Your number is, in fact, not bigger!
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 Vytron
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Your number is, in fact, not bigger!
A successor to this thread, but with different rules so it doesn't reach a deadlock.
1. You must be able to describe your number using computable functions.
2. Your number must be finite.
3. Your number must be welldefined.
4. Your number should be easy to understand for the layman.
4b Your complain against a number should be easy to understand for the layman.
5. You canNOT use meta references like "all the posted numbers so far +1" or "the highest number that will be posted on page 20 + 1" and such, only actual numbers count.
6. You can, however, just use the highest number posted on the thread, or steal a notation from someone else, and add 1, and it'll be valid.
4 is there because people tend to make very long posts with hard to read notation that makes it pretty much impossible to know if the new number is, in fact, bigger.
5 is there because, if you want to come to the thread and make a metadefinition like that, you actually have to take a look at numbers posted and make calculations for a valid entry.
6 is there so the thread can always have an influx of entries, you can come here and post a number that is, in fact, bigger than all the previous posted.
So I'll start with:
0
Beatable by any nonnegative number other than 0  let's see how long it takes us to beat the other thread
1. You must be able to describe your number using computable functions.
2. Your number must be finite.
3. Your number must be welldefined.
4. Your number should be easy to understand for the layman.
4b Your complain against a number should be easy to understand for the layman.
5. You canNOT use meta references like "all the posted numbers so far +1" or "the highest number that will be posted on page 20 + 1" and such, only actual numbers count.
6. You can, however, just use the highest number posted on the thread, or steal a notation from someone else, and add 1, and it'll be valid.
4 is there because people tend to make very long posts with hard to read notation that makes it pretty much impossible to know if the new number is, in fact, bigger.
5 is there because, if you want to come to the thread and make a metadefinition like that, you actually have to take a look at numbers posted and make calculations for a valid entry.
6 is there so the thread can always have an influx of entries, you can come here and post a number that is, in fact, bigger than all the previous posted.
So I'll start with:
0
Beatable by any nonnegative number other than 0  let's see how long it takes us to beat the other thread
Last edited by Vytron on Tue Mar 03, 2015 2:02 am UTC, edited 5 times in total.
Re: My number is, in fact, bigger!
10^{10}
 Vytron
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Re: My number is, in fact, bigger!
10^{10} + 10^{100}
Re: My number is, in fact, bigger!
10^{10}+10^{99.9}
 Vytron
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Re: My number is, in fact, bigger!
594/2375
Re: My number is, in fact, bigger!
1/4 + 1/1168
 Vytron
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Re: My number is, in fact, bigger!
Aha, that's in fact smaller than my last posted number, so I'm still champion with 594/2375
Fear those 8 symbols, the first number on the thread that failed to be beaten on first attempt!
Fear those 8 symbols, the first number on the thread that failed to be beaten on first attempt!
Re: My number is, in fact, bigger!
594/2375 = 0.25010526... = 1/4 + 1/9500
1/4 + 1/1168 = 0.25085616...
I challenge your challenge.
1/4 + 1/1168 = 0.25085616...
I challenge your challenge.
 orangedragonfire
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Re: My number is, in fact, bigger!
1 + 1/9999999999999999999
 Vytron
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Re: My number is, in fact, bigger!
Oh... god. Let it be known that I fail at math D:
1 + 1/999999999999999999
1 + 1/999999999999999999
Re: My number is, in fact, bigger!
(e+π)/5.859
Re: My number is, in fact, bigger!
e^{π/20000}
 orangedragonfire
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 orangedragonfire
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Re: My number is, in fact, bigger!
So, how exactly is a layman defined?
 Vytron
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Re: My number is, in fact, bigger!
Um, I guess it needs "on the fly definition", like, someone goes and say "Hello, I'm a layman and I don't understand your number" makes your number invalid (but you can make it valid by explaining in detail whatever you're doing), otherwise, it's fine.
100/99
100/99
Re: My number is, in fact, bigger!
layman, in this case, i'd say is a man(or a person of arbitrary gender/genders) who does not earn money or aspire to earn money from doing math, for example: engineers, researchers, math professors etc..
 orangedragonfire
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 orangedragonfire
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Re: My number is, in fact, bigger!
Let's go crazy and all the way up to
2
2
Re: My number is, in fact, bigger!
10^{100}
Re: My number is, in fact, bigger!
10^{101}
 Vytron
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Re: My number is, in fact, bigger!
Landmark! We have beaten 3 posts from the other thread!
10^{202}
10^{202}
Re: My number is, in fact, bigger!
10^10^10
 Vytron
 Posts: 429
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Re: My number is, in fact, bigger!
10^(10^10)
 orangedragonfire
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Re: My number is, in fact, bigger!
I think a tower of exponents is supposed to be solved from the top, so 10^10^10 = 10^(10^10)
Anyway, 10^^10
Anyway, 10^^10
Re: My number is, in fact, bigger!
11^(10^^9)
Re: My number is, in fact, bigger!
Nice try but what does ^^ mean?
 orangedragonfire
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Re: My number is, in fact, bigger!
It's using uparrow notation.
But since we are only using two arrows so far, you don't really have to understand it all yet. =D
a^^b means a tower of exponents a^a^a^a^...^a with b instances of a.
So 10^^10 = 10^10^10^10^10^10^10^10^10^10
But since we are only using two arrows so far, you don't really have to understand it all yet. =D
a^^b means a tower of exponents a^a^a^a^...^a with b instances of a.
So 10^^10 = 10^10^10^10^10^10^10^10^10^10

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Re: My number is, in fact, bigger!
10↑^{googol}10
O_o
O_o
Re: My number is, in fact, bigger!
what does ↑^{googol} mean?, by the way, what is a googol?
 Earthling on Mars
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Re: My number is, in fact, bigger!
We shall start with the number "Million"; that is 1,000,000. Now we shall multiply a Million by itself a Million times. This will be written 1000000^1000000 or 1000000^{1000000}, which is called "raising a Million to the Millionth power". Now we shall raise a Million to the Millionth power a Million times; that is 1000000^1000000^1000000... with a Million layers. This huge number shall be called Mega.
Now we shall multiply a Mega by itself a Mega times; that is Mega^Mega or Mega^{Mega}. And then we shall raise Mega to the Megath power Mega times; that is Mega^Mega^Mega... with a Mega layers. This number shall be called Mungo.
Now we shall raise a Mungo to the Mungoth power Mungo times; that is Mungo^Mungo^Mungo... with a Mungo layers. This number shall be called Humungo.
Now imagine a series of numbers, where each number made by taking the previous one and raising it to its own power that many times. The first number in this series is a Million, the second is a Mega, the third is a Mungo, and the fourth is a Humungo. We could continue this series with ever larger numbers.
My number is the Humungoth number in this series.
(I'm not sure if this will beat WaffleKirby's number, but it is more laymanlevel.)
Now we shall multiply a Mega by itself a Mega times; that is Mega^Mega or Mega^{Mega}. And then we shall raise Mega to the Megath power Mega times; that is Mega^Mega^Mega... with a Mega layers. This number shall be called Mungo.
Now we shall raise a Mungo to the Mungoth power Mungo times; that is Mungo^Mungo^Mungo... with a Mungo layers. This number shall be called Humungo.
Now imagine a series of numbers, where each number made by taking the previous one and raising it to its own power that many times. The first number in this series is a Million, the second is a Mega, the third is a Mungo, and the fourth is a Humungo. We could continue this series with ever larger numbers.
My number is the Humungoth number in this series.
(I'm not sure if this will beat WaffleKirby's number, but it is more laymanlevel.)
 orangedragonfire
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Re: My number is, in fact, bigger!
Your number is dwarfed by WaffleKirby's, but since somebody has challenged it's understandability, your number is still the biggest that follows the rules.
Let's see if I can simplify the notation above...
f(n) = n^^n
Million = 1000000
Mega = f(1000000)
Mungo = f(Mega) = f(f(1000000)) = f^{2}(1000000)
Humungo = f(Mungo) = f(f(f(1000000))) = f^{3}(1000000)
n'th number in this series = f^{n1}(1000000)
g(n) = f^{n}(1000000)
This means that your number is g(g(3)).
Now, let me introduce you to triple arrows. n^^^m is a tower of double arrows, n^^n^^n^^n^^n^^n^^...^^n of length m.
f(n) = n^^n = n^^^2
f(f(n)) = (n^^n)^^(n^^n) < n^^n^^n^^n = n^^^4
Each subsequent use of f doubles the number of n's in the tower, so
f^{k}(n) < n^^^(2^k)
This means that g(n) < 100000^^^(2^n)
g(3) < 1000000^^^8
g(g(3)) < 1000000^^^1000000^^^8
I can easily beat this with 4^^^4^^^4^^^4. Note that this number is still beaten easily by WaffleKirby's if anybody explains that number well enough for a layman.
Let's see if I can simplify the notation above...
f(n) = n^^n
Million = 1000000
Mega = f(1000000)
Mungo = f(Mega) = f(f(1000000)) = f^{2}(1000000)
Humungo = f(Mungo) = f(f(f(1000000))) = f^{3}(1000000)
n'th number in this series = f^{n1}(1000000)
g(n) = f^{n}(1000000)
This means that your number is g(g(3)).
Now, let me introduce you to triple arrows. n^^^m is a tower of double arrows, n^^n^^n^^n^^n^^n^^...^^n of length m.
f(n) = n^^n = n^^^2
f(f(n)) = (n^^n)^^(n^^n) < n^^n^^n^^n = n^^^4
Each subsequent use of f doubles the number of n's in the tower, so
f^{k}(n) < n^^^(2^k)
This means that g(n) < 100000^^^(2^n)
g(3) < 1000000^^^8
g(g(3)) < 1000000^^^1000000^^^8
I can easily beat this with 4^^^4^^^4^^^4. Note that this number is still beaten easily by WaffleKirby's if anybody explains that number well enough for a layman.
 firesoul31
 Posts: 93
 Joined: Wed Feb 27, 2013 10:30 pm UTC
Re: My number is, in fact, bigger!
Daggoth wrote:what does ↑^{googol} mean?, by the way, what is a googol?
As a layman, let me see if I understand it.
Firstly, a googol is 10^{100}.
Secondly, Knuth arrows are an alternate way to write exponentiation, right? So ↑ just means "to the power of". Two up arrows, however, mean something different; they mean you iterate the power that many times. So 5↑↑3 would be 5↑5↑5, or just 5^5^5. Three up arrows follow the same pattern but with ↑↑ instead of ↑, so 5↑↑↑3 would be 5↑↑5↑↑5. This continues, so for any given n up arrows, for the formula a↑^{c}b (the superscript indicates the number of up arrows), you get a↑^{c1}a↑^{c1}a↑^{c1}.... (continued for b times).
This means that Wafflekirby's number is 10 with 10^{100} up arrows connecting it to another 10, so following out all the calculations might be a bit timeconsuming, but this is at least how you would do it.
Pronouns: she/her/hers or they/them please.
 Vytron
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Re: My number is, in fact, bigger!
I could. But my options are explaining WaffleKirby's number, and beating it, or just beating ODF's.
Also, please, let's keep the definitions of ^^... and ↑↑... separated so the latter one is much stronger. Since we have to define symbols on the thread for the layman, we can use nonstandard definitions to make much larger numbers.
Now, I'll introduce, factorials.
Factorials are simple enough, you just, get some number, then, reduce 1 from this number to produce a second number, and multiply both numbers to get a third number. Then you reduce 2 from the first number and multiply it by the third number. Until the first number reaches 1.
5! = 5*4*3*2*1=120
My number:
(4^^^4^^^4^^^4)!
(4^^^4^^^4^^^4*((4^^^4^^^4^^^4)1)*((4^^^4^^^4^^^4)2)*((4^^^4^^^4^^^4)3)...)
NINJA: Damn, I refuse to accept that 5↑5↑5 is just 5^5^5, I hope we can define ↑s to be successor arrows, like:
3+3 = 3 (sum)
3*3 = 3+3+3 (iterated sums)
3^3 = 3*3*3 (iterated multiplication)
3^^3 = 3^(3^3) (iterated exponentiation)
3^^^3 = 3^^(3^^3)
...
3^^{n}3 = 3^^^^{n times}^^^3
a↑b = a^^{b}b
a↑↑b = a^{a↑b}b
a↑↑↑b = a^{a↑↑b}b
...
a↑^{n}b = a↑↑↑^{a↑^(n1)b times}↑↑↑b
With this definition:
3↑^{64}4 is already higher than Graham's number
10↑^{googol}10 goes higher than Graham(googol), but not higher than Graham(googol+1)
So, huh, I'll just go with:
(10↑^{googol}10)!
Also, please, let's keep the definitions of ^^... and ↑↑... separated so the latter one is much stronger. Since we have to define symbols on the thread for the layman, we can use nonstandard definitions to make much larger numbers.
Now, I'll introduce, factorials.
Factorials are simple enough, you just, get some number, then, reduce 1 from this number to produce a second number, and multiply both numbers to get a third number. Then you reduce 2 from the first number and multiply it by the third number. Until the first number reaches 1.
5! = 5*4*3*2*1=120
My number:
(4^^^4^^^4^^^4)!
(4^^^4^^^4^^^4*((4^^^4^^^4^^^4)1)*((4^^^4^^^4^^^4)2)*((4^^^4^^^4^^^4)3)...)
NINJA: Damn, I refuse to accept that 5↑5↑5 is just 5^5^5, I hope we can define ↑s to be successor arrows, like:
3+3 = 3 (sum)
3*3 = 3+3+3 (iterated sums)
3^3 = 3*3*3 (iterated multiplication)
3^^3 = 3^(3^3) (iterated exponentiation)
3^^^3 = 3^^(3^^3)
...
3^^{n}3 = 3^^^^{n times}^^^3
a↑b = a^^{b}b
a↑↑b = a^{a↑b}b
a↑↑↑b = a^{a↑↑b}b
...
a↑^{n}b = a↑↑↑^{a↑^(n1)b times}↑↑↑b
With this definition:
3↑^{64}4 is already higher than Graham's number
10↑^{googol}10 goes higher than Graham(googol), but not higher than Graham(googol+1)
So, huh, I'll just go with:
(10↑^{googol}10)!
 orangedragonfire
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Re: My number is, in fact, bigger!
In that case let's just start on the fastgrowing sequence.
f_{1}(n) = n+1
f_{k}(n) = f_{k1}^{n}(n)
f_{ω}(n) = f_{n}^{n}(n)
f_{ω+k}(n) = f_{ω+k1}^{n}(n)
f_{ω2}(n) = f_{ω+n}^{n}(n)
Examples:
f_{ω2}(f_{ω2}(50))
f_{1}(n) = n+1
f_{k}(n) = f_{k1}^{n}(n)
f_{ω}(n) = f_{n}^{n}(n)
f_{ω+k}(n) = f_{ω+k1}^{n}(n)
f_{ω2}(n) = f_{ω+n}^{n}(n)
Examples:
Spoiler:
f_{ω2}(f_{ω2}(50))
 Vytron
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Re: My number is, in fact, bigger!
Man, that's massive overkill. But still after looking at the other thread these numbers still look tiny.
I'm have to use my secret weapon: Conway arrows.
The thing with Conway arrows is that they have built in recursion, see, you had to add an extra "layer" of recursion for good measure, at the point you need to add an extra f_{ω2}(n) to your function to get to the next big number, Conway just adds 1 to the rightmost arrow.
Now, let's compare:
f_{ω}(n) will be dominated by 3 → 3 → n → 2
f_{ω+k}(n) will be dominated by 3 → 3 → 3 → 2+k
f_{ω2}(n) will be dominated by 3 → 3 → 3 → n → 2
Now, applying extra layers of f_{ω2} will not do much as Conway does that automatically at → n+1, so I can beat your number with:
3 → 3 → 3 → 51 → 3
Conway arrows for the layman:
a → b → c = a^^{c}b
a → b → c → d = a → b → (a → b → c1 → d) → d (when c reaches 0 decrease d by 1, when d reaches 1, kill it).
Also: Landmark! we have beaten the first page of the other thread, yes! We're going faster!
I'm have to use my secret weapon: Conway arrows.
The thing with Conway arrows is that they have built in recursion, see, you had to add an extra "layer" of recursion for good measure, at the point you need to add an extra f_{ω2}(n) to your function to get to the next big number, Conway just adds 1 to the rightmost arrow.
Now, let's compare:
f_{ω}(n) will be dominated by 3 → 3 → n → 2
f_{ω+k}(n) will be dominated by 3 → 3 → 3 → 2+k
f_{ω2}(n) will be dominated by 3 → 3 → 3 → n → 2
Now, applying extra layers of f_{ω2} will not do much as Conway does that automatically at → n+1, so I can beat your number with:
3 → 3 → 3 → 51 → 3
Conway arrows for the layman:
a → b → c = a^^{c}b
a → b → c → d = a → b → (a → b → c1 → d) → d (when c reaches 0 decrease d by 1, when d reaches 1, kill it).
Also: Landmark! we have beaten the first page of the other thread, yes! We're going faster!
 orangedragonfire
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Re: My number is, in fact, bigger!
hmm, I guess I'll have to go straight to f_{ω2}.
f_{ω2} (n) = f_{ωn} (n)
This should beat conwayarrow chains of pretty much any length.
f_{ω2} (7)
/edit: also, we should wait for a layman to tell us when we stopped making sense =D
f_{ω2} (n) = f_{ωn} (n)
This should beat conwayarrow chains of pretty much any length.
f_{ω2} (7)
/edit: also, we should wait for a layman to tell us when we stopped making sense =D
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