# Practice English Speaking&Listening with: Perfect Number Proof - Numberphile

Normal
(0)
Difficulty: 0

Matt: Okay, here we go, we have a link between Mersenne primes and perfect numbers

Matt: We're going to, well, I'm going to prove that link. So, here's how we're going to do it

Once again we're going to have n, and we're going to have 2^n - 1

But, I'm not going to bother with the ones that don't work. I'm only going to have the ones that work.

Which is 2, 3, 5, 7, and 13 and they give us the Mersenne primes

Ahh... 2 gives us 3 ... 3 gives us 7 ... ... 5 gives us 31 ... 127 ...8191 ...

and then they match up with ... some perfect numbers.

Which I'm gonna put over here. As "Perfect Number"

And so, 3 matches up with 6 7 matches up with 28

31 matches up with 496 127 matches up with 8128

and then this one matches up with ... something. We will get to that in a second.

Brady: Can there only be one Mersenne prime, among the factors of a Perfect Number?

Matt: Yes! There's only...

There's only ever one Mersenne prime in the factors of an even perfect number,

and every even perfect number has one Mersenne prime in there, somewhere.

The reason I keep saying even, and I'm gonna stop saying even from now on,

is because no one has ever found an odd perfect number.

But no one has ever managed to prove that odd perfect numbers don't exist, so..

For all we know, there could be an odd perfect number, but no one has come across it.

We do know, if we find an odd Perfect Number, it's not going to have a Mersenne prime as a factor.

So from now on when I say Perfect Number I mean even perfect number by default.

So, if you look at the link I showed you that these numbers were all factors of these perfect numbers

In fact I'm going to write those in.

So if you have 3, you multiply it by 2, you get 6.

If you get 7, and you multiply it by 4, you get 28.

If you get 31, and multiply it by 16, you get 496.

127 multiply it by 64, you get 8128.

So now, you should be thinking, "what's the pattern here?"

Right, so let's say I give you a new Mersenne prime. Let's say it's hot off the press.

Wow, I've just found 8191 as a Mersenne prime, you know what, I want to know what its perfect number is.

Because that's where the fun is.

Forget Mersenne primes, in fact Mersenne primes were first studied by the ancient Greeks

solely to get to the perfect number goodness.

And so all we want to know is what do I have to multiply that by to get over there.

And so people searched for all sorts of patterns. People go "Ah! It's obvious because if you square 2 you get 4.

If you square 4 you get 16.

That's the pattern", ignoring the fact that 16 squared is 256.

And the people go, "Oh no wait! Multiply by 4, because 4 times 4 is 16.

And 16 by 4 is 64" and they ignore that fact that 2 by 4 is not 4. So people find other patterns.

The actual pattern is...

this equals

2 to the 1

That equals 2 squared.

That equals 2 to the four.

That equals 2 to the 6.

And it's these numbers here, the 1, the 2, the 4, and the 6, that's where the pattern is.

That is always 1 less than the number over here.

So 1 less than 2, that's our 1. 1 Less than 3 that's our 2. 1 less than 5 that's our 4.

1 less than 7 that's our 6.

So this column is 2 to the n-1.

Brady: I reckon if you'd given me 2 weeks I wouldn't have figured that out.

Matt: So I use this to torment young people. I get highschool students to try and work out that pattern.

And basically I walk around being a professional jerk.

Because I won't give them the answer. And they come up with a pattern and I go, "does it work?"

and they'll try the next one and it won't.

And eventually one kid will work it out and they will be the most smug child you have ever met and justifiably so.

And they will happily retire

convinced that if you have a Mersenne prime

2^n -1

and you then multiply

that by 2^(n-1)

you always always always get a perfect number.

And they will go home and I will say, "Well how do you know it works?"

How do we know that's always going to work?

We haven't proven that. And they will confuse trying examples with proving something.

And I love trying examples. I don't want to rain hate on the trying example parade.

Because that's how you find patterns to start with but you then you got to prove them.

So we're going to prove that this always works.

And we'll be very careful, by the way because 2^n -1 sounds a lot like 2^(n-1).

And this subtract 1 can drift if you're not careful when you're copying things.

But what's quite fun if you get someone to do this, listen to how they pronounce this because people pronounce

superscripts in maths with rising inflections

and so you get kids to read this out and they go, "well this is two to the n minus one".

"times two to the n minus one"

and that, you know, means superscript.

Brady: Is that why Australians ... Matt: And that's why Australians are very good at maths

So that's why I put this at the end, right?

So I end on the inflection. If I put this at the beginning it messes with my - I can't do,

"Two to the n minus one, two to the n minus one." it's just not as much fun. So we're going to prove this.

And I'm going to prove it by doing an example first, then I'm going to do it in a general sense.

So let's be lazy and do it for 496.

And so a little while ago I impressed you and possible two or three other people

by randomly listing off, as if I had memorized them,

all the factors of 496.

All I was doing was starting at 1 and doubling.

Double 1 to get 2, double 2 to get 4, double 4 to get 8, double 8 to get 16.

If I double 16 I get 32,

which is 1 above a Mersenne prime.

So in fact to get any perfect number you start at 1, you keep doubling

until you can choose any Mersenne prime of your choosing

you get 1 above your Mersenne prime and instead of writing plus 32

I'm going to put plus 31 instead. I'm going to switch to the Mersenne prime.

And then I'm going to continue the doubling. I double that I'm going to get 62

I double that I get 124

I double that I get 248.

So in fact, I wasn't, I hadn't memorized those numbers

I just started a doubling sequence and then jumped off it to a Mersenne prime

Matt: I had memorized the Mersenne primes and then I doubled the Mersenne primes to finish the sequence

And because I am ... 'cos I know where we're going I'm going to switch back to adding in the number itself.

which you remember I did before. I said you could

add the number that the total was twice the original, I want to put that back in the mix.

So how could we add these up?

We could have just added them up by summing them.

Let's get slightly more excited. Because if you look at this

this sequence here is actually that sequence times the Mersenne primes.

So I could rewrite this, this equals 1+2+4+8+16

+ Mersenne prime 31 outside 1+2+4+8+16

And that's actually why I put this back on so we get that same 16 in there.

I can then make that 1 + 31 because it's 1

times that sequences + 31 times that sequence.

That's 1+31 times 1+2+4+8+16

That now equals, 32.

And that will always bump it back up to the next power of two so that becomes 2 to the 5.

And 1+2+4+8+16, if you actually adds them all together that equals 31

And if you multiply those together you get twice 496, you get 992.

So that's an unusual way to add them together

right, but now we're going to do the same thing in general

for this equation.

What I'm going to do, is I'm going to take

2 to the (n-1) times 2^n - 1.

And I'm going to show you if you add all the factors of this, they sum to give you that formula.

Or rather if you add all the factors, not just the proper factors they sum to twice that formula.

So what are the factors of that?

Well what have we got at the front here?

We've got 2^(n-1)

So that means the factors are definitely going to be

1+2^1

+2^2 +2^3 + all the way up to 2^(n-1)

So that's going to be 2^(n-2) +2^(n-1), it's going to be a whole number.

So this is obviously a big power of 2; there are all its factors.

This is prime; it's not going to have any factors.

But it's going to be this multiplied by all of these.

So now we're going to have this, we're going to have 2^n - 1

+ 2^1 outside of 2^n - 1

+ 2^2 outside of 2^n -1

+ all the way up... blah, blah, blah, + 2^(n-2) outside of 2^n -1

plus 2^(n-1) outside of 2^n -1

which is our original number again at the end of the list just like we had over here.

It's equivalent to the very same number on the end.

So that's the actual formula has appeared again.

So what we really want to show is, all the other ones add to the same as that.

So we're going to show that by showing that all of them together equal twice that.

Okay, so we can start to tidy this up a little bit

so that's going to equal

this top sequence can stay the same.

There's still going to be 1+ 2^1+2^2 + all the way up +2^(n-2) +2^(n-1)

and this one here I'm going to put that out the front, so that's going to be

2^n -1, so it's that prime number.

Okay, now, we're going to, because we have one lot of this, and that lot of that

so that's going to equal 1+that.

but there we got minus 1 so that's just going to equal 2^n

outside of, so what I've done, I've done same as over here

which was the Mersenne prime +1 gave us back to 2^5.

I've done this +1, the 1s cancel out and we're left with 2^n.

And this is going to be 1+2^1+2^2+ all the way up 2^(n-2)+2^(n-1)

Now here's our problem: How on earth can we add together a sequence of numbers, a series here

when we don't know how many there are? We don't know where it stops.

Somehow we have to add together a sequence where we don't know the end,

but we can do that with a little bit of cleverness.

Where we just say

let's assume we know the total, let's just say there's some total

And it equals whatever this all of this happens to add to but we currently don't know what it is.

But we've given it a name which is always the first step in maths

And what would be twice that total?

What if we had two times that total?

Well that would equal double all of this.

So that would then become our 2.

+ double that would become 2^2+ double that would become 2^3

+ all the way up double that would become double that would become 2^(n-1)+ double that + 2^n

And actually, that is the same as that

that term is going to be the same as that term, all the way down, that is the same as that, that is the same as that.

So fact, all of these match apart from the beginning one here and the end one here.

So we want to cancel all these out

if I take all of these and subtract all of these

I would just be left with this one subtract that one. So 2T

minus T, this row minus that row equals 2^n-1

And in fact 2T-T is just T, the total is 2^n-1.

And we call this the geometric series.

So that this, or it's a particular case of the geometric series where you can add together

any long list of sequence of numbers where each one is the previous one multiplied by a constant amount.

In this case we're multiplying it by 2 each time.

It works for other ratios.

So I can swap that in for all this mess becomes that.

So now this, bring that down, equals

2^n outside of 2^n -1.

And you're like, "Well we're practiaclly there." The only thing we're missing is

we expect a minus 1 there and we haven't got it here. Where's that gone?

Well like I said this is twice the original number.

So we want to prove this equals twice what we started with.

So we go with that equals twice times what's left if you take out a 2.

Well that's 2^(n-1) outside of 2^n-1.

And that's it.

And so that is the proof that if you had a Mersenne prime 2^n-1

and you multiply it by 2^(n-1)

you'll guarantee to get a perfect number

because if you take all the factors of that expression

and then you add them all together, you get back to twice your original expression

which is our definition of the perfect number.

And so people talk about the fact that there's a link between Mersenne primes and perfect numbers

but now we have managed to prove it.

That's - I'm so pleased I'm going to put the lids back on to both of the pens.