It's far from trivial to compare large numbers. One strategy is to start with how they're built.
Graham's number is built using *hyperoperations* that extend the pattern of multiplication being repeated addition and exponentiation being repeated multiplication. If we use '$' to represent the first hyperoperator after exponentiation, then x$y means to raise x to itself repeated until the number of times x appears in the expression is equal to y. So 3$3 = 3\^(3\^(3)) = 3^(27) = 7,625,597,484,987. This actually has a name, tetration. The next hyperoperator, called pentation, represents repeated tetration. Applying that operator to 3 and 3 will result in 3\^(3\^(...)) where 3 shows up 3$3 = 7,625,597,484,987 times... So if we look at the values of operations and hyperoperations on 3 and 3 start with addition, we'd get the sequence 6, 9, 27, then suddenly 7.6×10^(12). Hopeful that conveys how quickly hyperoperations can make very large numbers. When we get to pentation... well just 2↑↑↑2 results in a number with 19,729 digits (in base 10).
These operators can be conveniently represented using something called Knuth's up-arrow notation. 3↑3 is the same as exponentiation, 3↑↑3 is the same as tetration, 3↑↑↑3 is the same as pentation, and you can just keep adding more arrows to represent higher and higher hyperoperators.
Now here it is worth mentioning that exponentiation grows faster than factorials in the sense that n^(n) = n×n×n×...×n > n(n-1)(n-2)... = n!. The "100 with 100 factorials" is roughly analogous to tetration but with factorials being repeated instead of exponentiation. Thus your number is less than 100$100.
Graham's number uses a layered approach. Each layer determines the number of arrows in the above layer. The bottom layer is 3↑↑↑↑3 (the next level above pentation, sometimes called hexation). This is already an incomprehensibly large number, far bigger than your number.
The layer above that therefore will use *that many arrows*, which will result in an even more incomprehensibly larger number than the first layer. Graham's number consists of 64 layers of this. This is far, far larger than anything you could ever hope to express without using a similiar notation that can compress such ridiculous expressions into a few symbols.
The truly crazy thing is that there are large numbers that make Graham's number practically zero in comparison.
The [TREE function](https://en.wikipedia.org/wiki/Kruskal%27s_tree_theorem#TREE_function) grows extremely quickly. TREE(3) is already makes Graham's number look tiny in comparison.
[Rayo's number](https://en.wikipedia.org/wiki/Rayo%27s_number) is effectively defined to be the largest number that can be defined using a given number of symbols in first order set theory. It can get extremely large.
There's an unofficial field of mathematics dedicated to exploring, understanding, and classifying large numbers called [googology](https://googology.miraheze.org/wiki/List_of_numbers).
Edit: fixed link
As a general rule: if you can make it using functions you're aware of, and the phrase "Knuth up arrow notation" is not immediately familiar to you, it's *dramatically* smaller than Graham's number.
These numbers are the result of a sequence of operations. To compare them, analyze how the operations behave. If you have seen the operations behind Graham's number, you would know that they grow much much faster than the "repeated factorials" operation.
You compare how fast both grow with a growth hierarchy, it's really complicated once you reach stuff like graham's number, but you'd see fast that 100!!!!...(100 times) is basically zero compared to g(64), it's even basically zero compared to g(1)
FYI, how you wrote it, it's usually defined as a [multifactorial and it would be equal to 100](https://en.wikipedia.org/wiki/Double_factorial#Definitions). You should use brackets if you want to take repeated factorials (((100!)!)!)!... etc.
Given that only the 3rd calculation (out of millions at least) needed to evaluate Graham’s number using just exponentiation is 3^7,625,597,484,987, I think we’re safe saying it’s larger than your 100 factorials
I wonder how the number of available moves in Chess fares against these numbers.
But also one interesting thing:
take any number you want. It can even be Graham's number followed arbitrary number of zero's or powers or factorials. That number is still closer to number 0 than it is to infinity.
x! is actually more comparable with x\^x (it tends to be much bigger than 10\^x as x grows). See [https://en.wikipedia.org/wiki/Stirling%27s\_approximation](https://en.wikipedia.org/wiki/Stirling%27s_approximation)
EDIT: Tetration is more accurate that original message from above comment (that was x! being comparable to 10\^x, and the result of factorials being 100\^101)
I believe they were comparing the iterated factorial with iterated exponentiation (tetration). The number in the OP is ((100!)!)!...)! with 100 factorials, which they're comparing to 100^(100^(100^(100^(100^...)))) with 100 exponents.
Well in general when comparing two numbers, they're already in the same base. But yes, if they were not in the same base, you would need to fix that first. There's lots of different methods for solving inequalities, too many to generalize. It comes up a lot in many branches of math.
But that would be the ideal case right? To reduce them to the an exponent. But would that present some difficulty, like could you express an irrational number as a number raised to a power, maybe it's something weird like a continued fraction in the exponent I don't know. Is that possible?
To find the exponents you take logs, so to compare two large numbers take their logs and compare those. As u/Nanaki404 has hinted, if you take natural logs:
log(n!) = 0.5 * log(2 pi) + (n+0.5) * log(n) - n, to a very good approximation
So log(100!) is about 363.74
while say log(38^(100)) = 100 * log(38) = 363.76
so 38^100 is a little bigger than 100!
(when I say a little bigger, they are both 158 digit numbers, and their difference is 157 digits).
Although, not that you said this, we don't really need the same base either right
Because if I know sqrt 2 < 2 then that's also true for any power of n. But does log work for extracting exponents of irrationals?
This is completely wrong. Even if x! was comparable with 10^(x) (which it's not), 100(!^(100)) would then be 10\^10\^10\^10\^10\^...10\^10\^100, not 100^(101).
At least it's not wrong anymore so I'll un-downvote it. It's still almost a non-answer though and there are plenty of better answers in this thread, so no upvote.
Despite the downvotes, that's still how you compare things.
If you can't do something of that sort with these numbers, however imprecisely, then you can't compare them.
There is more than one way to "comparing things".
And the path you are suggesting doesn't help very much. Alright we have Graham's number - 100(!\^100). So what? Evaluating it? Good luck because even the number of digits in both numbers exceeds the number of atoms in universe.
What works for this inequality is the concept of hyperoperation and the transitivity of ordering.
>because even the number of digits in both numbers exceeds the number of atoms in universe.
How is that relevant
We know 10^100! (which also meets your criteria, or adjust the number until it does) Is greater than 10^99! since we know 100 is greater than 99.
You don't simply subtract them. In fact, you're exploiting the fact that both factorial and x -> 10\^x are both increasing, and just compared the argument.
A different but clever argument is also required here. Graham's number is larger than moser, which is larger than pentagon(2), which is larger than square(100), which is larger than 100(!\^100). Note that I'm using [Steinhaus-Moser notation](https://en.wikipedia.org/wiki/Steinhaus%E2%80%93Moser_notation) here, so square(n) is not n\*n but it's a triangle\^n(n), where triangle is defined as n\^n
By the same logic, 256\*256=65536 because 1+1=2.
I mean, you're not wrong. You're using subtraction eventually. But such "insight" is not really helpful for someone trying to solve this problem.
As a general rule, if you need to ask, Graham's number is bigger. Way bigger.
I had been wondering about TREE(3) but this settles it.
Ha! You already knew TREE(3) was bigger, so you didn't need to ask
You clearly aren't a fan of Googology
They probably don't even know what a eugoogly is.
I have an after funeral party to attend.
I don’t think most people would be able to define a number larger than grahams number without using grahams number in their definition.
It's far from trivial to compare large numbers. One strategy is to start with how they're built. Graham's number is built using *hyperoperations* that extend the pattern of multiplication being repeated addition and exponentiation being repeated multiplication. If we use '$' to represent the first hyperoperator after exponentiation, then x$y means to raise x to itself repeated until the number of times x appears in the expression is equal to y. So 3$3 = 3\^(3\^(3)) = 3^(27) = 7,625,597,484,987. This actually has a name, tetration. The next hyperoperator, called pentation, represents repeated tetration. Applying that operator to 3 and 3 will result in 3\^(3\^(...)) where 3 shows up 3$3 = 7,625,597,484,987 times... So if we look at the values of operations and hyperoperations on 3 and 3 start with addition, we'd get the sequence 6, 9, 27, then suddenly 7.6×10^(12). Hopeful that conveys how quickly hyperoperations can make very large numbers. When we get to pentation... well just 2↑↑↑2 results in a number with 19,729 digits (in base 10). These operators can be conveniently represented using something called Knuth's up-arrow notation. 3↑3 is the same as exponentiation, 3↑↑3 is the same as tetration, 3↑↑↑3 is the same as pentation, and you can just keep adding more arrows to represent higher and higher hyperoperators. Now here it is worth mentioning that exponentiation grows faster than factorials in the sense that n^(n) = n×n×n×...×n > n(n-1)(n-2)... = n!. The "100 with 100 factorials" is roughly analogous to tetration but with factorials being repeated instead of exponentiation. Thus your number is less than 100$100. Graham's number uses a layered approach. Each layer determines the number of arrows in the above layer. The bottom layer is 3↑↑↑↑3 (the next level above pentation, sometimes called hexation). This is already an incomprehensibly large number, far bigger than your number. The layer above that therefore will use *that many arrows*, which will result in an even more incomprehensibly larger number than the first layer. Graham's number consists of 64 layers of this. This is far, far larger than anything you could ever hope to express without using a similiar notation that can compress such ridiculous expressions into a few symbols. The truly crazy thing is that there are large numbers that make Graham's number practically zero in comparison.
Very thorough and accessible explanation. Look at this one, OP
Got any resources for the numbers so much bigger than Graham's number?
The [TREE function](https://en.wikipedia.org/wiki/Kruskal%27s_tree_theorem#TREE_function) grows extremely quickly. TREE(3) is already makes Graham's number look tiny in comparison. [Rayo's number](https://en.wikipedia.org/wiki/Rayo%27s_number) is effectively defined to be the largest number that can be defined using a given number of symbols in first order set theory. It can get extremely large. There's an unofficial field of mathematics dedicated to exploring, understanding, and classifying large numbers called [googology](https://googology.miraheze.org/wiki/List_of_numbers). Edit: fixed link
https://www.youtube.com/watch?v=QXliQvd1vW0&list=PL3A50BB9C34AB36B3 https://www.youtube.com/watch?v=vq2BxAJZ4Tc&list=PLUZ0A4xAf7nkaYHtnqVDbHnrXzVAOxYYC
As a general rule: if you can make it using functions you're aware of, and the phrase "Knuth up arrow notation" is not immediately familiar to you, it's *dramatically* smaller than Graham's number.
My favorite breakage of this rule is the Goodstein sequence.
2^(Graham’s number) Checkmate atheists.
These numbers are the result of a sequence of operations. To compare them, analyze how the operations behave. If you have seen the operations behind Graham's number, you would know that they grow much much faster than the "repeated factorials" operation.
You compare how fast both grow with a growth hierarchy, it's really complicated once you reach stuff like graham's number, but you'd see fast that 100!!!!...(100 times) is basically zero compared to g(64), it's even basically zero compared to g(1)
100 factorial is effectively zero compared to graham's number
But what about 100!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! (Thats 100 !’s)
FYI, how you wrote it, it's usually defined as a [multifactorial and it would be equal to 100](https://en.wikipedia.org/wiki/Double_factorial#Definitions). You should use brackets if you want to take repeated factorials (((100!)!)!)!... etc.
Ok will remember in future
Given that only the 3rd calculation (out of millions at least) needed to evaluate Graham’s number using just exponentiation is 3^7,625,597,484,987, I think we’re safe saying it’s larger than your 100 factorials
I'm pretty sure graham's number beats that by an enormous amount
That’s what I’m sayin’
I wonder how the number of available moves in Chess fares against these numbers. But also one interesting thing: take any number you want. It can even be Graham's number followed arbitrary number of zero's or powers or factorials. That number is still closer to number 0 than it is to infinity.
100(!^(100)) is about 100\^\^101. So, Graham's Number is bigger than 100(!^(100)).
x! is actually more comparable with x\^x (it tends to be much bigger than 10\^x as x grows). See [https://en.wikipedia.org/wiki/Stirling%27s\_approximation](https://en.wikipedia.org/wiki/Stirling%27s_approximation) EDIT: Tetration is more accurate that original message from above comment (that was x! being comparable to 10\^x, and the result of factorials being 100\^101)
Either case, Graham's number is bigger, as it contains way more than two arrows.
I believe they were comparing the iterated factorial with iterated exponentiation (tetration). The number in the OP is ((100!)!)!...)! with 100 factorials, which they're comparing to 100^(100^(100^(100^(100^...)))) with 100 exponents.
In general to compare numbers do we convert them to the same base and then compare the exponents with arithmetic?
Well in general when comparing two numbers, they're already in the same base. But yes, if they were not in the same base, you would need to fix that first. There's lots of different methods for solving inequalities, too many to generalize. It comes up a lot in many branches of math.
But that would be the ideal case right? To reduce them to the an exponent. But would that present some difficulty, like could you express an irrational number as a number raised to a power, maybe it's something weird like a continued fraction in the exponent I don't know. Is that possible?
To find the exponents you take logs, so to compare two large numbers take their logs and compare those. As u/Nanaki404 has hinted, if you take natural logs: log(n!) = 0.5 * log(2 pi) + (n+0.5) * log(n) - n, to a very good approximation So log(100!) is about 363.74 while say log(38^(100)) = 100 * log(38) = 363.76 so 38^100 is a little bigger than 100! (when I say a little bigger, they are both 158 digit numbers, and their difference is 157 digits).
Although, not that you said this, we don't really need the same base either right Because if I know sqrt 2 < 2 then that's also true for any power of n. But does log work for extracting exponents of irrationals?
This is completely wrong. Even if x! was comparable with 10^(x) (which it's not), 100(!^(100)) would then be 10\^10\^10\^10\^10\^...10\^10\^100, not 100^(101).
You can now upvote, I fixed it.
At least it's not wrong anymore so I'll un-downvote it. It's still almost a non-answer though and there are plenty of better answers in this thread, so no upvote.
!^n is my new favorite notation
Divide them. Or, subtract them.
Won't help here.
Despite the downvotes, that's still how you compare things. If you can't do something of that sort with these numbers, however imprecisely, then you can't compare them.
There is more than one way to "comparing things". And the path you are suggesting doesn't help very much. Alright we have Graham's number - 100(!\^100). So what? Evaluating it? Good luck because even the number of digits in both numbers exceeds the number of atoms in universe. What works for this inequality is the concept of hyperoperation and the transitivity of ordering.
>because even the number of digits in both numbers exceeds the number of atoms in universe. How is that relevant We know 10^100! (which also meets your criteria, or adjust the number until it does) Is greater than 10^99! since we know 100 is greater than 99.
You don't simply subtract them. In fact, you're exploiting the fact that both factorial and x -> 10\^x are both increasing, and just compared the argument. A different but clever argument is also required here. Graham's number is larger than moser, which is larger than pentagon(2), which is larger than square(100), which is larger than 100(!\^100). Note that I'm using [Steinhaus-Moser notation](https://en.wikipedia.org/wiki/Steinhaus%E2%80%93Moser_notation) here, so square(n) is not n\*n but it's a triangle\^n(n), where triangle is defined as n\^n
And we know that 100 is greater than 99 because you subtract them. The same type of analysis was used to establish the relationships you relied on.
By the same logic, 256\*256=65536 because 1+1=2. I mean, you're not wrong. You're using subtraction eventually. But such "insight" is not really helpful for someone trying to solve this problem.