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« **on:** November 08, 2019, 01:46:09 am »
Graham's number

Let’s have some fun with really big numbers.

Suppose you took the four vertices of a square. It may be obvious but important that we're dealing with two dimensions so it has four vertices (2^2). Draw lines connecting all possible pairs of every vertex; you'll find there are 6 in total (four on the 'outside' and two diagonals).

Now you can draw each of the 6 lines one of two colors (red or blue in my example). And there are a bunch of different combinations you can have there. You could draw many of these square representing each combination. See example A.

So what? Well there are some interesting configurations you could create... but not with this square; we need more dimensions. A three dimensional cube has 8 vertices. And now there are 28 possible line segments. Now color each line one of two colors same as before.

The whole point of this exercise is to avoid the following configuration:

A square (contained somewhere within this cube) with all 6 vertices the same color. See example B.

In three dimensions (a cube) it is possible to avoid this. Is it avoidable in larger dimensions? There is the 'four dimensional cube' or a Tesseract. See the below picture (Which I believe is correct but keep in mind only a representation). Here there are 120 different line connections. How many different ways can you color these?

A simple calculation but a not so simple answer: 2^120= 1,329,227,995,784,915,872,903,807,060,280,344,576

Already a number that's difficult to express by computers. To put this number into perspective here's a list of some 'ordinary numbers'

one thousand = 1,000

one million = 1,000,000

billion = 1,000,000,000

trillion = 1,000,000,000,000

quadrillion = 1,000,000,000,000,000

quintillion = 1,000,000,000,000,000,000

sextillion = 1,000,000,000,000,000,000,000

septillion = 1,000,000,000,000,000,000,000,000

octillion = 1,000,000,000,000,000,000,000,000,000

Astronomers by the way, estimate the number of stars in the known universe is around 70 sextiliion...

The above number is larger than all of these. But back to the problem; the question is can you avoid a flat (in a plane) square of all connected vertices with the same color within the tesseract? The answer is yes. It is known that it's avoidable in fact in 5 dimensions, all the way up to 12. So is it always avoidable?

The answer is no.

If the dimension is large enough you cannot avoid this configuration.

Let's take 13 dimensions. In a 13 dimensional cube you'd have 8192 vertices.

8192*8191/2 = 33,550,336 line segments. So possible configurations of red/blue lines in this example would be 2 to the power of 33,550,336... A number that most computers cannot compute. The answer to 13 is not presently known. But the point at which it must happen that it cannot be avoided, that is, the upper bound, is known; Graham's number.

How big is Graham's number? It's pretty big. So big that a different type of notation must be used to help describe it. Arrow notation describes powers of powers. Here’s how it works:

3↑3 means 3 to the 3rd power (3 cubed) = 27

3↑↑3 means 3 to the power of three, to the power of three (in other words 3 to the 27 power) = 7,625,597,484,987. Every new arrow adds another exponent on top of the exponent, a tower of exponents so to speak.

So what is 3↑↑↑3? Simple; we already know two arrows mean 7 trillion, so this means 3 to the power of 7 trillion. What number is that? Well the answer contains 3.6 trillion digits. So I can’t write it here. There is no specific name for this number or any numbers in this realm for that matter. This kind of number cannot even be written accurately in scientific notation (x*10^n). Already this is a number that is not really comprehendible by humans because we have no point of reference for such numbers. We’re used to dealing with single and double digits in our lives. We may even struggle with mere triple digits. And 3↑↑↑3 isn’t even close to the realm of Graham’s number.

Now take 3↑↑↑↑3. Once again we can skip steps because we already know that this essentially means 3 to the power of ((3↑↑↑3)a 3.6 trillion digit number). So this number has many, many, many times more digits than the previous 3.6 trillion digit number. For every single arrow added to this operation it puts you in a whole other world of size so to speak. And every arrow raises it by a degree far greater than the previous increase. And still, we’re not even in the ballpark of Graham’s number.

3↑↑↑↑3 represents the number of arrows in between another pair of threes. This gives you another number of even greater earth shattering epicenes. Remember just 3↑↑↑3 is a number that cannot be written down. The number we’re looking at is gotten to by doing an operation of 3 followed by a number of arrows equal to (3↑↑↑↑3) So; insane orders of magnitude greater than 3↑↑↑↑3. This is by the way far greater than a googol or a googolplex.

The answer to that then gives you another ridiculous number. This then is the number of arrows to use in the next step, between another pair of threes. Repeat this process 64 times. This is Graham’s number.

This number is so large that (based on current knowledge) there is not room in the finite universe to store the number via digital data nor time in the universe’s lifetime to write it out. In other words there are more particles in the known universe than digits in this number.

Amazingly while most of the actual digits are unknown the last several are (…….195387); because any power of three’s final digits are very predictable.

Even getting to Graham’s number is mind-boggling. Of course it’s nothing compared to infinity. But that’s a topic for another entry. What’s your favorite gigantic number?