I believe that for 3↑↑↑↑3 the shorthand hexation tower would actually be the height of the result of 3↑↑↑3, not just 7.6 trillion high. Another way to write Grahal (G1) is 3↑↑↑(3↑↑↑3). When doing this you no longer have something as simple as 3↑↑↑3. You have 3↑↑↑x, where x=(insanely big number). To evaluate that you beak it down further as 3↑↑(3↑↑(3↑↑....3↑↑). Remember that in up arrow notation the trailing number is number of times the leading number is operated. Thus, to reach 3↑↑↑↑3 you must think of it as 3↑↑↑x where x= the number of 3's you get from calculating 3↑↑↑3. Put another way 3↑↑↑↑3 is a 3 "pentated" 3↑↑↑3 times. Now let me say something I often don't see when doing my research. The parentheses matter. When looking at 3↑↑(3↑↑(3↑↑....3↑↑) or something similar, each parentheses from left to right is generating a next tower. So 3↑↑(3↑↑) for example, is 3↑↑(7.6 trillion). We understand that of course. That's 3↑↑↑3. Cool. To reach Grahal we start working from right to left evaluating 3↑↑(3↑↑(....)↑↑3) an incredible number of times, and in doing so this means that every time you calculate left you are building a tower of 3's within paranthases, which will give you the height of tower to calculate as you move further left. Note that you are of course moving down your power towers right to left as normal (as you did when evaluating the tower of 3's 7.6 trillion high, etc), but after calculating a power tower you then go back to 3↑↑(3↑↑(....)↑↑3), having solved a term of just one of your parentheses, and you now move one step left, using the number you just calculated to give you the numbers of 3's in the tower you are working on next. As you can imagine, 3 or 4 steps in you reach crazy tower heights but this operation has to be done many many more times as the tower building nature of pentation demands. Finally, when you do this tedious right to left tower, then parentheses, then back to tower and so forth operation 3↑↑↑3 times you have G1. Back to your shorthand. First off, you do a great job explaining maths and I commend you. I am also not as familiar with left hand exponent type shorthand as I am with up arrow notation. Still, if hexation is repeated pentation, then what you wrote as pentation in shorthand (a left tower 3 3's tall) should be repeated 7.6 trillion times. Hexation is vastly more powerful than just a shorthand tower of 3's 7.6 trillion tall.
Thanks for the detailed explanation though it took some time for me to digest it...I too thought that something was not quite right for pentation notation too.. Won't it actually be a left 3s power tower of 7.2 trillion as pentation too is repeated tetration..
I have limited experience, so I may be off, but the general rule is that an n-arrow operator expands into a right-associative series of (n−1)-arrow operators. So I believe you are correct. I've seen pentation represented with a subscript left variable. A pentation of 3 to the 3 will have a left power tower of height ~7.63x10^12, a Hexation of 3 to the 3 would have a shaft of 3's descending to the left for the same height. If you wanted to express Hexation with a left power tower instead of a shaft, then it wouldn't be a tower of 3's ~7.63x10^12 high, but a left power tower ~7.63x10^12 high, composed of right power towers of 3 ~7.63x10^12 high.
Thank you! I commented on a previous video of yours that if anyone could explain Graham's Number to me in a way I could understand, it would be you. I was right! Though the number itself is completely unimaginable, you've made its derivation understandable. No mean feat! Thank you once again!
I'd like to suggest a sensible limit. There is a minimal amount of energy associated with each bit of information. ( 2.9×10^−21, Landauer's principle) If we assign one bit to each unit in the number, then which number has minimum energy exceeding the energy of the known universe? Which number has the minimum energy needed to create a black hole, etc. Apply that math.
One could substitute the G in Graham for the G in God ...the Creator of the finite universe. He's the only one Who knows what lies beyond infinity. That boggles my mind
Your number is between 3^^3 and 3^^^3. The latter is so unfathomably large that nothing you measure in the universe will dwarf in comparison, including the number of possible states of the universe, and number of possible universes (assuming finite universe).
Love your videos for their educational value, thought provoking and engaging. Your narrative is cogent and easy to follow. You’re an excellent explainer. I wish you would produce a video on “…gravity is not a force and we are not being pulled down…”.
@@PrimeNewtons You are that smart. More important is your ability to explain complex subjects in a way that others can understand. Please, don't sell yourself short. You have talent and you should explore expanding the breath of your subject matter.
Love your enthusiasm and amazement in how these numbers grow. Easier to say G2 has G1 arrows between the 3s, G3 has G2 arrows, etc. To draw out 3^^^^3 (G1), you construct a sequence of growing power towers. Start with 3^^3...that number says how many 3s are in the next power tower....THAT number says how many 3s are in the next power tower...etc. etc. and you iterate this process 3^^^3 times !! G1 is insanely large. Adding an arrow each time the number gets larger in a way that is incomprehensible...and yet G2 has G1 arrows! G64 is beyond insanely large. And yet TREE(3) is larger than G64 in a way that is completely insane...
❤Insane number and calculation which already been mentioned in Hindu Vedas ❤ Love your explanation and enthusiasm 😊😊😊😊 keep learning keep smiling ☺️☺️☺️
A better way to think about the subsequent steps in reaching the number is to take the previous number and put that number of arrows between 3 and 3 to get the next number. For example, G(64) is equal to 3 (insert G(63) number of arrows here) 3.
Thank you very much sir professor, but what good are these hyperoperations with astronomical numbers, with which only quantum computers can work easily, in classical everyday life, what useful applications can they have?
Well if you're talking about TREE(3) then no we don't even know it's last digit and the only 2 things we know are it's finite and it would take you 2^^10³ symbols to prove that TREE(3) is finite using finite arithmetics but if you're talking about tree(3) then yeah it has a representation but it's crazy😊😅😊
I'm have no interest in advanced maths but I watch all your videos about big numbers and complex maths operations because, being a great teacher, you leave me feeling like I now know😊 That being said, it would be interested to watch you 'explain' mathematical proofs of some huge numbers that we hear or that we, laymen, easily throw around everyday, such as: how far away the sun, the moon and nearest galaxies are from the earth; the age of the earth/sun; the surface temperature of the sun; the diameter of the earth, the sun, and the moon; the speed of light, and so on.
I made a picture of a formula for the 8th hyper operation before for fun. It alternated between “going up” and “going right” and was massive. Lots of “b times” subtext in it. It’s interesting that H_n(2, 2) = 4 always. Everything collapses
What’s evenly more insane? G(64), an insanely huge number, is just an insanely small portion of the way to get to infinity. We’re maybe small, but we’re the salt of the Earth. 🙏
I think you made a mistake there. In hexation, the height of the tetration-tower is the result of the previous level. So in this case the height will not be 7.6 tril (3^^3), but actually 3^^^3.
@@spinothenoooob6050 The first thing he said wrong, was within the first 3 minutes. He said the simplest hyperoperator was "incrementation", and claimed it has to do with "decimal places". That's not even an operation. The simplest, is called "succession", and is "adding 1", or "increasing the actual number itself, by 1.
Grahams Number might be my all time favourite "Big number" I know there are some very much larger numbers than G64 But i love the concept and the idea! Its soo amazing to see how fast this number grows and also we can ""somehow"" describe and somehow "imagine" it how extremly big its gonna be Even G1 is out of scale.. nothing in our observerble Universe comes even close to it, and you couldnt even write that number down (except with the arrow notation or some other methods, but not the full number) Even if you could write the numbers in the smallest possible unit, you still wouldnt even have Enaugh Space in our Universe to write G1 out..... And now just imagine that the number just ^startet^ Its so awesome to wrap your head around it until you get headache haha Very grest video!
Well actually if you write down a zero in each atom of the observable universe, even then you still wouldn't be able to write down a googolplex Let alone G1 man
Fun fact: if you imagine the string the digits of Graham's number in your head then your head will collapse to form a black hole the size of your head😊
14:45 i think you meant the tower will be 7.6 trillion 3's high, 3^^^3 wohld be 3^^7.6 trillion but 3^^^^3 is best visualised as a list 7.6 trillion items long, where the first entry on the list is a power tower (3³) 7.6 trillion 3's long, and that number is the number of 3's in the next power tower (entry 2) and the 7.6 trillion'th entry on the list is 3^^^^3 which is so unfathomably large it can't really be put into perspective in a single tetration
There was one mistake in hexation. 3↑↑↑↑3≠3↑↑[3↑↑(...(3↑↑3)], which is repeated tetration 7.6 trillion times. 3↑↑↑↑3 can be written as 3↑↑↑(3↑↑↑3) Which is 3↑↑[(3↑(3↑(...(3)]
10:48 A question Can i write the pentation of 3 as 3 to the superpower 3 raise to 3 raise to 3 As in, 3 pent 3^3^3 !? I hope you get what i intend to write 😢
Yeah you can as long as you put the parentheses and compute it properly(and by computing I mean trying to write it as a tower as obviously the string of digits is huge like there is not enough space in the observable universe write this number even if you're writing in every plank volumes)
12:10 - you are wrong - The height of the tetration tower is 3↑↑↑3, not only 3↑↑3=7625597484987 In other words: 3↑↑↑↑3=3↑↑↑3↑↑↑3, but you wrongly present 3↑↑↑3↑↑3.
He was trying to solve a combinatorics problem of all the possible configuration of colouring all the line segments of an nth dimensional cube so that all the segments of same colour doesn't lie in the same plane. But the proof was long and recursive. This number is the upper bound for the problem and it took these many steps to solve it 😊
Bro you don't know the truth Well yn it's a really small number 😊 Graham's number is approximately 0 to TREE(3) which in itself is tiny compared to BB(10¹⁰⁰) and that's too small to even notice compared to RAYO(10¹⁰⁰) And fun fact we know last 500 digits of Graham's number I memorised like the last three ...387 We know them because if you take a big enough power of three the digits are quite predictable And here we only things we know about the other numbers is the fact that they are finite and we know the mathematical symbols required to prove TREE(3) is finite( iam saying TREE(3)as it has the smallest and the easiest to memorize)is 2^^10³ using finite arithmetics and that is just to prove it's finite Have a great day 😊😊😊
Well p-adic should be more preferred in a lot of calculation but I don't think we can as I don't think so as yn we would still need the string of digits in the number😊
Thank you for this video, it's mind blowing! I just have one query - u say 3^^^^3 = 3(sup^)3(sup^)3(sup^)... with there being 7.6 trillion threes However, 3^^^^3 = 3^^^(3^^^3) = 3^^^ [ 3(sup^)3(sup^)3 ] so in fact, 3^^^^3 = 3(sup^)3(sup^)3(sup^)... with there being 3(sup^)3(sup^)3 threes , which is way way bigger!
Think that's a big number? TREE(3) is so much bigger, it makes G64 look MICROSCOPIC in comparison. Watch Numberphile's video where they compare the two.
@@spinothenoooob6050, when you use common sense and logic, TREE(3) must contain many millions of each digit. Also, I have read where they have calculated the last 16 MILLION digits of Graham's number.
@@pacman52280bro I meant the that Graham's number is big but we still know digits of the number but TREE(3) is just so huge that we don't even know what the digit are exactly, like if you ask someone what is the 3rd digit of TREE(3) they can tell you anything as you can't refute but for Graham's number you know immediately it's a 3
It was used to set an upper limit on a particular number that is the solution of a very, very complicated problem in an esoteric branch of mathematics. Have a look on the Wikipedia entry for Graham's Number for a bit more of an explanation, though it is quite tricky and I completely fail to grasp its meaning :)
Great videos. Great explanations. I love how when you put something crazy on the board you look at the camera like your looking at our reaction.
I appreciate that!
You are right," never stop learning inorder to live"
I believe that for 3↑↑↑↑3 the shorthand hexation tower would actually be the height of the result of 3↑↑↑3, not just 7.6 trillion high. Another way to write Grahal (G1) is 3↑↑↑(3↑↑↑3). When doing this you no longer have something as simple as 3↑↑↑3. You have 3↑↑↑x, where x=(insanely big number). To evaluate that you beak it down further as 3↑↑(3↑↑(3↑↑....3↑↑). Remember that in up arrow notation the trailing number is number of times the leading number is operated. Thus, to reach 3↑↑↑↑3 you must think of it as 3↑↑↑x where x= the number of 3's you get from calculating 3↑↑↑3. Put another way 3↑↑↑↑3 is a 3 "pentated" 3↑↑↑3 times. Now let me say something I often don't see when doing my research. The parentheses matter. When looking at 3↑↑(3↑↑(3↑↑....3↑↑) or something similar, each parentheses from left to right is generating a next tower. So 3↑↑(3↑↑) for example, is 3↑↑(7.6 trillion). We understand that of course. That's 3↑↑↑3. Cool. To reach Grahal we start working from right to left evaluating 3↑↑(3↑↑(....)↑↑3) an incredible number of times, and in doing so this means that every time you calculate left you are building a tower of 3's within paranthases, which will give you the height of tower to calculate as you move further left. Note that you are of course moving down your power towers right to left as normal (as you did when evaluating the tower of 3's 7.6 trillion high, etc), but after calculating a power tower you then go back to 3↑↑(3↑↑(....)↑↑3), having solved a term of just one of your parentheses, and you now move one step left, using the number you just calculated to give you the numbers of 3's in the tower you are working on next. As you can imagine, 3 or 4 steps in you reach crazy tower heights but this operation has to be done many many more times as the tower building nature of pentation demands. Finally, when you do this tedious right to left tower, then parentheses, then back to tower and so forth operation 3↑↑↑3 times you have G1.
Back to your shorthand. First off, you do a great job explaining maths and I commend you. I am also not as familiar with left hand exponent type shorthand as I am with up arrow notation. Still, if hexation is repeated pentation, then what you wrote as pentation in shorthand (a left tower 3 3's tall) should be repeated 7.6 trillion times. Hexation is vastly more powerful than just a shorthand tower of 3's 7.6 trillion tall.
I thought that as well
If only I wrote this much for my essay
Thanks for the detailed explanation though it took some time for me to digest it...I too thought that something was not quite right for pentation notation too.. Won't it actually be a left 3s power tower of 7.2 trillion as pentation too is repeated tetration..
I have limited experience, so I may be off, but the general rule is that an n-arrow operator expands into a right-associative series of (n−1)-arrow operators. So I believe you are correct. I've seen pentation represented with a subscript left variable. A pentation of 3 to the 3 will have a left power tower of height ~7.63x10^12, a Hexation of 3 to the 3 would have a shaft of 3's descending to the left for the same height. If you wanted to express Hexation with a left power tower instead of a shaft, then it wouldn't be a tower of 3's ~7.63x10^12 high, but a left power tower ~7.63x10^12 high, composed of right power towers of 3 ~7.63x10^12 high.
Immediately saw that - not 7.6 trillion threes high - but a tower 3^3^3^….. 7.6 trillion threes high, high.
Thank you! I commented on a previous video of yours that if anyone could explain Graham's Number to me in a way I could understand, it would be you. I was right! Though the number itself is completely unimaginable, you've made its derivation understandable. No mean feat! Thank you once again!
I'd like to suggest a sensible limit. There is a minimal amount of energy associated with each bit of information. ( 2.9×10^−21, Landauer's principle) If we assign one bit to each unit in the number, then which number has minimum energy exceeding the energy of the known universe? Which number has the minimum energy needed to create a black hole, etc. Apply that math.
You would reach that very early on with Graham’s number
One could substitute the G in Graham for the G in God ...the Creator of the finite universe.
He's the only one Who knows what lies beyond infinity.
That boggles my mind
The number of God is 3.@@bobwineland9936
G1‘s brother Tritri {3,3,3} maybe
Your number is between 3^^3 and 3^^^3. The latter is so unfathomably large that nothing you measure in the universe will dwarf in comparison, including the number of possible states of the universe, and number of possible universes (assuming finite universe).
Love your videos for their educational value, thought provoking and engaging. Your narrative is cogent and easy to follow. You’re an excellent explainer. I wish you would produce a video on “…gravity is not a force and we are not being pulled down…”.
I wish I was that smart. Thank you for the kind words.
@@PrimeNewtons You are that smart. More important is your ability to explain complex subjects in a way that others can understand. Please, don't sell yourself short. You have talent and you should explore expanding the breath of your subject matter.
@@PrimeNewtons BTW, I say his video and even Ron Graham, didn't do a good enough job explaining his number.
Love your enthusiasm and amazement in how these numbers grow. Easier to say G2 has G1 arrows between the 3s, G3 has G2 arrows, etc. To draw out 3^^^^3 (G1), you construct a sequence of growing power towers. Start with 3^^3...that number says how many 3s are in the next power tower....THAT number says how many 3s are in the next power tower...etc. etc. and you iterate this process 3^^^3 times !! G1 is insanely large. Adding an arrow each time the number gets larger in a way that is incomprehensible...and yet G2 has G1 arrows! G64 is beyond insanely large. And yet TREE(3) is larger than G64 in a way that is completely insane...
❤Insane number and calculation which already been mentioned in Hindu Vedas ❤
Love your explanation and enthusiasm 😊😊😊😊 keep learning keep smiling ☺️☺️☺️
Thanks for your excellent explanation. I am in Nairobi Kenya.
In the context of hyperoperations, incrementation is strictly adding 1 to n, thus allowing addition to be repeated incrementation.
Love your enthusiasm!
Best explanation I have ever seen on this subject; thank you.
man, Graham's number is not insanely large,it's this Graham guy who's insanely insane tetration
I love your videos.
A better way to think about the subsequent steps in reaching the number is to take the previous number and put that number of arrows between 3 and 3 to get the next number. For example, G(64) is equal to 3 (insert G(63) number of arrows here) 3.
Yes I was surprised this was not mentioned
Never in my life i imagined i will be this scared of maths at 12:31 of night
Thank you very much sir professor, but what good are these hyperoperations with astronomical numbers, with which only quantum computers can work easily, in classical everyday life, what useful applications can they have?
Don’t know if it’s possible but I have to say, I’d love to see the proof that could use such a number. I’m sure it be insane.
Nice video, how about tree(3) ? Thanks.
Is the tree(3) more insanely tall in it's index representation ?
Well if you're talking about TREE(3) then no we don't even know it's last digit and the only 2 things we know are it's finite and it would take you 2^^10³ symbols to prove that TREE(3) is finite using finite arithmetics but if you're talking about tree(3) then yeah it has a representation but it's crazy😊😅😊
I'm have no interest in advanced maths but I watch all your videos about big numbers and complex maths operations because, being a great teacher, you leave me feeling like I now know😊
That being said, it would be interested to watch you 'explain' mathematical proofs of some huge numbers that we hear or that we, laymen, easily throw around everyday, such as: how far away the sun, the moon and nearest galaxies are from the earth; the age of the earth/sun; the surface temperature of the sun; the diameter of the earth, the sun, and the moon; the speed of light, and so on.
Perfect. Exactly what I was talking about
I made a picture of a formula for the 8th hyper operation before for fun. It alternated between “going up” and “going right” and was massive. Lots of “b times” subtext in it. It’s interesting that H_n(2, 2) = 4 always. Everything collapses
Grandiosa explicación
This is pure entertainment for me ❤
best video about g64 so far
good and nice😉
Yet there's a number out there that makes Graham's number look very close to zero, isn't there?
Hii sir.....I am your big fan.....
I will always try to learn what you teach.....it's nice and awesome.....thank you for teach us....
How about more details on the referenced cube proof that used this number?
What’s evenly more insane? G(64), an insanely huge number, is just an insanely small portion of the way to get to infinity. We’re maybe small, but we’re the salt of the Earth. 🙏
Even for the value "2", G(1) is hard to compute. But I believe it is possible with modern tech.
For 2, it's just 4, lol
No matter how many hyperoperations you apply with 2 on both ends, you'll get just 2² which is 4
Mind blowing 🤯🤯. My brain is freez when 7.6 trillion of tetration you show
I think you made a mistake there. In hexation, the height of the tetration-tower is the result of the previous level. So in this case the height will not be 7.6 tril (3^^3), but actually 3^^^3.
Yeah noticed that too
Yep, and even on G(2), 3 isn't tetrated by 7.6 trillion it's tetrated by itself 7.6 trillion times or 3 Pentated by 7.6 trillion
To be fair half of what he said after that is wrong 😢😢😢
But still no probs everyone makes mistakes 😊😊😊
@@spinothenoooob6050 The first thing he said wrong, was within the first 3 minutes.
He said the simplest hyperoperator was "incrementation", and claimed it has to do with "decimal places". That's not even an operation.
The simplest, is called "succession", and is "adding 1", or "increasing the actual number itself, by 1.
@@scmtuk3662yeah that's what I was thinking as well😅
Can you do a video about tree 3. Thanks
Just a question: does this kind of operation have any practical application?
It made my head hurt, so you succeeded
great video
🎉🎉🎉
Grahams Number might be my all time favourite "Big number"
I know there are some very much larger numbers than G64 But i love the concept and the idea!
Its soo amazing to see how fast this number grows and also we can ""somehow"" describe and somehow "imagine" it how extremly big its gonna be
Even G1 is out of scale.. nothing in our observerble Universe comes even close to it, and you couldnt even write that number down (except with the arrow notation or some other methods, but not the full number)
Even if you could write the numbers in the smallest possible unit, you still wouldnt even have Enaugh Space in our Universe to write G1 out.....
And now just imagine that the number just ^startet^
Its so awesome to wrap your head around it until you get headache haha
Very grest video!
Well actually if you write down a zero in each atom of the observable universe, even then you still wouldn't be able to write down a googolplex
Let alone G1 man
@@spinothenoooob6050 yea but you could "theoretically" write down in the smallest possible unit, the Planck lenght
And that doesnt apply for G1
Fun fact: if you imagine the string the digits of Graham's number in your head then your head will collapse to form a black hole the size of your head😊
What kind of chalk did you use in this video?
Why stop at G(64)?
Why didn’t I have a teacher like you when I was at grammar school -it was 1960!
14:45 i think you meant the tower will be 7.6 trillion 3's high, 3^^^3 wohld be 3^^7.6 trillion but 3^^^^3 is best visualised as a list 7.6 trillion items long, where the first entry on the list is a power tower (3³) 7.6 trillion 3's long, and that number is the number of 3's in the next power tower (entry 2) and the 7.6 trillion'th entry on the list is 3^^^^3 which is so unfathomably large it can't really be put into perspective in a single tetration
You're right that hurt my brain
There was one mistake in hexation.
3↑↑↑↑3≠3↑↑[3↑↑(...(3↑↑3)], which is repeated tetration 7.6 trillion times.
3↑↑↑↑3 can be written as
3↑↑↑(3↑↑↑3)
Which is 3↑↑[(3↑(3↑(...(3)]
10:48
A question
Can i write the pentation of 3 as 3 to the superpower 3 raise to 3 raise to 3
As in, 3 pent 3^3^3 !?
I hope you get what i intend to write 😢
Pentation as in 3↑↑↑3 = 3↑↑(3↑↑3) = 3↑↑(3↑(3↑3)) so I guess you could write it, in your notation, as 3 tetr 3^(3^3)
Yeah you can as long as you put the parentheses and compute it properly(and by computing I mean trying to write it as a tower as obviously the string of digits is huge like there is not enough space in the observable universe write this number even if you're writing in every plank volumes)
How can 8 be written as a tetration when 2 is considered as a base ?
Hey, where is yours "let's get into the video"?)
My only dumb question here is " Does infinity play any role in Graham's number"? or does it redefine what infinity is.
No Graham's number is approximately 0 to TREE(3) 😅😅😅
And TREE(10¹⁰⁰)
RAYO(10¹⁰⁰) is approximately 0 to infinity
As you can take f(x,n)=RAYO^n(x)
And plug large values for x and n to get even bigger numbers😊😊😊
12:10 - you are wrong - The height of the tetration tower is 3↑↑↑3, not only 3↑↑3=7625597484987
In other words: 3↑↑↑↑3=3↑↑↑3↑↑↑3, but you wrongly present 3↑↑↑3↑↑3.
But why the Grahams numbers stop at the 64th level?
idk, he likes 2^n maybe
Because it was sufficiently large for his proof.
Because he was trying to make it as small as possible as it will help him towards solving the real problem but it took this much 😊😊😊
He was trying to solve a combinatorics problem of all the possible configuration of colouring all the line segments of an nth dimensional cube so that all the segments of same colour doesn't lie in the same plane. But the proof was long and recursive. This number is the upper bound for the problem and it took these many steps to solve it 😊
Why isn't infinity minus 1 the largest number?
What about infinity minus 0.000000000000000000000001?
How can 8 be as a tetration when 2 is taken as a base ?
Oh my god, this is so crazy number!
Bro you don't know the truth
Well yn it's a really small number 😊
Graham's number is approximately 0 to TREE(3) which in itself is tiny compared to BB(10¹⁰⁰) and that's too small to even notice compared to RAYO(10¹⁰⁰)
And fun fact we know last 500 digits of Graham's number
I memorised like the last three
...387
We know them because if you take a big enough power of three the digits are quite predictable
And here we only things we know about the other numbers is the fact that they are finite and we know the mathematical symbols required to prove TREE(3) is finite( iam saying TREE(3)as it has the smallest and the easiest to memorize)is 2^^10³ using finite arithmetics and that is just to prove it's finite
Have a great day 😊😊😊
Heptation, octation, it never ends.
Can we calculate it 10-adically?
Well p-adic should be more preferred in a lot of calculation but I don't think we can as I don't think so as yn we would still need the string of digits in the number😊
11:40 I got goosebumps
Graham's number G(64) is so tiny compared to a number called TREE(3).
While this is what I call mind-boggling, I have to ask...why?
Thank you for this video, it's mind blowing!
I just have one query - u say 3^^^^3 = 3(sup^)3(sup^)3(sup^)... with there being 7.6 trillion threes
However, 3^^^^3 = 3^^^(3^^^3) = 3^^^ [ 3(sup^)3(sup^)3 ]
so in fact, 3^^^^3 = 3(sup^)3(sup^)3(sup^)... with there being 3(sup^)3(sup^)3 threes , which is way way bigger!
Now hear me out, What about
G(64)↑↑↑↑↑↑↑↑.......↑↑↑↑G(64)
[_________________]
G(64)
11:35 sir you made a mistake it is no 7.625*10¹² high no it is 1.6*10³⁶*¹⁰^¹¹ high
I do not want to know the steps following G(2)....and it is still far from infinity. I better have a drink and do not think further.
What are the applications of this concept ?
Combinatorial research
We just saw 3!2 lol at 12:14
Wdym 7.6 b times at 12:59
Think that's a big number? TREE(3) is so much bigger, it makes G64 look MICROSCOPIC in comparison. Watch Numberphile's video where they compare the two.
Fax🎉🎉🎉😊❤❤❤
The fact we know the last 500 digits of Graham's number but not even a single digit of tree(3) is the most beautiful thing for me ❤🎉
@@spinothenoooob6050, when you use common sense and logic, TREE(3) must contain many millions of each digit. Also, I have read where they have calculated the last 16 MILLION digits of Graham's number.
@@pacman52280bro I meant the that Graham's number is big but we still know digits of the number but TREE(3) is just so huge that we don't even know what the digit are exactly, like if you ask someone what is the 3rd digit of TREE(3) they can tell you anything as you can't refute but for Graham's number you know immediately it's a 3
12:01 XD
This is useful how.......?
It was used to set an upper limit on a particular number that is the solution of a very, very complicated problem in an esoteric branch of mathematics. Have a look on the Wikipedia entry for Graham's Number for a bit more of an explanation, though it is quite tricky and I completely fail to grasp its meaning :)
Something to do with something called “Ramsey Theory” about colored vertices in higher dimensional cubes?
Everything after Hexation is Vexation. 🙂
Septation
Octation
Nonation
Decation