Hexation and Graham's Number

Hexation and Graham's Number

[Majan-vK]

You are right, " never stop learning inorder to live"

[nharvey]

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.

[Video_Montage]

Never in my life i imagined i will be this scared of maths at 12:31 of night

[kidkante]

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.

[KevinJB]

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!

[eqab]

man, Graham's number is not insanely large, it's this Graham guy who's insanely insane tetration

[ukdavepianoman]

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...

[fanlessfurmark]

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.

[senshtatulo]

In the context of hyperoperations, incrementation is strictly adding 1 to n, thus allowing addition to be repeated incrementation.

[potawatomi]

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…”.

[Ron_DeForest]

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.

[SonOfKorahMusic]

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.

[mosesmuchina]

Thanks for your excellent explanation. I am in Nairobi Kenya.

[OpPhilo]

Mind blowing 🤯🤯. My brain is freez when 7.6 trillion of tetration you show

[xyz.ijk.]

Best explanation I have ever seen on this subject; thank you.

[namo]

❤Insane number and calculation which already been mentioned in Hindu Vedas ❤
Love your explanation and enthusiasm 😊😊😊😊 keep learning keep smiling ☺️☺️☺️

[coreymonsta]

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

[bastardferret]

Yet there's a number out there that makes Graham's number look very close to zero, isn't there?

[loganv]

It made my head hurt, so you succeeded

[GhostZeroGZ]

Perfect. Exactly what I was talking about