## viewpoint

### Large numbers

Emma Farnan | Thursday, December 7, 2017

What is the largest finite number that you can think of? Maybe some of you will know what a googol is, which is 10^100 so basically a one with 100 zeros after it. Some may even have heard of a googolplex which is 10^googol. Yes this is a huge number but it can be conceived of still, at least how many digits it has can be conceived. But what if I told you that even this is tiny?

So, basically there is a number called “Graham’s Number” that is so large, it is genuinely impossible to describe how many digits it even has. It once held the record for the largest number ever used in a mathematical proof, but I can explain how to build it.

First, we need a new notation, so we will use a ↑ as this notation. This basically will say how many times to raise something to a power. So 3↑3 is 3^3 = 27 and 3↑↑3 is 3↑(3↑3) so 3^27 which is over 7.6 trillion. Another way to say this is having a “tower” of threes that is three tall because it would be 3^3^3. Now 7.6 trillion might already be hard to imagine, but let’s do it again. 3↑↑↑3 would be 3↑↑(3↑↑3). So now we have three raised to the third power 7.6 trillion times. Now, our tower is 7.6 trillion threes tall (remember that a tower only three tall is that 7.6 trillion). At this point, it would be easy to say that we should stop because we are so beyond what people can think of realistically.

Unfortunately, we haven’t even remotely approached Graham’s Number — and by remotely, I don’t mean that we are halfway there. I mean that we haven’t even really started on our journey yet. At this point, we have pointed out how unimaginably much larger each arrow makes the number, remembering 3↑3 is 27 and 3↑↑3 is 7.6 trillion. So one more step, 3↑↑↑↑3? Huge! The height of this tower of threes has the equivalent to a tower of 7.6 trillion threes as the number of threes in it. And this is the real starting point of Graham’s Number, because we will define 3↑↑↑↑3 as equal to g_{1}.

Now, you see how we described it as g_{1}? What would g_{2} be? Interestingly, that would be 3↑↑↑ … ↑↑↑3 where there are g_{1} arrows. I am not saying there are g_{1} threes in the tower, I am saying that there are g_{1} arrows. Do this again, and g_{3} is 3↑↑↑ … ↑↑↑3 with g_{2} arrows. I cannot stress how unbelievably gargantuan these numbers are. It is impossible to even attempt to describe how many digits there are. The number of atoms in the observable universe is only something like (3↑↑3) raised to the sixth power, and we were far beyond this with simply 3↑↑↑3.

Back to Graham’s Number. Remember g_{3}? Continue that same process of creating a number higher of “g” until you get to g_{64}, and you have finally reached Graham’s Number. At this point, Graham’s Number seems like it may as well be infinite. When counting to infinity, we usually just go one, two, three, … ∞. But alas, Graham’s Number is farther from infinity than one is from Graham’s Number. By definition, even g with the subscript of Graham’s Number is smaller than infinity. So, next time you think about “infinity,” think about how large finite numbers can get and know infinity is honestly unimaginably further away than that.

*The views expressed in this column are those of the author and not necessarily those of The Observer.*