Like when Bill Bailey and Rich Hall were doing that Exploration episode on QI, where Stephen asked how native Americans and the founding fathers communicated - it just makes it that much more funnier
@@slowfreq When measured against infinity, ANY finite set of numbers means you have excluded "almost every number". You have excluded an infinite amount of numbers. Regardless of how big the finite set, it will always pale in comparison to the infinite amount that is left.
Calculations estimate the number of photons in the observable universe to be 10^89. Still alot missing to get to G64 Its quite simple. Even if you wrote the digits as small as the plank distance the entire size of the observable universe is not enough to write G64 down.
for those who would like to understand the question: imagine a 2 dimensional cube (also called a square). now try to connect the corners with diagonal lines of two colors(4 corners, two possible diagonal lines), without all the four corners beeing connected with the same color. (quite easy: one red diagonal \ and a blue / diagonal) now, in 3 dimensions you have 8 corners, and 16 diagonals. and still the same task: try to draw all possible diagonals with only two colors without getting a square that has all it's corners in thesame color. (harder than in 2 dimensions, but still possible) now, the question is, at how many dimensions will it become impossible to accomplish.
You didn't state the problem very well, unless perhaps you know a simplification of the problem that I don't. Firstly, the way you said it, it sounds like the corners of the cube somehow receive a unique colour ("without getting a square that has all it's corners in thesame color"). But each edge can be coloured independently, so talking about corners confuses the problem. But more important is that you need to colour the usual edges of the cube too, not just the diagonals between the corners. It has nothing to do with "corners", per se. A better way to say it is that you consider _all_ possible edges between _every_ pair of vertices of the n-dimensional cube (in 3d, the original 12 edges, the 12 diagonal edges on each face, and the 4 edges between diagonally opposite vertices of the cube). You need to colour each such edge with one of two colours so that, whenever you pick four vertices in one plane (e.g., the four vertices of a face of the 3-cube, or 4 along a diagonal slice across the 3-cube), then both colours occur on an edge between some of those vertices. The theorem says that for sufficiently large dimensions you can't do this, but unfortunately the gap between where we know you can do it, and where we know you can't (at dimension Graham's number), is beyond cosmically large.
I already knew about the Grahams number. What Stephen didn't mention is that this is the biggest number that has ever een used in a calculation. It's also rediculously much bigger than a googolplex.
Something that escaped me the first time I seen this, but now that I'm thinking of it: I'm sure most people who see this assume the problem is "really hard" (and in the sense of solving it, it certainly is). To actually be able to understand what the problem means and to at least begin to understand how one would solve it, however, really is quite trivial once one knows basic algebra/geometry AND what is actually going on in this sort of math. It's simple layers built upon even simpler layers.
+lsrwLuke Here's my proof that that claim is false. Consider Graham's number + 1. It is greater than grahams number by one, and is used as the proof of this argument.
The question is actually really easy to grasp, they're just presenting it in the most obtuse way possible. It's literally drawing squares with diagonal lines. Children can do that. The proof is the hard part.
Grahams number is insane so i think this is how it goes correct me if i'm wrong The number uses arrow notation so it starts as 3^3 which is 27 then it is 3^^3 which is basically 3^(3^3) this equals 7.6 trillion then you get 3^^^3 which is 3^(3^(3^3) and this is a huge number i can't type out then it goes to 3^^^^3 which is another ridiculously big number now here's where it gets insane the ridiculously big number which is the answer to 3^^^^3 is now the number of arrows and the answer to that is the number of arrows for the next one and you do this 64 times until you get to grahams number
Actually, it's much, much worse than that. Start from 3^^3 which is, yes, about 7.6 trillion. A double arrow expression basically makes a power tower of the first number with a height of the second number. Next, 3^^^3 becomes 3^^(3^^3) or 3^^(~7.6T) which makes 3^3^3^3^3^3^3^....7.6 TRILLION TIMES....3. In other words, 3^^^3 is power tower of 3s which is over 7.6T high. Next, 3^^^^3 becomes 3^^^(3^^^3) which then equals 3^^3^^3^^3^^....3 where the number of threes is 3^3^3^....3 where THAT number of threes is 7.6T. Insanity ensues after only adding a single arrow each time, it's much MUCH worse than just exponentiation. That number, 3^^^^3, is called g1. The number 3^^^^^....^^^^^3, where the number of ARROWS is g1, is called g2. 3^^^^^....^^^^^3 with g2 arrows is g3, and so on. Grahams number (the more popular version at least) is g64, or big G. It's absolutely incomprehensible.
Graham's number is significant because it was the largest number every actually used, sure you can think of bigger numbers but we haven't found a use for them yet. So there is another large number called a googolplex 10^(10^100). If you filled the observable universe with grains of sand, a googolplex is very roughly equal to the number of ways you could arrange the grains of sand. The relative difference between 1 and a googolplex is basically the same as a googolplex to Grahams number.
@@JaneDoe-ci3gj "The relative difference between 1 and a googolplex is basically the same as a googolplex to Grahams number" Not even close to a comparison. Graham's number is so much bigger than that. Also, if you want an even bigger number, look up tree(3).
So he read out the question and my mind froze over.. I listened to the words in a fog of 'wtf' and then thought "6 !" I was totally amazed when norton said 6 and even more so when steven said that 6 had been most peoples answer. Lucky guess or passive abilities hidden by stupidity? Wrong nonetheless, but curious now..
It wasn't really what people (graph theorists) thought was the answer. It was a lower bound on the answer - they didn't know the answer, but they knew it couldn't be less than six. Now it's known that it cannot be less than 13.
Interesting, the fellow on the far left reminds me of a UK version of Timmy from WKUK. I have to say I find the answer to the question hilarious. It's like saying "It's somewhere between a little bit, and a bit less than infinity".
that people with actual intelligence instead of actors could solve it? now, please don't misunderstand: i don't mean that actors can't be intelligent, but Fry and Mitchell are not as intelligent as they are perceived to be. and what's even more important to this question is, that both of them have next to no idea about natural sciences or mathematics. (especially fry... he said so many extremelly false things on the show)
+andrew7taylor my anwer was to arthur, not you... but: does he? in hi wiki page it's only written, that he studied it. not that he actually got the degree. but yes, this would give him much more chances to understand this question as he actually knows, without having to think too much about it, what an n-dimensional hypercube is.
Can a mathematician imagine a __-dimensional hyper-anything without knowing the number of dimensions (Forgetting the ~difficulty~impossibility of imagining a number of spatial dimensions greater than 3 for a moment) or do they manipulate words? Either way, how do they do it?
No, he hasn't divided it by infinity. He's calculated the limit of (Graham's number / x) as x approaches infinity, which is 0. You might want to read up on limits of functions.
Graham's number is unimaginably larger than other well-known large numbers such as a googol, googolplex, and even larger than Skewes' number and Moser's number.
Am I missing something? (no sarcasm intended, I'm confused) Graham Norton did understand the question, even though he answered wrong... I know David Mitchell is smart, and Dara O'Briain is mathematical genius, but why doesn't anyone mentions Graham, who (almost) answered correct?
While generally true (and that's pretty funny actually ^.^), the more I have to use math programming various things the more I find much of what mathematicians do are quite useful. Granted sometimes they don't know it when they're doing it: while watching various college course videos from UNSW here on YT I've seen multiple equations/formulas/functions that were worked out hundreds of years before they were needed in computers to solve data sorting/seeking, the internet's encryption, ect.
Yeah, there's often more than one way of proving something, and which is better or more elegant depends on personal taste. I happen to like this proof because most people could understand it, even if it's by contradiction, which one of my maths lecturers once said was the last resort in proofs.
The not enough ink thing isn't about how big a number is. Stick a decimal point in after the first digit and it's then quite a small number with a lot of kerfuffle. And also in need of ever so slightly more ink for the decimal point.
"It is so big that all the material in the universell couldnt make enough inc to write it out". Thats a googolplex. One exponent multiply 3 with itself 1 googolplex times. This number is the amount of times 3 needs to multiply with itself. Repeat for about 7 trillion times. Thats only 3 arrows in. To make the 4th we even need to make the same procedure but with rows instead. And 5th arrow is repetitions of this structure. But the amount of arrows are the number we got from arrow 4. And we Repeat that 64 times
He has a degree in history. Just because he went to Cambridge doesn't make him an authority on all things academic. Same as Stephen Fry, he's a well read intelligent man but when it comes to the science things on QI he's just reading the autocue and only has a superficial understanding of whats going on. His maths ability will be roughly the level of someone who did GCSE maths. Brian Cox on Would I Lie to You made the same mistake talking about physics to David assuming he would understand.
You've misunderstood g1. 3↑↑↑↑3 = 3↑↑↑(3↑↑↑3) But what you've taken it as is to be 3↑↑(3↑↑↑3). To understand the quadruple up arrow, you begin with 3, and that describes the height of a power tower, which gives us 3^(3^3) = 7.6 trillion, then that number describes the height of the power tower 3^^^3, and then that number describes the height of the next power tower (this is where you stopped) but then that number describes the height of the next... and again and again... repeat this 3↑↑↑3 times.
Graham's number IS huge. Let's start off: so 3↑3 is 3^3=27. 3↑↑3=3↑(3↑3)=3^(3^3)=3^27=7.6 trillion. 3↑↑↑3=3↑↑3↑↑3=3^3^3^3^3... where the stack is 7.6 trillion threes high. Now imagine 3↑↑↑↑3 (where 3 is raised to itself 3↑↑↑3 times), and set that equal to g(1). g(2) is 3↑↑↑...↑↑↑3 where there are g(1) arrows. g(3) is 3↑↑↑...↑↑↑3 where there are g(2) arrows. Keep doing this until g(64), and that's Graham's number--much, much larger than googolplex.
I just want to start this comment by saying nobody thinks you're stupid or ignorant, just relax a little. What you've said is true, but Cantor had another, more convincing proof in which an infinite series of real numbers, each of infinite length was used to construct a new infinitely long real number which was different to all the ones in the set, thereby proving that you could have different sizes of infinity. See here (remove spaces): math. bu. edu/ people/ jeffs/ cantor-proof
OK, I'll play. Infinity is limitless and indefinable, because any assigned value can be endlessly increased. Georg Cantor, who I think you're referencing, reasons that there's an infinite number of natural numbers but also an infinite number of real numbers between each successive natural number, so the tally of real numbers is an infinite multiple of infinity. I suspect the loophole was saying that natural numbers represented an infinite quantity when they were only a subset of a larger group.
I could watch the video all the way through and lie that I knew the answer in advance or I can be honest and watch it up until he finished the question and then post my comment but I would still get the same response either way dubbing me as a "liar" . There is no way of telling if a person genuinely knew it unless he/she had proof that they are as smart as they say they are be it on their channel, on facebook or google plus or any other site.
@JaxWeb Yeah it drops very very gradually, but then over long enough time (see: infinity)... ^.^ Anyhoo, I think we've well established now that infinity does all sort of crazy nonsense. The sum of the all integers in the series -1, 2, -3, 4... to ∞ isn't an integer is probably my favorite, but there really is no need to pick on one when there are (pardon the pun) nearly countless oddities when working with ∞. That aside: how sad is it that ∞'s ASCII number is an embarrassingly finite 236? XD
Grrr, RU-vid won't accept my pseudocode so I'll just write it out in basic logic: For every prime (example: 2), the total percentage of whole numbers beyond this prime that are also prime is divided by that prime and all previous primes and added to the sum of all primes divided by previous primes (example: after "2", half of the whole numbers beyond it can't be prime because they are divisible by two (ie all even numbers); then for "3", the next prime, 3/2 of future numbers are now also out).
@Ensirum Yeah all sorts of interesting logical things happen near infinity. I couldn't help notice a function I was working on the other day, checking the percentage of prime numbers given a set of numbers (n); the percentage of prime numbers approaches 0 as n approaches infinity (so "near infinity" the prime numbers will run out; thankfully we can never reach that point; I gotz to have me some primez ^.^).
Way bigger. you know how 2^3=2x2x2? Well 2^^3 = 2^2^2. 2^^^3 = 2^^3^^3 or 2^3^3 and keep doing ^3 7,625,597,484,987 times. Getting an idea of how large we are getting? Now Graham's number starts with 3^^^^3. That's g1. g2 is 3 and then g1 ^'s then a 3. g3 is 3 then g2 number of ^'s and then a 3. Grahams number is g64. As you can see, this number is just ridiculously large.
So is Grahm's number (don't know how to spell his name) bigger than a Googelplex? A Googelplex is a trillion times a trillion a trillion times, which is a Googel, then a Googel times a Googel a Googel times. A Googelplex is so big that even if you somehow wrote a 0 on every particle in the universe there would not be enough.
I know we can never get to infinity, however your observation is that as you approach infinity the percentage of prime numbers reduces to 0 in the set, surely you need to show that the number is never 0 pre infinity. Limits were never my interest/strong point and I haven't done proper maths in ages.
well, your observation is that as we approach infinity there are 0 primes in a set of numbers n. The conclusion you say you reach from this is that there are infinitely many primes, just with the statement "we can never get there". I don't see how this is a proof, maybe I'm too dense.
Suppose there are finitely many primes. Multiply them all together. Add 1. The resulting number does not factorise into any of the other primes, therefore it must also be prime, which is a contradiction; hence the original statement must be false, i.e. there are infinitely many primes. QED
@NCaradoc2008 No, not infinity + 7 XD. Anything finite (as Graham's number is) is an infinite distance away from infinity, so it's as far away from infinity as 1 is. It's just incredibly, incredibly large. "Large" isn't applicable to things like infinity, despite some people's notion of it.
Of course Georg Cantor didn't reason in that kind of mushy and nonsensical terms. He simply said that sets have the same size if there is a bijection between them, and proved there is no bijection between the integers and the reals. Suck on that.
I don't think you understand the concept of infinity. It definitely doesn't mean the same as an inconceivably large number, which is what Graham's number seems to be. If your idea of infinity can be conceptually enlarged in any way, then it isn't infinity.
All I can suggest is looking up /wiki/Asymptote ^.^ (or, again, just ignoring my original post since I was talking about reaching infinity, and that does all sorts of crazy stuff that I couldn't properly explain outside of 6 months =D)
@Truthiness231 Actually you're wrong there :P There's an infinite amount of prime numbers. They aren't running out, they're just spread further out. You were perfectly right with "approaches 0" but then, how many decimal 0's can you fit before a 1? Infinite.
The video never describes the process of calculating the quadruple up arrow, and also, take a look at one of the top comments. Plus, he's incorrect specifically at the part where he explains what g1 is. The rest is fine.
Tiny doesn't even do the size difference it justice. If you could write a googleplex on every particle in the universe, and then multiplied all them together it would still be tiny compared to Graham's number.
Well, theoretically, you can in vision Googolplexian in your head. Graham's number, id visualize, would, in all serious, turn your head into a black hole. That's not joke either, it would turn your head into a black hole.
I just invented a number twice as big as Graham's number, which is the largest number of dimensions required to make TWO of what Stephen said. It ends with a 4. It shall henceforth be called Maximilian's Number.
@Truthiness231 In the same sense, if you have people in a square formation, each an infinite amount of distance from each other, standing on an infinitely radial disc. They are all relatively at the center of the disc.
I don't think Graham Norton understood the question, he just went for 6 because it's the number of faces on a cube and perhaps he thought it was relevant or perhaps he was just being humorous.
Since we only have 500 characters to put numbers is the answer is no. We couldn't even put a googolplexian in this small space and a googolplexian is 1 compared to Graham's Number
It would probably just try to boil very hard but eventually fail to perform in the pot due to fear of what psychopaths might think of it if it doesn't boil and boil to perfection
I doubt anyone would actually get that by themselves. But anyone could say "yeah I got it too" even if they didn't. No point in asking really since most people troll
@Alexbrainbox Oh, In that case still no. Grahams number is so massive you couldn't write it out even if you able to write on each individual atom in the observable universe.
Well yes, but even so one is more likely to understand graph theory with a background in science than history. Of course, none of them were going to get it.
@starofcctv94 I mean, actually physically write the numbers smaller, like zoom in far so you would use less ink, rather than writing it mesoscopically?
