"Even if we had a googol googol googol googol googol ... ... ... ... ... ... ... ... ... ... ... ... googol googol googol googol (say 10^31 times) observable universes you still wouldn't have enough particles to write out Skewes' Number!!! Now THAT is a proper description." ~ Sbiis Saibian (source) Introduction
We've looked a variety of large numbers so far: numbers that exist within the real world, numbers encountered in science, and numbers used to tell the world about large numbers. But now, it's time to look at two famous numbers that appeared in mathematics: the Skewes' numbers.
The Skewes' numbers are large upper-bounds to the solution of a problem whose answer is still not known, and they were named after Stanley Skewes who proved them to be upper-bounds. The smaller bound assumed the Riemann hypothesis to be true, and the larger one about 20 years later did not assume it. Although the upper-bounds have been improved, the numbers have gotten quite some fame due to their size. In this article I'll talk about the math behind Skewes' numbers, then the numbers themselves and how big they are, and finally the further history of the problem in the years following Skewes' proof.
Skewes' Problem
The problem Skewes was working on was this question:
What is the smallest natural number x such that π(x) > li(x), where π(x) is the prime counting function and li(x) is the logarithmic integral function?
Both π(x) (π is the Greek letter pi) and li(x) are simple functions which serve as approximations of each other. First we'll take a closer look at what each of them are.
The Prime-Counting Function
π(x) is the prime counting function, a function represented with the Greek letter pi (π). Here, the letter pi is not used to represent the number equal to about 3.14159. π(x) is defined as the number of prime numbers less than or equal to x. For example, π(5) = 3, since there are 3 numbers less than or equal to 5 which are prime, and they are 2, 3, and 5. Here are the first 20 values of π(x):
π(1) = 0
π(2) = 1
π(3) = 2
π(4) = 2
π(5) = 3
π(6) = 3
π(7) = 4
π(8) = 4
π(9) = 4
π(10) = 4
π(11) = 5
π(12) = 5
π(13) = 6
π(14) = 6
π(15) = 6
π(16) = 6
π(17) = 7
π(18) = 7
π(19) = 8
π(20) = 8
And here are some larger values of π(x), according to Wikipedia[1]:
π(1000) = 168
π(1,000,000) = 78,498
π(1,000,000,000) = 50,847,534
π(10^12) = 37,607,912,018
π(10^15) = 29,844,570,422,669
π(10^18) = 24,739,954,287,740,860
π(10^21) = 21,127,269,486,018,731,928
π(10^24) = 18,435,599,767,349,200,867,866
Hold on now. How is it possible to calculate π(x) for such large inputs?! You may be thinking, don't you have to calculate every prime number below x, and even that is quite hard, so how can you even do it? You don't need to do it by brute force (i.e. checking every prime); there are methods for calculating π(x) using some mathematical tricks. We won't worry about how exactly you can do that, just that it is possible to calculate such values.
Besides the algorithms for calculating π(x) (which do have a limit for which inputs you can realistically use), there are functions to approximate π(x), which is where the second part of Skewes' problem comes into play.
Approximating π(x)
The distribution of prime numbers has fascinated mathematicians since ancient times, largely since that distribution has no apparent regular patterns. However, over the years mathematicians have proved some useful facts relating to it. The most important one is the prime number theorem, which states that the limit of π(x)/(x/ln(x)) as x approaches infinity is 1. In other words, π(x) and x/ln(x), as x approaches infinity, will be approximately the same value. ln(x), as you probably know, is the function defined as the number a such that ex = a, where e is the important mathematical constant with a large amount of special properties equal to about 2.781828... which is most commonly defined as the limit of (1+1/x)x as x approaches infinity.
To see that π(x) and x/ln(x) are generally about the same value, let's compare π(x) with x/ln(x) for some small inputs:
With these results it looks like π(x) will generally be larger than x/ln(x). But these small examples aren't nearly enough to make good conclusions, so let's get some larger examples:
Notice that x/ln(x) always tends to lag a bit behind π(x), though the ratio between x/ln(x) and π(x) becomes closer to 1 as x gets larger. It turns out that when x ≥ 11, x/ln(x) will always be a little smaller than π(x).
There are better approximations to π(x) than x/ln(x). The approximation we're focusing on here is the logarithmic integral function, noted as li(x). If you know your calculus, think of the function as the integral:
x
/\
|
| 1/ln(t) dt
|
\/
0
Otherwise, just think of li(x) as the area between the graph of y = 1/ln(x) from 0 to x and the x-axis. The graph of y=1/ln(x) looks like this:
 and the graph of li(x) looks like this:
 The reason why at some points the area is negative is because area of a function below the x-axis is considered negative in mathematics, while area above it is considered positive.
The logarithmic integral function is a much better approximation for π(x) than x/ln(x) is. Observe some values of each of those functions for large values of x:
li(x), as you can see, is quite a good approximation! Take a look at π(x) (black), li(x) (red), and x/ln(x) (blue) graphed together, and see how much better of an approximation for π(x) li(x) is than x/ln(x):  (graph taken from [3] but modified)
However, li(x) seems to always be a little bigger then π(x). For quite a while most mathematicians thought that was always the case. However, if you think about the erratic behavior of the distribution of prime numbers, making such a claim seems a lot like jumping to conclusions. Therefore, it was a kind of mission to see if there is a point where li(x) is less than π(x).
In 1914 a mathematician named John Edensor Littlewood proved that there does indeed exist a number x such that li(x) is less than π(x)[4]—in fact, there exists an infinite amount of x's where li(x) is less than π(x). Since there exists such a number x, that means that there must be a smallest value of x such that li(x) is less than π(x). Finding that smallest x is where Skewes himself comes into play.
Skewes' Proof
Stanley Skewes was a South African mathematician (born 1899, died 1988) who was a student of Littlewood's at Cambridge University. According to Wikipedia[5] he is not known for much other than discovering his eponymous numbers.
In 1933 Skewes made his attempt at solving the problem of the smallest number x where π(x) > li(x). He did not fully solve a problem, but he did give an upper-bound to the solution, i.e. a number the solution can be shown to be no greater than. He arrived at the upper-bound:
eee79
a number which can be approximated in base 10 as:
10108.852142197*1033
and it has an 8.85-decillion-digit number of digits (more on its size later). The number falls between a googolplex and a googolduplex, and it easily broke the record of the largest number used in serious mathematics.
However, there's a little twist: he solved the problem assuming the Riemann hypothesis to be true. What exactly is the Riemann hypothesis? It is a hypothesis that's somewhat more complicated than the other things we've learned about so far, and that's what we'll examine in detail next.
The Riemann Hypothesis
The Riemann hypothesis is one of the most important conjectures in mathematics. It can be thought of as a hypothesis so notable that it has an indirect relevance in the field of large numbers, as it makes the difference between Skewes' number and the second Skewes' number, which are two of the "classic" large numbers right up there with the googol, the googolplex, Graham's number, and the like. The hypothesis was proposed by Bernhard Riemann in 1859, and to this day it remains yet to be found out whether it is true or not. According to Wikipedia[6] the hypothesis is the subject of several books, and finding whether or not it's true is part of both Hibert's 23 problems and the Millennium prize problems.
But what exactly is the Riemann hypothesis? It has to do with a well-known function called the Riemann zeta function, a function denoted with the Greek letter zeta (ζ). It's a function that can input any real or complex number. A complex number is any number of the form a+bi, where a and b are real numbers, and i is the imaginary unit defined as the square root of -1. In a complex number a is the real part and b is the imaginary part, and real numbers are a subset of the complex numbers where the imaginary part is 1. Here is how the Riemann zeta function is defined. When the real part of the number inputted is greater than 1, ζ(x) is defined as the infinite sum, 1/1x + 1/2x + 1/3x ..., which can be noted with the sum notation as:
∞
Σ 1/nx
n=1
It can be extended to other complex numbers using analytic continuation of that series. The zeta function is known for having fascinating behavior. Here are some values of the function:
ζ(-1) = 1/1-1 + 1/2-1 + 1/3-1 .... = 1+2+3+4+5 ... Although this sum diverges to infinity with the usual calculations, using infinite series it can be derived to be equal to -1/12. We won't go into detail on how exactly that paradoxical sum can be derived, but this serves as an example of an unusual value of the function.
ζ(0) = 1/10 + 1/20 + 1/30 + 1/40 ... = 1/1 + 1/1 + 1/1 ... = 1+1+1+1 ..., which can be strangely derived to equal -1/2. Once again we won't go into detail about this strange sum, other than that it is strange. N.H. Abel famously said about such series that "divergent series are the invention of the devil".
ζ(1) = 1/11 + 1/21 + 1/31 + 1/41 ... = 1 + 1/2 + 1/3 + 1/4 ... This sum, like the previous two, diverges to infinity, although this time the reason why is not straightforward: since the terms get smaller and smaller you'd expect it to converge to a finite number, but in fact it once again diverges to infinity. The difference between the first x terms of this sum and ln(x), as x approaches infinity, is a mathematical constant known as the Euler-Mascheroni constant, denoted with the Greek letter gamma (γ), which is equal to about 0.57721566... Even stranger, unlike ζ(-1) and ζ(0) a finite value for this sum cannot be derived with infinite series. However if it is interpreted as a limit:
 then that limit takes on the value of the Euler-Mascheroni constant, which is just as strange.
ζ(2) = 1/12 + 1/22 + 1/32 ... ~ 1.64493007.... = π2/6. This sum is notable because it gives an example of pi showing up in a place you wouldn't expect it to. The sum was proven to be equal to π2/6 by Euler in 1734.
ζ(3) = 1/13 + 1/23 + 1/33 ... ~ 1.2020561... This number is known as Apéry's constant, and unlike ζ(2) it has no known compact expression like π2/6 for ζ(2). However there are many infinite series that are equal to this constant.
ζ(4) = 1/14 + 1/24 + 1/34 ... ~ 1.082323... = π4/90. Like ζ(2) this is an example of pi cropping up in a place you don't expect it to, and it has a nice compact expression.
However, plugging in real numbers into the zeta function only scratches the surface of what the zeta function has to offer. If you plug complex numbers into the function things get wild, and that's where the Riemann hypothesis comes into play. What the hypothesis asks is best explained visually. To start off, here's a graph of the zeta function with any real or complex inputs:
 This graph is a somewhat unusual graph. Here, the x and y axes represent the inputs of the function, while the colors represent the outputs. The x-coordinate of a point represents the real part of the input number, while the y-coordinate represents the imaginary axis of the number. For example, the point (1, 1) which is one unit up and one unit right of the origin would represent 1+i, whose real part and imaginary parts are both 1.
Now the colors are a bit more complicated. The coloring is based on the complex plane, which is the 2-dimensional coordinate plane mapping each complex number to a point on that plane: the x-coordinate is the real part and the y-coordinate is the imaginary part. If the imaginary part is 0 then the number in the complex plane is just a real number. Then the coloring is defined based on the number's position in the complex plane: the darker a color is the closer it is to 0, and light colors denote values far from 0. Black is zero and white is infinity whether it be positive, negative, or whatever else (such as the limit of a*2+a*i when a approaches infinity). Then the hue of the color denotes which angle the number is from 0 in the complex plane. Red denotes positive real numbers (such as 1), cyan denotes negative real numbers (such as -1), yellow-green denotes positive multiples of i (such as i), and violet denotes negative multiples of i (such as -i). And that's what all those colors mean.
The Riemann hypothesis focuses on what the zeroes of the function are, points where ζ(x) = 0. In this graph the zeros are point of pure black. The dark zone in the middle of the graph isn't quite black, but rather representing numbers quite close to 0. The exceptions are -2, -4, -6, etc.—those are the zeta function's trivial zeros. The only known non-trivial zeros of the zeta function are black points in this graph which all line up on a vertical line in the middle of the graph. Those zeros in this graph are all complex numbers with the real part equal to 1/2. But are there any non-trivial zeros of the zeta function that do not have the real part equal to 1/2? That's where we get to just what the Riemann hypothesis is—it says that the only non-trivial zeros of the Riemann zeta function are where the real part of the input is equal to 1/2. As of yet nobody has been able to prove whether this is true or not.
So how does the Riemann hypothesis relate to Skewes' problem? This part is more interesting. Many of the formulas for the prime counting function make use of the zeros of the zeta function. The reasons for this relation are rather complicated, but they relate to this curious alternate formula for the zeta function, discovered by Euler:
ζ(x) = 1/(1-2-x) * 1/(1-3-x) * 1/(1-5-x) * 1/(1-7-x) * 1/(1-11-x) ...
With the notation for infinite products this can be written as:
∞
Π 1/(1-p-x)
p prime
This alternate formula for the zeta function is a big reason why the Riemann hypothesis relates at all to the distribution of prime numbers. Hopefully you now get the idea that the Riemann hypothesis is notable to Skewes' problem.
The Second Skewes' Number
In 1955, Stanley Skewes went to again find an upper bound for when π(x) is first greater than li(x), but this time he did not assume the Riemann hypothesis to hold true. Not assuming the Riemann hypothesis made Skewes' problem more difficult, and he arrived at an upper-bound bigger than Skewes' number, equal to:
eeee7.705
This number is bigger than a googolduplex, and it has roughly a 3.3*10963-digit-number of digits. In base 10 it can be approximated as:
10103.299943221*10963
The number dethroned the original Skewes' number as the largest number used in serious mathematics. The two Skewes' numbers became honored as the largest numbers used in mathematics until the much larger Graham's Number came along in 1977, and in subsequent years Graham's number has itself lost that title, but let's not get ahead of ourselves.
Now that we've reviewed the two Skewes' numbers, we're now ready to talk about how big they are.
How big is Skewes' number?
To get an idea of how big either Skewes' number is, we must first convert to base 10. Doing this isn't as hard as you may think. You can make use of the common logarithm log(x), defined as the number a with the property that 10a = x. We start with the expression for the first Skewes' number:
eee79
Then, by converting e to 10log(e), since by definition 10log(x) = x, we can change this expresion to:
(10log(e))ee79
To simplify the expression we use one of the laws of exponents, (ab)c = ab*c, and so we obtain:
10log(e)*ee79 Next we convert another e to 10log(e):
10log(e)*(10log(e))e79
and use the laws of exponents again:
10log(e)*10log(e)*e79
To get rid of the leftmost log(e), we need to convert it to 10log(log(e)):
1010log(log(e))*10log(e)*e79
Then we use another law of exponents, this time ab*ac = ab+c:
1010log(log(e))+log(e)*e79
The bold expression, log(log(e))+log(e)*e79 is an expression that evaluates to a small enough number for any calculator to handle. Plugging that expression into the Keisan online calculator[2] with 50 digits of accuracy gives us the result:
8,852,142,197,543,270,606,106,100,452,735,038.5506795094512173
Thus we obtain the estimate for Skewes' number:
10108,852,142,197,543,270,606,106,100,452,735,038.5506795094512173
which can be compactly approximated as:
10108.852142197*1033
Just how big is that number?! It's certainly hard to give an analogy of its size. Let's consider just the difficulty of writing Skewes' number, or more specifically, the integer part of Skewes' number. The integer part of any real number is just the number with everything after the decimal part cut off. For example, the integer part of 441.93 is 441, and the integer part of -3.16457 is -3.
To find the number of digits in a large number, just take its base-10 logarithm, add 1, and round it up to the nearest integer. Finding the base-10 logarithm of a large number when written out in base-10 exponent form is easy: just chop off the bottom 10. In our case the base-10 logarithm of Skewes' number is about:
108.852142197*1033
And adding 1 makes basically no difference to the number anyway, ditto for rounding up or down. So there we have the number of digits in Skewes' number, and now we can consider how hard it is to write it out.
First off, imagine you could fit a digit into each Planck volume in the observable universe (from here on out I'll just say "universe" instead of "observable universe"). Not that you could actually do that, so this is an EXTREMELY generous estimate of how many digits could fit in the universe. Fitting a digit in each Planck volume would give you:
3*10185 digits — not even close!
So we need to take things up a notch. How about we make a second-order universe, which is a sphere of as many universes as there are Planck volumes in the universe! Another way to put this is shrinking the universe down to the size of a Planck volume, and fitting as many of those shrunken-down universes as you can into our universe, then expanding all those universes to normal size—that's pretty insane alright! And fitting a digit into each Planck volume in the second-order universe gives you ...
9*10370 digits — still not even close!!
Well let's then make a third-order universe, which would be like shrinking down our second-order universe to the size of a Planck volume, and then filling us our universe with all those shrunken-down second-order universes. That would be equal to the number of Planck volumes in the universe cubed, which gives you:
2.7*10556 digits — >:(
How about we make a sphere of third-order universes with as many spheres as there are Planck volumes in the universe, which is a fourth order universe? Or a sphere of the same amount of fourth-order universes, which is a fifth-order universe? Or a sixth, seventh, eighth, ninth, tenth, etc, order universe? How about a millionth-order universe, how many digits will that give us? No conventional calculator could calculate that number of digits, so we'll need to use Robert Munafo's online calculator called Hypercalc[8]. Hypercalc can handle numbers far larger than any ordinary calculator. It can handle any number we'll encounter in section 1 of this site, and many of the numbers we'll encounter in section 2! Raising 3*10185 to the millionth power, which is how many digits the millionth order universe would give us, equals:
1.8*10185,477,421 digits
This number has 185 million digits, but the number of digits in Skewes' number has 8.85 decillion digits. In terms of more familiar -illions that is 8.85 billion trillion trillion digits!! So we're still far from being able to write Skewes' number!
How about a trillionth order universe? With Hypercalc would give us about:
101.85477*1014 digits
We're definitely getting somewhere, but this still doesn't cut it. Even by multiplying this number by itself a quintillion times you wouldn't get to the number of digits in Skewes' number! So what order universe would we need to write Skewes' number?
We would need a 48 nonillionth order universe to make it happen. A nonillion is a million trillion trillions! It's near impossible to even comprehend that, but we can try:
Imagine starting with the plain old observable universe, and scaling it up by a factor of 3*10185 (the number of Planck volumes in the universe) every second. In 5 seconds we'd already reach past 101000 digits, and in a day we reach about 1015,951 digits, and in a year we reach a 31-millionth order universe which could fit about 105,800,000,000 digits! But we wouldn't reach a universe vast enough to write Skewes' number until 1.52 septillion years! That's about a quintillion times the age of the universe. To get an idea of how long of a time that is, consider that even a quintillion seconds is twice the age of the universe! Could things get any more insane?!
As it turns out, we've only gotten to the difficulty of writing Skewes' number! We haven't even gone close to getting a picture of how big Skewes' number itself is—we've only been blown away by its number of digits!!
To get a feel of how big Skewes' number really is, we need to take things up a notch. First off, let's get back to the 48-nonillionth-order universe that could fit the digits of the integer part of Skewes' number. Let's call that universe a Skewes' universe. But now, let's not just have as many of those universes as there are Planck volumes in the universe. That's only going one extra step in universe orders from the 48 nonillion steps we took to get to the Skewes' universe. Instead, let's make as many Skewes' universes as there are Planck volumes in the Skewes' universe! Call that a second-order Skewes' universe. Likewise, as many second-order Skewes' universes as there are Planck volumes in the Skewes' universe would make a third-order Skewes' universe. Continue with a fourth-order, fifth, sixth, seventh ... tenth, 100th, 1000th ... millionth, billionth, trillionth ... decillionth ... centillionth ... it takes about as many steps up as there are Planck volumes in the Skewes' universe to get to a universe with Skewes' number of Planck volumes!
It's so easy to get utterly completely blown away by numbers this big ... and yet, this isn't the end of the insanity. Consider the base-10 representation of Skewes' number:
10108.852142197*1033
This can also be approximated as:
10101033.94704
Now consider a common approximation for Skewes' number:
10101034
This looks like quite a good approximation, right? It's actually A LOT BIGGER than Skewes' number. How much bigger? First, let's redefine a Skewes' universe as not just a universe big enough to fit the digits of the integer part of Skewes' number, but as a universe that is made of a Skewes' number of Planck volumes! Now, let's not just have as many Skewes' universes as there are Planck volumes in the universe, or as many as there are digits in Skewes' number. Instead, let's have Skewes' number of Skewes' universes (under our new definition)! Call that a second-order Skewes' universe. Then of course, Skewes' number of second-order Skewes' universes would be a third-order Skewes' universe. Then we can have a fourth-order, fifth-order, sixth-order, seventh, eighth, ninth, tenth ... 100th ... 1000th ... millionth-order universes ... and so on ...
So then does it takes 48 nonillion scalings up from Skewes' universe to reach 10101034 Planck volumes? Nope, it's much much worse. With 48 nonillion scalings up we would not even reach a universe with 10101033.94705 Planck volumes (in contrast with 10101033.94704, which is approximately Skewes' number)! Hold on now, how can that be right?! Let's take a closer look at what's going on:
Each time we scale up our nth-order Skewes' universe to a (n+1)th order Skewes' universe, we multiply the number of Planck volumes by Skewes' number. A nth-order Skewes' universe would therefore have Sn particles, where S is Skewes' number. Let's see what happens with a 48-nonillionth order Skewes' universe:
(10101033.94704)4.8*1031 — that's Skewes' number raised to the 48-nonillionth power.
= 10(101033.94704)*(4.8*1031) — just using the laws of exponents here. ~ 10(101033.94704)*(1031.6812)
= 1010(1033.94704+31.6812)
Now let's consider 1033.94704 + 31.6812. This is itself much smaller than 1033.94705 which is about 1033.94704 + 3*1028. All this is completely counterintuitive—and yet we haven't gotten close to the sphere of 10101034 Planck volumes!
Let's continue the nth-order Skewes' universe idea now. After the 48-nonillionth order, let's continue through the orders, with a decillionth-order Skewes' universe, a vigintillionth-order Skewes' universe ... googolth, centillionth, 101000th, 1010,000th ... 101,000,000th ... 101012th order ... and so on ... when we reach a 101.15*1033th order Skewes' universe, that would be made up of about 10101034 Planck volumes. Absolutely mind-blowing!!!
As you can see, getting to the number of digits in Skewes' number is already utter insanity, and then getting to Skewes' number itself is far more insane. Unbelievably, it takes just as long to get from Skewes' number to what looks like a good approximation for Skewes' number! As you can see Skewes' number is an utterly unfathomable number, with some pretty mind-bending properties. However, it's nothing compared to the even more monstrous second Skewes' number, which is the next number for us to examine.
How big is the second Skewes' number?
The second Skewes' number, which is less known than the original Skewes' number, is equal to:
eeee7.705
Converting that to base 10 gives us:
(10log(e))eee7.705
= 10log(e)*eee7.705
= 10(10log(log(e)))*(10log(e))ee7.705
= 1010log(log(e))*10log(e)*ee7.705
= 1010log(log(e))+log(e)*ee7.705
Now, what to do with that log(log(e))? Adding it to log(e)*ee7.705 will have almost no effect on it, since log(e)*ee7.705 is so big. Therefore we can chop off and still be able to approximate the second Skewes' number. Thus we continue with:
1010log(e)*ee7.705
= 101010log(log(e))*(10log(e))e7.705
= 101010log(log(e))+log(e)*e7.705 Now log(log(e))+log(e)*e7.705 is actually small enough to calculate with an ordinary calculator. Plugging it into the Keisan online calculator gives us 963.5185064675691205137. Therefore the second Skewes' number is about:
101010963.5185064675691205137
This number is not only much much larger than a googolplex, it's also larger than a googolduplex which is:
101010100
Hopefully you are starting to get the idea that the second Skewes' number is HUGE ... but how much larger is it than Skewes' number? Let's return to the idea of Skewes' universes. First there is the Skewes' universe of Skewes' number of Planck volumes, then the second-order Skewes' universe made of Skewes' number of Skewes' universes, then third-order, fourth-order, etc. But let's not stop at the 101.15*1033th order, but go to however many scalings up it takes us to get to the second Skewes' number.
Even after a googolplex scalings we wouldn't reach the second Skewes' number. It would take about 1010963.518 scalings to reach it. To get a feel of how many scalings that is, let's use a time analogy.
Imagine a universe much much more vast than ours, whose volume in Planck volumes is, at any moment, denoted by a number N. In the beginning of that universe, its volume in Planck volumes is Skewes' number. Now imagine that every Planck time from there on out the universe's volume multiplies by Skewes' number. This means that one Planck time into the universe's time it would be equivalent to a second-order Skewes' universe, and in the next Planck time after that it would be a third-order Skewes' universe, and so on. After just one second, it would be equivalent to roughly a 1043th order Skewes' universe!
Now imagine going forward in time from this beginning, up to 13.7 billion years (the age of our universe). At that point that universe would be as old as our universe is. Call that time T1. But now, imagine zooming waaaaaaaay past T1, to the point where the span between the Big Bang and T1 is just one span amongst as many spans as Planck times since the Big Bang! In other words, we've now dwarfed the age of the universe by a factor of the age in Planck times! Call that point in time T2, the point where the universe is that old.
So at this mysterious strange time T2, will that universe be made of at least 10101034 Planck volumes? Nope, not even close. The age of the universe in Planck times is about 8.03*1060 [9A], and multiplying that by itself gives you an age of 6.45*10121 Planck times for the point in time, T2. This means that that universe at that point in time is equivalent to a 6.45*10121th order Skewes' universe, aka Skewes' number to the power of 6.45*10121 Planck volumes. As with the 48-nonillionth-order Skewes' sphere the number is much less than something like 10101033.94705. Simply observe:
(10101033.94704)6.45*10121
= 10(101033.94704)*(6.45*10121) ~ 10(101033.94704)*(10121.81)
= 1010(1033.94704+121.81)
So at what point in time would we reach 10101034 Planck volumes? We'll need to scale things up more. Imagine scaling the age of the T2 universe up by a factor of the age of the T1 universe in Planck times. The point in time where we've reached that point, T3, is about 5.18*10182 Planck times after the start of the universe. Then we can have T4, which is scaling up the time from the start of the universe to T3 by a factor of 8.03*1060, then T5, T6, T7 ... T10 ... T100 ... T1000 ... ... T1,000,000 ... ... T1012. Turns out that we finally reach 10101034 Planck volumes at T1.9*1031, or after 19 nonillion scalings up. Now think about that for a minute. You may think that a nonillion scalings up is not entirely unimaginable, but think again. If you do one scaling up per second starting from the Big Bang of our universe, in the present day you'd be under a trillionth of the way there. And guess what ... we aren't even close to the second Skewes' number!
After a googol scalings up (i.e. T10100) we'd get a universe with about 10106*10101 Planck volumes. That's over a googolduplex. Now a googol is MUCH LARGER than the age of our universe in Planck times. That means that if you go start with T1 at the Big Bang, then do one scaling up (e.g. from T35 to T36) every Planck time, and continued to the present day, you still wouldn't reach a universe even close to large enough to fit a googolduplex Planck volumes, let alone the second Skewes' number of Planck volumes.
At T5.15*10961, we FINALLY REACH a second Skewes' number of Planck volumes. Now 5.15*10961 (that's more than a centillion centillion centillions) is a number that is itself far far far beyond human comprehension. And that's the number of scalings up of a device that was itself used to get an idea of all the scalings up, needed to reach the second Skewes' number!
Is that mind-blowing? Good, hopefully you realize now that the Skewes' numbers are exotically huge. They exhibit utterly counterintuitive behavior we don't see with numbers we use in day-to-day life at all; and yet, they can be written quite easily using standard mathematics! These large numbers are nothing like the puny millions, billions, and trillions we encounter in day-to-day life—they're utterly exotic numbers that they can't even compare to!!!
The funny thing is that the most commonly given example of the size of numbers this big is that all of their digits wouldn't fit in the universe; however, this doesn't come near capturing a proper idea of their size, since even a number like 1010200 (which is much smaller than either Skewes' number) would be GUARANTEED not to fit, even with fitting each digit in a Planck volume! To actually get an idea of how amazingly big these numbers must be, we need to say goodbye to even considering realistically writing all their digits, and take much more radical methods! This is just what I did, a rare example of an attempt to get a true feel of how big Skewes' numbers really are (Sbiis Saibian's article on Skewes' numbers[4] is another example). It's actually a little annoying that it is so popular to simply say that [INSERT NUMBER BIGGER THAN A GOOGOLPLEX HERE] has so many digits that they wouldn't fit in the universe; in a sense, this represents shying away from the enormity of these monsters, when it isn't too hard to use something much much more mind-blowing.
But before we get carried away into a discussion of people not even wanting to get a full picture of the magnitude of these giants, let's switch gears and go into the further history of Skewes' numbers after Skewes' proofs.
Better bounds for Skewes' problem
In addition, Lehman's proof took away Skewes' numbers' status as practical upper-bounds, and from there on out they became recognized instead as footnotes or mathematical curiosities, as former upper-bounds that ended up surprisingly huge. Nonetheless Skewes' numbers retain their recognition, as "legitimate" numbers due to their use other than just for the sake of largeness, and for the next five years they were still recognized as the largest numbers in mathematics.
Skewes' numbers were dethroned from that title in 1971 by a far far larger number used as an upper-bound in an unrelated problem in Ramsey theory by mathematicians Ronald Graham and Bruce Rothschild. That number is a whole different kind of number from Skewes' number, as it can't be represented with mainstream mathematical notation and needs a whole new notation to express! An alternate, larger version of that number which was devised in 1977 by mathematics writer Martin Gardner as an easier-to-explain version of Graham and Rothschild's number, and that number is today known as Graham's number. To this day Graham's number is still regularly honored as the largest number used in serious mathematics, although numbers that are themselves FAR LARGER than Graham's number have since then taken that title, such as TREE(3) and SCG(13), which aren't upper-bounds at all but rather extremely large solutions to problems in mathematics! But let's not get ahead of ourselves. We'll discuss Graham's number in detail in section 2, and for now let's get back to the further history of Skewes' problem.
H.J.J. te Riele in 1987 proved a still better upper-bound to the problem, equal to ee27/4, or about 8.17*10370 [11][9B]. This bound is equal to roughly eighty thousand vigintillion centillion. This means that it's quite a lot smaller than the previous bound, having about a third as many digits as Lehman's bound. Then in 2000 the bound was improved to 1.39822*10316 by Bays and Hudson[12]. It may be tricky to reember that this bound is about a septendecillion (that's a quintillion quintillion quintillions) times smaller than Riele's. After 2000 a few small improvements were made to the upper-bound. The most recent such improvement was in 2010 by Stefanie Zegowitz[13], reducing the upper-bound to e727.951332973, which is about 1.39716*10316. This is very slightly smaller than the previous bound 1.39822*10316, being only 0.07% smaller. Robert Munafo incorrectly claims that this value is the solution to Skewes' problem[9B]. To this day, the solution to Skewes' problem remains unknown and an open question. However, it seems likely that the solution is around 1.39716*10316, considering that nobody has been able to improve the bound very much from the slightly larger 1.39822*10316. But we still don't know the solution for sure. In addition there is a current lower bound for the solution to Skewes' problem, a number that the solution is known to be no smaller than. If a number X is a lower bound for the solution to a problem, that means that we know that the solution must be a number greater than X. For Skewes' problem there must be some lower bound: we can check each individual number up to some point for whether or not π(n) > li(n). However after a certain point doing that checking will become difficult. Therefore proving a lower-bound for this particular problem is harder than finding an upper-bound/ For proving an upper-bound you just need to prove that some number particular number n less than a number x has the property π(n) > li(n), but for proving a lower-bound you need to prove that π(n) > li(n) does not hold true for all n less than your lower-bound x. As of 2015 the best known lower-bound for Skewes' problem is 1014, or one hundred trillion. Tadej Kotnik proved this lower bound in 2008[11][14]. While I can't really comment on how he proved that lower bound, that allows us to say that, we currently know that: 1014 < N < 1.39716*10316
where N is the smallest positive integer such that π(N) > li(N). As you can see, Skewes' problem is still yet to be solved, though we're certainly closer to an answer than we were in Skewes' time. Conclusion Today, Skewes' numbers continue to have a well-earned spot among the classic large numbers, and are given an honorable mention in the large number discussion. However, I doubt you've heard their real size explained to such detail as I did! These numbers are pretty insane and head-spinning numbers, and they continue to be noted as mathematical curiosities, and as examples of really really big numbers. But as it turns out, these numbers aren't even close to being the biggest numbers ever, not even the biggest numbers in the popular imagination. They are just barely starting the endless numberscape of googology! In any case, the Skewes' numbers are among the most famous large numbers these days, and therefore are worth discussing in detail in an article like this. Up next we'll look at the extensions to the -illions way past Skewes' number or anything like that! Sources
[1] Wikipedia's article on the prime counting function (link)
[2] Keisan online calculator, used to calculate large values of li(x) and other things (link)
[3] thatsmaths.com's post on the prime number theorem (link)
[4] Sbiis Saibian's article on Skewes' numbers (link)
[5] Wikipedia's article on Stanley Skewes (link)
[6] Wikipedia's article on the Riemann hypothesis (link)
[7] Wikipedia's article on the Riemann zeta function (link)
[8] Robert Munafo's Hypercalc (link)
[9A] Robert Munafo's Notable Properties of Specific Numbers, page 19 (link)
[10] 1966 paper proving a better upper-bound to Skewes' problem (link) [11] Wikipedia's article on Skewes' number (link) [12] Bays and Hudson's 2000 paper on Skewes' problem (link) [13] Zegowitz's 2010 paper on Skewes' problem (link) [14] Kotnik's 2008 paper on Skewes' problem, only the beginning viewable for free (link) |