movie-consumption marketplace as it was much cheaper than using the theatrical model
So that way you can easily count with them. Say you have a set of apples {🍎, 🍏}, and you want to know how many apples are in the set. Intuitively you know it's 2. But how would you prove it. Well, if we say that two is {∅, {∅}}, then we can line up each item of the set of apples with each item of Two™ and they will have the same cardinality (amount of items), {🍎: ∅, 🍏: {∅}}, and so we can declare that that is what it means for there to be two apples. If two was defined as {{∅}}, we wouldn't have that advantage.
I have one really big question: How can you discuss the impossibility of doubling the cube in the framework of plane geometry? It seems like a strange non sequitur that it's part of the standard discussion of the subject.
EDIT: My previous post is still valid but I made a mistake in this one. Like... If you're allowing that you can construct a cube in the first place, then you can draw a line between the opposite vertices of a cube and thereby construct the cube root of 2.
To construct cube with volume 2, what is needed is the ability to construct ³√2, which could act as the side length of the cube. That's what's impossible.
This is such a fantastic video. So well done, accessible to different levels of math background, very approachable, funny... About 2 days ago i suddenly became obsessed with the more... shall we say, insane aspects of set theory. I jumped in the deep end and started with the continuum hypothesis, and then eventually I hit the axiom of choice. I am good at math and know a lot of stuff about a buncha things because i love it and I've done research, but i'm only in high school and I've never had any formal instruction in set theory. Pretty early into the research that started two days ago, i found this concept referenced in a longer video. It was explained well enough that i *understood* it as in i knew what they meant, but I was far from *comfortable* with the fact. I'm still not entirely adjusted, but this has been a huge help even though i already knew where it was all going. Thank you!
I don't discuss fields in this video but if you have a field F and extension E, then the set of all automorphisms of E which preserve F is the Galois Group of the extension. To find out more about field extensions, I examine them in my latest video!
You're such a fantastic educator, the way your enthusiasm comes through even with a carefully scripted video is always engaging! Lots of educational content can be hard to absorb for those of us with ADD, but you've turned what could be boring lectures into my favorite math youtube channel <3
Nice intro to Steiner systems. As a combinatorial design theorist, this was a beautiful presentation, and I even learned some new things I didn't know before.
There's a step in the proof of Gauss's Lemma I've always been bothered by because it was so often was glossed over, yet never obvious to me. It is this: Fix a prime p and polynomials f(x) = a(0)+...+a(m)x^m and g(x) = b(0)+...+b(n)x^n with integer coefficients. Claim: If p|fg then (p|f or p|g). In other words, if p divides all coefficients of fg then it must divide all coefficients of at least one of f or g. Here's a proof (of the contrapositive): Suppose neither p|f nor p|g holds. Then there must be a *first* term a(r)x^r of f and a *first* term b(s)x^s of g where p divides neither a(r) nor b(s). So (p|a(k) ∀ 0≤k<r) and (p|b(k) ∀ 0≤k<s). Now look at the coefficient of x^(r+s) in fg: ... +a(r+2)b(s-2) +a(r+1)b(s-1) +a(r)b(s) +a(r-1)b(s+1) +a(r-2)b(s+2) +... Note that all terms with the exception of a(r)b(s) are divisible by p, so that the coefficient of x^(r+s) in fg is *not* divisible by p. Therefore fg is not divisible by p. ∎ (Quantifying the contrapositive over all primes p yields the less explicit "the product of primitive polynomials over ℤ is primitive".)
When I had access to AutoDesk ~10 years ago, I did play around with some of those methods of making shapes without defining lengths and radii, but with making lines dependent on each other. But I believe I probably would have never come to use circles to create lengths only available with Pi. (it just looks so obvious now xD)
It seems like something is missing here. We assume that A+B is equal to B+A but what happens when A and B are not like units? It seems that you can still add them together but you lose something about the initial components by combining the units. 2in+3cm is not 5in or 5cm. And if you say that it’s 5 (cm and inches). If we walk back 3 inches and then 2 cm you don’t return to the same spot, even though we have used up 5 (cm and inches). Also I struggle with the sizes of infinity. If we just look at the infinity of the counting numbers and say it is one size, what is the size of the number of ways you can rearrange that set? Would that not also produce a counting number? But if the counting number or rearrangements is larger than the integers it just means we haven’t reached the end point. It also seems as there is no unique infinity…for there to be infinite the unit by which you must be counting is 0. I don’t mind the bijection, but I am not sure that is really the same thing or it is counting size. And you can make the diagonal argument just the same with any partial list of integers.
This is called high quality video , discussing Maths , I think with no beautiful animations this video is still at the level of 3b1b or greater than it ... Thanks for the video broo , keep making more Also I made a video about a new calculus, "discrete calculus" Can you make a video on it in your style ?
Just stumbled upon this randomly and am mightily impressed - I don't generally comment on videos. Thanks very much for this! I'm just diving back into Python, and I'm going to enjoy following your thought process!
"We're still a while away from seeing AI have a go at some of the unsolved problems across mathematics" Give it three to five years I guess. Unless some devs decide we need to solve them for ourselves, because otherwise our self-esteem will suffer.
