All right, so what I have is basically a case of you can't believe your eyes. So,
you know in a lot of the stuff that we look at in maths
we look for patterns?
Right?
And we try it for a few examples and if we see a pattern then we think, oh maybe this continues on, right? But it's
not like a proof. It's not a convincing argument.
And so what I want to talk about today is an example of when the pattern just completely
misleads us. If you take, like, a random number,
is it more likely to have an odd number of prime factors or an even number of prime factors?
That's the basic question. Say 10 - the way that it breaks down into prime numbers is that it's the product of 2
which is prime, and 5 which is also prime. So it has two prime factors,
okay, so it has an even number of prime factors. But of course if I take any prime number,
like say 5 or 2 or whatever then it only has one prime factor,
and so it has an odd number of prime factors.
You're gonna see when I write it down in a minute that I'm gonna ignore, totally, the numbers that have repeated prime factors.
So I'm gonna define a function and I'm gonna use this notation, because all the mathematicians will yell at me if I don't!
[laughs] (Brady: What is that?)
So this is the letter mu, and what this function is gonna do is it's gonna take in a number, and it's gonna output
1 if n - my number - has an even number of prime factors.
So like 10, we talked about that example, because 2 & 5 are its prime factors, but also none repeated.
It's gonna give me the value minus 1 if all this stuff is still true;
except I have an odd number of prime factors.
Any prime number would be given the assignment of the value minus 1
under this function. We'll give it the value 0 if a prime is repeated when we factor the number.
Let's do some hands-on work here, because this is kind of a complicated definition.
(Brady: Does it get a zero if the prime - 'cause sometimes you get a factorisation where a certain factor is repeated?)
- Yeah - (Brady: And another one just appears once.) - Ah, yeah
(Brady: Like, you know, 2 times 2 times 2 times 5.) - Yeah. Yeah. - (Brady: That would count as a zero?)
So that would count as a zero - if any time we have a repeat then we assign zero to this thing.
Yeah, let's look at some, some concrete examples, because I think it's easy once you start working with it,
but hard to talk about abstractly.
Okay, so here's my numbers, and I'll look at the values of this function, mu. So let's forget about zero,
I don't want to even think about it. Let's start with 2.
Okay, that's a prime number. It has one prime factor, which is an odd number,
so we assign it a minus 1. 3, same deal. 4,
this is the bad news situation where we have a repeated prime factor, because it's 2 times 2.
So we assign it 0.
5 is prime, it gets a minus 1. 6 factors into 2 times 3. That's two; number of prime factors,
and so we assign it a 1. So let me just fill in a few more here.
8 takes zero,
9 is a square, it takes zero. 10 - we already talked about it has two prime factors. And you can keep going, right?
Alright, so if I want to know which happens more often
all I have to do is add up the values of this function.
'Cause it's negative when I have an odd number of prime factors,
it's positive when I have an even number of prime factors -
what I should do is just add up the minus 1s and the plus 1s and just see if I get something positive or negative
(Brady: And which side of the ledger are you on?) - Which side of the ledger am I on. And this will tell me what's going on.
What do I get when I add up these numbers?
So in the first initial computation,
there's a few more numbers with an odd number of prime factors than an even number of prime factors.
So I don't have an answer for you, [laughs]
about like what happens more often because, it's a little - you have to be a little careful to formulate what you mean.
Because you're sort of doing this trade-off going back and forth between
minus 1 and plus 1, and - you can always sort of find another prime number
so there's a minus 1. But then okay, you can multiply that 1 by 2,
so then there's a plus 1. So you keep going through all of the whole numbers
and you're gonna get lots of minus 1s and you're gonna get lots of 1s. So there's a lot of cancellation.
Ok? And so what I mean is that this sum should always be pretty small.
Minus 2 is smaller than 7, by a decent amount. You can imagine as you keep going on the numbers get really big
but the sum doesn't get so big. - (Brady: Is it gonna hover around zero, like?) - So that's the question.
So first of all, you can show that it's not gonna hover around zero. It's gonna zigzag like crazy actually.
But, the conjecture,
the thing that people thought was true for a long time,
was that it's never gonna be larger than the square root of the number that you've gone up to,
So if I do this and I go up through this process of adding together all these minus 1s and plus 1s, and say I
stop at a million;
then the thought is well, okay, whatever number I get, it's some number between minus
1,000 and positive 1,000; 'cause that's the square root of a million. - (Brady: But it could be either side of zero?)
It could be either side of zero. And in fact, it is known that it switches back and forth,
infinitely often. So, that's like a really complicated like thing to think about, yeah! - (Brady: That's amazing too!)
But - - (Brady: So it could get really high? It could get up into the thousands and then it could drop to the minus a thousand?
That's right, yeah,
and just zigzagging back and forth, sort of living around zero but in a really complicated way.
And so, the thought was that - okay, it's complicated but, because I have all this cancellation it can never be too big;
it has to be between minus the square root of the number, and the square root of the number.
So this is known as Mertens conjecture, and because this is called Mertens function - this sum that I'm adding up. And it was thought to
be true for a long time and it turns out that it's not true. But it's crazy because if you do this,
I mean we can now compute really far, and you go billions and billions and billions of
numbers out - and it still satisfies this square root bound. What is now known is that there is a
counter example to this conjecture. You eventually zigzag out past the square root of the number, but we also know that
we have some terrible terrible bounds on how big the number has to be to actually do this.
So it's something like 10 to the 10 to the 40. Something like that.
(Brady: So do we know the exact number where it first breaks the rule?)
Well,
I mean if we knew it we could never write it down, because we would need all of the atoms in the universe to write it down
[laughs] - (Brady: Aww yeah)
So no, the answer is no we don't know. Like we'd have no way to describe the first time that it happened
(Brady: But we know it exists?) - But we know it exists.
So, despite all computational evidence that could ever be produced in the history and future life of the universe;
we know that it still fails to be true. So
what we're looking at is a picture of this function, where we're adding up these minus 1s and plus 1s and going along with the
integers. And so, as you can see, what's happening
is that the - as we add more and more terms to the sum,
we go back and forth - here's zero - we go back and forth around zero - we start kind of more negative, by the way.
That's kind of fun to notice, because there's lots of prime numbers that are small.
So, we go back and forth around zero and, but, no matter how much back and forth we do
we're not getting anywhere close to this boundary given by the square root of the number.
So we zigzag back and forth positive and negative, positive and negative. But no matter how large we get
we don't even come close to approaching that boundary.
(Brady: You can't draw that graph for the point where it gets nearer the square root because numbers is too big?)
Yeah, that's right. The numbers are too big
I mean,
they just can't even be written down much less computed.
(Brady: But one day that that's zigzag, that Alps mountain range will break free of that square root loop)
It will, but I don't think we're gonna be around to see it.
[laughs]
Unfortunately. - (Brady: Does it *just* break away from that square root rule? Or does it blast past it?)
So that's a really complicated question. We don't know the answer to that we know it doesn't go too far past it
but actually the reason why people were initially - or one
interesting thing initially noticed about this question: is that if it had been true, if this false thing had been true, it would imply the
Riemann Hypothesis. Which is this really big, open question in mathematics that's thought to be like very difficult and very strong.
So understanding the growth of the, the size of this Mertens function is actually really, intimately connected to some of the most difficult
problems in mathematics. And I think there was at some point some hope of proving the Riemann hypothesis by proving this conjecture,
so when it was shown to be false
we lost that hope.
(Brady: I love that there's some number floating out there, when the rule first gets broken, but we could never even write it down.)
Yeah, that's right, I mean
we could never even arrange - I
keep going back to atoms in the universe, because that's what makes me sort of happy and blows my mind.
Right, we could never even arrange all of the atoms in the universe to describe the - like they could never be lined up in any
computable way. But we know it's there - (Brady: And it's like finite? It's an exact number?) - Yeah. Yeah,
it's - the only way we can describe it as the first number that fails Mertens' conjecture.
(Brady: Has that got a name? Is it, is it called like his - his number or something?)
No, that's a good point. I don't think it has a name. Maybe we should name it
(Brady: That's my new favourite number)
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[Preview of next video] And,
when I say that, I'm going to approximate pi in the Mandelbrot set,
I don't mean that I'm going to approximate pi on this picture here.
I mean that I'm going to cook up a bunch of numbers that have to do with the Mandelbrot set, that approximate pi.