Today I'm going to try and show that e is an irrational number. So e is this fantastic number, shall I just write it,
it's 2.718 and it goes on and on and on and on and on; and it plays a role in just about every aspect of physics, it seems to be everywhere in maths, some staggeringly important number. (Brady: What do you call it
amongst yourselves? Like,) (what- do you just call it e? Or do you
call it the-) I call it e, yeah, just friendly.
- (And if someone) (at the pubs came up to you and said,) (Hey Ed I know you mathematicians and) (physicists have this number called e,) (and you all really like it and it's) (important - what actually is e?) (What would you say to them?)
- Oh defining e is a- the way I like to define e is it's the only number, which if I differentiate it, I get back to itself again. So let me try and show you. So this is e, but if I raise e to some power x say, so x could be 1 it could be 2, so it could be e squared or it could be three e cubed;
then if I plot this as a function of x then it does something like the following. When x equals zero, e to the zero is 1, so it begins at 1. And then as I increase x this grows, so e is roughly roughly 3 so when x equals 1 this has got a value of roughly 3. When x is 2 e's got a value roughly, you know, it's e squared which is roughly 3 squared, right? Which is of order 9; so you can see it's rising up fairly rapidly. Now I could ask this- ask the question, what's the slope of this function, right? You know, if I- a slope we learn at school is drawing a tangent, at any point, will tell me the slope. So that would be the tangent at that point there, and this would be the tangent at that point there. And if I plot the tangent then I actually get - maybe tangent should be in red perhaps - I will basically get this. So that's the tangent. Imagine e to the x is telling me about the distance I've traveled as a function of time, x here is time, so these are maybe seconds; one second, two seconds, three seconds. And then I decide, actually it's not just the distance I need, it's the actual velocity I need. I need the rate of change of the distance, so that's the slope. Well the rate of change of the distance is exactly the same function. Okay, so then I say, well all right actually what I really need, I need the acceleration. I need to know how fast this velocity is changing; and if I do the same again that's that's the slope of the velocity, so it's the same curve again. So the velocity curve, which has to be- the acceleration curve, which has to be in black, is exactly the same again. So these are- this exponential function, e to the x, is the only one that if I differentiate it, so if I work out d by dx of e to the x then I get e to the x, so it's the same. And then if I work out the next differential of it I get e to the x, and it goes on and on and on and on; and that's the beauty of this this particular function, that all of its derivative- all of its slopes match the actual function itself. (Brady: And why is that useful?) I don't know why it's useful in particular. I don't know what it is about that property that means it plays such an important role. I mean, in terms of calculations
it's it's it's it makes things incredibly straightforward, because you can just do this; but in terms of its actual physical or mathematical properties I'm not sure what it is about that. It means it carries on forever, okay, because the slope just keeps- no matter how far you go, whereas other functions will die off if you differentiate them. And the fact- but the fact it's keeping exactly the same behaviour is is one of the most intriguing things. So what we want to do
today is to demonstrate that this number, which goes up goes on and on and on and on forever and ever and ever, cannot be represented as the ratio of two integers, that's what an irrational number is. A rational number can always be represented as the ratio of two integers-
- (A fraction?) A fraction. In fact they're made up of relative- in relative prime
integers but that's just the technicality, here we're going to demonstrate that e cannot be done that. It's a proof due to Fourier - absolutely brilliant mathematician. I think it's okay, I think we'll be able
to do it. (Okay, so you're going to prove that e is
irrational.) So it's- so let's do it - the proof e is irrational. And I wish it was my proof but it's not, it's due to Joseph Fourier. (I've been to his grave.)
- Have you? - (Yeah.) Does it mention this in it? He did so many things-
- (Hmm it's in Paris, it's) (in the famous cemetery in Paris.) (It's a cool grave,) (he's- he looks like Voldemort. There's) (like a bust of him on on it and he looks) (like Voldemort, just saying.) Spooky! So let's let's do it. So what what do I mean by that?
I mean that e cannot be represented as the- by the ratio of two integers, as a fraction as you say. So p and q are known as relative prime
integers. You know, no matter how big your integers are you will not be able to represent e. And the- so the proof is done
in the following way: you assume that this isn't the case, it's it's like James did with the square root of- showing the square root of 2 was irrational. And you assume that actually you can work it, and then you show there's an inconsistency. So you assume that e can be represented as the ratio of two integers. And the one thing that I've I've I'm hoping you will allow me to write down is, there is a representation of e that I can write down. So we can write
that e has an infinite series expansion, it's 1 plus 1 over 1 factorial, and over 2 factorial; so 1 over 3 factorial; and it goes on and on and on.
- (This is just common knowledge?) This is common knowledge,
you do this at school. And my daughter's done AS level, she's she's been doing this. There's this infinite series expansion of e where 3 factorial, just to remind everyone, is 1 times 2 times 3, that's all it is. So now what I want to do is; let me call this equation 1 all right? And I'll call this equation 2. So I'm going to multiply equation 2 by q factorial. So there's
this q here okay? And so what I'm going to do is I'm going to multiply this expression by q factorial. So the left hand side becomes q factorial times e, and what does the right hand side become? Well it becomes q factorial, plus q factorial over 1 factorial, plus q factorial over 2 factorial, plus q factorial over q factorial - because I've eventually reached the qth term - but then there's a whole series of other terms which- for now let's just call them the rest, right? There's- there'll be a q factorial over q plus 1 factorial, and a q factorial of a q plus
2 factorial - it just goes on forever. But we've just- we know from our assumption that e equals p over q, okay? And that then means that q factorial times e, this is equal to - what is that equal to? That's equal to q minus 1 factorial, times q, times e - because that's what q factorial is, it's q minus 1 factorial times q. But we all- we already know that q times e equals p, so this is just q minus 1 factorial times p. So this is an integer, that's the key point here. q minus 1 factorial is just an integer, p is just an integer, so the product of them is just an integer. So q factorial times e, the left hand side of this expression, is an integer, okay? Now let me look at the right hand side of this expression. Well hopefully you can be convinced that this thing that I've under- underlined here, this is also an integer; q factorial is just an integer, this is q factorial over 1, well that's just an integer, this is q factorial over 2 factor- factorial where the 2s cancel. So this here is also an integer. This is an integer, this is an integer; this then implies that the rest must be an integer, right? I can't have an integer here, an integer here, and have this not being an integer. Otherwise it would- imagine
this was a half, then I'd have an integer equals an integer plus a half and that's not right. So so the rest must be an
integer. Okay, so now I'm going to concentrate on the rest, right, so what is the rest? Well we also have that the rest, let me call it r, right, for the rest, what is it?
It's q factorial, because that's what I multiplied everything by, into- and it's all these other expressions. So it's 1 over q plus 1 factorial, plus 1 over q plus 2 factorial, plus - well let me do
the next one - q plus 3 factorial, and it goes on forever and ever like this, okay? I can now divide out, because, you know, this- imagine q was 6: this would be 6 factorial divided by 7 factorial, so it would be 1 over 7, that would be all that's left. So this is just equal to 1 over q plus 1, plus 1 over- and this is 1 over q plus 1 times 1 over q plus 2, and this is just you - get the
idea - 1 over q plus 1 times 1 over q plus 2 times 1 over q plus 3, plus dot dot, okay. And what I can do, I can simplify this. I can bring out a 1 over q plus 1, into the front and so this becomes 1 plus 1 over q plus 2, plus 1 over q plus two times q plus three, okay and it goes on forever and ever and ever. So the key point now that I want to make is that this- this is what the rest is, it's exactly this, but the the proof- the the key thing of the proof now comes in the fact that I know that this has to be less than the following: well I keep the 1 over q plus one, I keep the 1. Now 1 over q plus 2 is less than 1 over q plus 1 right? Imagine q was 6 again, then 1 over 8 is less than 1 over 7. Similarly here this is less than 1 over q plus 1 squared, right? Because this would be 1 over 8 times 1 over 9, which is clearly less than 1 over 7 squared. And it just goes on like that, right, that this is less than this. Now this number here is less than 1, because q is- q is going to be bigger than 1, this is less than 1. There is a- there's a beautiful little proof due to Gauss, and I'm going to prove it for you so that we can then just finish this off. Gauss apparently did this when he was about three right? All right so so there's the following: if x, some number, is less than 1 then we have the sum of n terms which is 1 plus x plus x squared plus x cubed plus..up to x to the n. You can see what's going to happen right? My x's are going to be my 1 over q plus 1s, because these will be the things that's less than 1, that's where- Now let me do the following: Gauss did this. x times that, right, is equal to - and then it becomes I multiply the 1, I get the x, I get the x squared, I get the x cubed, I'm going to get the term that's x to the n, and then if I multiply x to the n by x I get one last term which is x to the n plus 1. And I've shifted them all along so that you can see that if I take one of these away from the other all these terms cancel, leaving me with just the first term and the last term. So in particular it turns out that Sn into 1 minus x - so I've just taken the second term from the first term - equals, well, there's nothing to take away from the 1. They cancel, they cancel, they cancel; everything cancels, that cancels and I'm just left with that last term: 1 minus x to the n plus 1. Now the series we've got is an infinite series, this- this series ends after n terms. This series goes on to infinity. So what happens as n goes to infinity? Ss n goes to infinity, because x is less than 1, x to the n plus 1 goes to zero. Imagine a half. If I keep- a half times a half is a quarter, times a half is an eighth, times a half is a sixteenth - it just gradually goes towards zero. So the x to the n plus 1 goes to 0.
We're nearly there Brady, hang in there. So this means that S infinity is 1 over 1 minus x. So this sum I can now do, because x is- I'm just replacing x by 1 over q plus 1. This S infinity for me becomes 1 over 1 minus 1 over q plus 1, and that is equal to q plus 1 over q right? That's just rearranging
this. So if I come back to the all-important remainder, the rest, this thing here, that I'm trying to bound, remember I've said it - remember remember remember - I've said it
must be an integer. So the remainder, I've now- I know it has to be less than 1 over q plus 1 times by this sum, well this is this sum, so it's times q plus 1 over q. So the q plus 1s cancel. So this is r must be less than 1 over q, okay? Right we're nearly there now, because we already know that r has to be positive, because it's just the sum of a positive- of positive terms. So r is greater than zero, which means that this- for this proof to work r is bounded between zero and 1 upon q. But q is bigger than 1, so this is a fraction less than 1 - i.e. r is less than 1. Which means it cannot be an integer. Therefore e cannot be represented as a rational number, so it has to be irrational. Ta-da! [Laughter]
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Thatβs a really neat proof. The explanation of what e is and why we care about it was kind of terrible though.
I like this proof. Nice and elementary.
e that shall not be named
This is the first youtube video ever where looking at the comments was not a mistake:
Assume that e is rational.
Then it would be e-rational
Contradiction.
Edit: Joke isn't mine
This is the most "classic" numberphile video published from some time ago.
I shuddered when Ed said, 'e can be defined as the only number which, if I differentiate it, I get back to itself again'.
Dude, for one thing, if you differentiate a constant, you get zero. Also, if you differentiate the function Aex for any constant A, you get Aex, so the self-derivative property is not unique to ex.
Please, numberphile, don't get physicists to explain mathematical concepts.
Sorry i must have missed it but what does this have to do with Voldemort?
Im taking calc this semester and they jus showed us the proof useing tailors and remainder formula!