We were talking about dimension. So the last thing, so I reviewed the last few things we did. So one thing we have proven is that if x is a variety and u in x is open and stands, so that means non-empty, then the dimension of x is equal to that of u. So particular means that the dimension of variety is a birational invariance. And then if x is a variety and u is an open subset, it follows that it's irreducible. That's a relatively easy exercise, I mean, which, OK. OK. And so using this, we had shown that, say if x is a variety, x in a n is an affine variety and f is a polynomial, which does not vanish on x. Then if I take then every irreducible component of x intersected 0 set of f, its dimension precisely n minus 1, if the dimension of x is n, its dimension dnx minus 1. OK, so this was more or less the last thing we did. And now we wanted to talk about dimension of fibers, so we had this theorem. Well, I call it theorem. So let f from x to y be a morphism of varieties, morphism. Assume there's a non-empty open subset u in y, such that the fibers over all points in u have dimension n for all p in u. Then the dimension of x is equal to the dimension of y plus n. So if you have a morphism, the dimension of the source is that of the target plus the dimension of the fiber. And this also holds if the dimension of the fiber is only constant or an open subset, so a dense open subset. The rest can never be so large as to change the dimension of x. We want to, I had also stated, I think, a kind of what could be converse, I mean a stronger form of a converse. So that's the theorem on the dimension of fibers. So this says, so let f from x to y be a surjective morphism of varieties. And we take n to be the difference of dimension. We know that the dimension of x is here, so OK. So then we have first, for all points in y, the dimension of the fiber is bigger equal to the difference of dimensions. So it can never be that the dimension of the fiber is, so to speak, too small. So it can only, on an open, on some subset, it will be equal to n, but otherwise it will be bigger. But it will never be, it can never be that some fibers are smaller than most of them. And the second one is that there exists an open subset, and there exists an open, and therefore non-empty, and therefore dense subset u in y, such that the dimension is equal to n. And actually, I didn't state that before. In addition, we can assume that the fiber is irreducible. x and y are both irreducible, and the claim is that also the fiber has to be. But this, we will not prove. This is more difficult. This is something like a converse to this theorem, but a bit stronger. And it's considerably more difficult to prove than this. We will prove this, but not this, without proof. OK, maybe before we do have it. OK, so let's try to prove the first theorem. I mean, this one. So we know that the fiber will be a closed subset, so the statement makes sense. We can talk about the dimension, which will be the maximum of the dimensions of the irreducible components. So is this the wrong? OK. So we put v to be f to the minus 1 of u, which is an open subset of x. So we can now, we know that the dimension of u is equal to the dimension of y, because u is an open subset in y. At the dimension of x is equal to the dimension of v. So the statement does not change if we replace y by u and x by v. So we can assume that u is equal to y. So that means we can assume that for all the points, the dimension of the fiber's n. So replacing y by u and v x by f to the minus 1 of u, so by v, we can assume that the dimension of f to the minus 1 of p is equal to n for all points p in x. So this story that the assumption is only true for an open subset of y can be immediately removed. So we can as well assume that all fibers have this dimension. OK. So now we want to prove this somehow by induction over the dimension of y. So prove the statement. But maybe we have to first see. Prove the statement by induction over the dimension of y. So if y is a point, so if dimension is 0 is a point, there's nothing to show. Because then x is equal to f to the minus 1 of that point. It has dimension n. So that's kind of trivial. So the statement is trivial. It's actually. So we want to make induction over the dimension of y. So for this, we first make another simplification. We want to assume that both x and y are fine. Because then we can cut down by hyper planes to reduce the hyper surfaces to reduce the dimension. So replacing y by an open affine subset, which still has the same dimension as y. And x by an open affine subset of f to the minus 1 of this affine subset, we can assume that x and y are both fine. And if fine means, after all, that it's isomorphic to a close subset of some affine space. But dimension is invariant at the isomorphism. So we can, as well, assume that x and y are close subvarieties of some affine spaces. In fact, x in, I don't know how I wanted it, a l and y in a m close subvarieties. So we do this because now we can cut down with hyper surfaces and we know what happens. So given, so now f is now a morphism between subvarieties of affine space. So it's given by polynomials, as we have learned. So we can write f equal to, say, f1 to fm, as many as the dimension of the space in which y lies, where the fi are some polynomials in the variables on the space in which x lives. That was how morphisms of affine varieties were described. So now, let g be a polynomial here in kx1 to xm. Such that if I take the 0 set of this polynomial intersected with y, we want that this is neither empty nor the whole of y. So z of g does not lie in the ideal of y. We can certainly do that. And this shot, and we can also assume that this will not vanish on this. We can always find the polynomial which doesn't vanish at any given point, and so we find that. So now we know that this intersection has dimension 1 less than y, so we want to use that to prove our result. So let y prime be, say, equal of the 0 set of g intersected x, and we put x prime equal to the f to the minus 1 of y prime. Then we know, how do we get this? So this is just f to the minus 1 of the 0 set of g intersected with y. So that means we obtain it by putting f, it is just the 0 set of g composed with f, which means x prime is equal to, say, x intersected the 0 set of g of f1 to fm, the coordinates replaced by these polynomials. But this itself is, again, a polynomial in kx1. So this thing, so g of f1 to fm is a polynomial in kx1 to xl. So we take the intersection of x with the 0 set of such a polynomial. And this polynomial does not lie in the ideal of x, because otherwise, already here, we would have that this is equal to the whole of y. Because I mean, the 0 set is not equal to the whole of x, because that would mean that here the 0 set is equal to the whole of y. And so we know that by our previous theorem that the dimension of this thing is the dimension of y minus 1. The dimension of this thing is the dimension of x minus 1. So we find that the dimension, which we, how would we want to write it here? So the dimension of x is equal to the dimension of x prime plus 1 by this previous theorem, which is equal by induction. Now we can apply the induction hypothesis to the map. So we apply induction hypothesis. So for any irreducible component of x prime, so x tilde of x prime mapping to corresponding irreducible component y prime. So every irreducible component of x, so y tilde of y prime, every irreducible component of x, we map to some irreducible component of y, because they are irreducible. So it cannot map to more than one. So for any such, we know that the dimension of, so that by induction, the dimension of x tilde is equal to the dimension of y tilde plus n, because now the dimension of this thing is smaller, it's n minus 1. And so we have this. And so thus we can do this. This dimension of x prime plus 1, which is the dimension of y prime plus n plus 1. The dimension of y prime plus 1 is the dimension of y. And so this proves it. So this is relatively simple. We just, I mean, we make this number of reduction steps, but then afterwards we just have to cut down by a hypersurface, and then the thing is, and then we know it for the hypersurface, and by induction we get it. OK, so we give a number of simple applications, which it's nice to know that this is true. For instance, the dimension of the product is the sum of the dimensions of the factors, which by definition does not follow. If you look at these chains of things, there's no reason why. So the example dimension of x times y for varieties x and y is equal to dimension of x plus dimension of y. And for this, we just apply this result to the projection to one of the factors. So we take p from x times y to y, say, then the dimension of the fiber is the dimension of x, and the dimension of the target is the dimension of y, and then the theorem applies. Similarly, we can look at the affine cone of a projective variety. So let x in the n be a projective variety. So we can look at the affine cone, so cx. So you maybe remember this is the set of all points in the n plus 1 whose class in pn lies in x and plus the 0.0. And so then the claim is dimension of cx is equal dimension of x plus 1. Well, if I take cx without 0, this is certainly an open subset, an open-dense subset in x. At least I've assumed that x is not empty. It's an open-dense subset of cx. And we have a morphism from cx without 0 to x, namely just we take a point x0 to xn in a n plus 1, and we map it to its class x0 to xn in pn. And you can show, so it doesn't follow directly from the criteria for what we have, for what the morphism is. But you can also just prove it directly from the definition that this is straightforward to see that this is a continuous map, inverse image of something closed is closed. And you can also easily see if you pull back a regular function, then it becomes a regular function here. I mean, in particular, if instead of x you take here, say, pn and here an minus the origin, and then the general thing will be the restriction of that. What? Is this a homework? Oh, you got this as a homework. OK, I forgot that. But anyway, so then you certainly know how to do it. OK. And so you have a, so there's no worry about this. I also didn't attempt to prove it. OK. And so, and then obviously the fibers, all fibers. So what are the fibers? These are all the n tuples, say, a0 to an, which have the same class as 0 to an. But the equivalence class is by multiplying by non-zero constant. So the fibers are c with, you know, k without 0. The fibers are isomorphic to a1 without 0. No, I mean, you can check this. No, the fibers, if you just look at what the fibers are, you see an isomorphism with a1 without 0.  And this is an open subset. So the dimension of cx is dimension of x plus 1. So now we want to get some kind of, maybe that's the last result we want to prove in this form about dimension.  Namely, a formula, a statement about the dimension of closed sub varieties in an, of the dimension of intersections of sub varieties in an and pn. So how the dimension behaves when you intersect. And the statement is that the dimension will go down, well, I write it down, then I can say. So theorem, so what happened? So let x, so there's two parts, first the fine case. x and y in an closed sub varieties. Then every irreducible component of the intersection has dimension at most, at least, say, bigger equal to the dimension of x plus the dimension of y minus n. We'll kind of say something about that in a moment. And the second statement is the one for projective space. So let x and y in pn be closed sub varieties. So the first statement is the same as before. So every irreducible component, x intersected y has dimension bigger equal to the dimension of x plus the dimension of y minus n. But there's more, namely, if this number here is positive, is non-negative. So if dimension of x plus dimension of y is bigger equal to n, then the intersection is non-empty. Here, there's no prediction about the intersection being empty or not, because I just say something about every irreducible component. If the intersection is empty, there is no irreducible component. Therefore, the statement is completely void. And in fact, you can certainly have two hyperplanes in an, which are parallel, and the intersection is empty. So this can certainly happen regardless of how big the dimension of n is. But here, you have the same, first you have the same set about dimensions, but if there's not too much room for them, if the dimensions add up to at least the dimension of the ambient projective space, you get it's non-empty. OK, so first, this statement, I mean, it's not so complicated, but it's somehow slightly easier to remember if instead of talking about dimension, you talk about co-dimension. So if x in n, so this is a remark, has a dimension, say, n minus k, we also can say the dimension, the co-dimension of x in pn, but I just said co-dimension of x, is equal to k. Now, the definition of co-dimension is just the dimension of the ambient space minus the dimension of the thing. So then in this form, the statement in this form, the statement is the co-dimension of x intersected y is smaller equal to the co-dimension of x plus the co-dimension of y. So when you intersect things, the co-dimension is at most the sum. We had seen the special case for hyperplanes. If you intersect the hyper-surface, if you just intersect the sub-variety of the hyper-surface, the dimension drops at most by 1. So and this statement here, that the intersection is non-empty, is something which is very special for projective space. Where is it? Did I actually write this? So the fact that x intersected y is not empty if the co-dimension of x plus the co-dimension of y is small or equal to n. So it's the same as that dimension. So it's not so exciting anyway. So this means this is special for projective space. So we know it's false for a fine space because we can have parallel lines in A2 and we can have parallel hypersurfaces in NEAN. But even we, for instance, have as an example, we find that, for instance, we can use this to prove that p1 times p1 is not isomorphic to p2. So it's not. And this is just very simple. So if we take, so if they were, so I maybe write numbers up, if p1 times p1 is isomorphic to p2, then we would have that this statement holds. So then for any one-dimensional sub-variety, any sub-varieties say x and y in p1 times p1, we have that the intersection is non-empty. We get x intersected y is different from the empty set because this is true in p2 and this statement about intersection in isomorphism will preserve that. But this is obviously not true because if p and q are two different points in p1, then we can look at p times p1 and certainly and q times p1. And this is just the product. So p is different from q. So p times p1 is disjoint from q times p1. You have just two different fibers. This is empty. Obviously, these are isomorphic to p1, each of them. So have dimension 1 and this has dimension 2. So this is the case where this does not hold. So now let us try to prove the theorem. Well, it's also you can also do so if you take a fine space, you can take a point. You can do it like that. It's more than many more examples in the fine space where things don't intersect. But you can also make this. I mean, a fine space is an open subset of that and you can restrict to that. That is also OK. So now we come to the proof of the theorem. So it's not so clear how to do it. But we will use a trick. We know how to intersect with hypersurfaces. So we want to reduce to the case of intersection with hypersurfaces. That's not quite clear. It's not true that if x has co-dimension n, it can be taken as the intersection of n hypersurfaces. This is not true. But we can still reduce it to intersection of hypersurfaces by bringing in the diagonal. So let's see how that works. So use the diagonal. So the trick, so we do one trick, use intersection, use the diagonal. So reduce to intersection, actually with hypersurfaces, but even more especially with hyperplanes. So with linear hypersurfaces. And so how we use the diagonal, so we are in the first case. So let delta in an times an be the diagonal. We have seen that delta is isomorphic to an. We have delta, small delta, from an times an, the diagonal morphism, which maps an to delta. So it maps the point p to the point pp. It maps and it is an isomorphism onto its image. OK. And now we see, well, I expect the statement you can remember. It's not so complicated. So the remarkable thing that you want to use is if you take the inverse image under delta of, say, x times y, or if you want of, which is the same as the inverse image under delta of x times y intersected delta, so now with this delta, which is an isomorphism, this thing, I claim, is just x intersected y. So if you just think of it, this is the set of all points p in an such that it gets mapped to x times y under the diagonal map. So such that p comma p is an element in x times y. But that means that p is an x and p is also in y. OK, so this is just obviously the case. And so we see, so thus, intersected y is isomorphic to x times y intersected delta under the diagonal morphism. And this intersection takes place in A to N. So well, now I claim that delta can be written as an intersection of hyperplanes. So let's see how. So let's see. So let x1 to xn be the coordinates on the first factor, first an, and y1 to yn the coordinates on the second factor an. So how do I describe the diagonal? Well, I claim that then delta is a 0 set of x1 minus y1, x2 minus y2, and so on, until xn minus yn. Because if I have a point p equal to a1 to an, then the ai are just the value of this. And the statement that you have a point pp means precisely that the ith coordinate in the first thing is equal to the ith in the second. So that's just obviously the case. So what you see, this is the intersection of N hyperplanes, so N polynomials of degree 1. Now we had this theorem. So by a previous theorem, we have that every irreducible component of, say, a variety, whatever, c in an with a hypersurface, say, z of f in an has dimension in big n, at most one less than the dimension of c, bigger equal to the dimension of c minus 1. Namely, there are two possibilities. Either z is contained in the 0 set of f, in which case the dimension is equal to the dimension of z, or it's not contained in 0 set of f, then we know the dimension is one less. So if z, you have dimension of c is equal to the dimension of z set at cf. If z is not contained, big dimension of c, I don't know why I do it right in this way around, but anyway it's OK. So we can do induction. So if we now would intersect our z with n different hypersurfaces, the dimension would drop at most by n, because at each step it drops at most by 1. So if you apply this to our situation, so by induction, we have that dimension of x intersected y, which is equal to a dimension of x times y intersected delta, is bigger equal to the dimension of x times y, which is the dimension of x plus the dimension of y, minus the number of equations, which are n. And this is actually true not only for this, but for every irreducible component, because the statement says every irreducible component has that dimension. So you see that by induction, every irreducible component, say z of x intersected y, has dimension of z is equal to the dimension of the corresponding irreducible component here, has at least, OK, the notation is not so great. But anyway, so which is this, OK? Well, it doesn't always have to be that on the, I think it's obvious, even though it's true. OK. So now we come to part two. So we want to, now this argument to the diagonal doesn't work for projective space. You cannot say that it's like that. But instead you can reduce the state. So we want to prove this by reducing to the case of a fine space, which is a bit funny because the statement that you want to prove in the case of projective space is actually stronger. It's kind of, it's a bit amazing that you can prove something stronger to reduce to a case where that stronger statement is false. But we will see it can be done all the same. So reduce to 1 by using a fine cone. So we have if x and y in pn are closed sub varieties. And by definition, if I take this section of Cx with Cy, this is the same as the cone. Just look at the definition. This is obvious. So everything is compatible. We know that the dimension of Cx is equal to the dimension of x plus 1. Same for x, same for y, and x intersected y. So we have this very simple relation. And so therefore we can just say, so let Z be an irreducible component of x intersected y. And then C of x of Z is an irreducible component of C of x intersected y. And we know that this thing has the correct kind of co-dimension. We know, and the dimension of Z is by what we had seen as an example, equal to dimension of the cone over Z minus 1. And we had, on the other hand, seen that this is bigger equal to the dimension of the ambient of the dimension of Cx plus dimension of Cy minus the dimension of the ambient defined space, which is now an n plus 1. And this is the dimension of x plus 1, this dimension of y plus 1. And here however I forgot to subtract 1. The dimension of Z is equal to the dimension of Cx plus the dimension of Cy minus the dimension of the ambient space. And I have to subtract the 1. Otherwise it doesn't work out. So minus n, so minus, well I can say minus n plus 2. And then you see this is just equal to the dimension of x plus dimension of y minus n. So besides the state S k. So this is kind of trivial. Now for the other statement, we want to show that if this number is non-negative, then the intersection is non-empty, which is something which is not true in the affine case. So assume the dimension of x plus dimension of y is bigger equal n. We want to show that the intersection is non-empty. Well, it's not true such a thing if you take affine varieties, but it's always if you take any such cones, the intersection will always be non-empty because every cone contains a 0.0. So we know that Cx intersected Cy is not empty because the origin lies in Cx intersected Cy. So every cone, by definition, contains the origin. So we know that every irreducible component of Cx intersected Cy has dimension. What is it? Bigger equal to 1. No, the dimension of Cx plus dimension of Cy minus n plus 1, which is one more, the dimension of x plus dimension of y minus n. And so this is at least bigger equal to 1. So it follows that Cx. So we know from this story that the intersection is non-empty. So there is an irreducible component. So the statement that every irreducible component has dimension bigger equal to 1 is a non-empty statement. We know it has this dimension. So it's somehow. But the dimension is not just bigger equal to 0, which would mean that it contains 0.0, but it actually is bigger equal to 1. So it does not only consist of the 0.0. So thus Cx intersected Cy is different from the 0.0. And we know that, which, after all, is just C of x intersected y. So if the cone of something is not just the 0.0, then the something is not empty. So thus, x intersected y is different from the empty. So it's this trick with a cone that they always have to intersect. So the thing that in general was kind of not such a big deal that every irreducible component has to have a certain dimension does not necessarily mean that any such irreducible component exists. But if you are talking about cones, then it will have to exist. And so it always is a non-empty statement. OK, so this was as much as I wanted to say about this way, about dimension. But there is another way of looking at dimension, which is also about as commonly studied, used as the one that I used. I mean, a different definition of dimension, which is equivalent to the one we use. And I want to introduce this dimension and briefly explain this new other definition of dimension and briefly explain why it's equivalent to the one that we have used. But this will be very sketchy. In particular, we introduce some concepts using some theorems, which I do not prove. But anyway, it's maybe also a nice, brief repetition of some of the field theory that you had or something. But anyway, it's dimension and transcendence degree. So I mean, just as a motivation, so we know that the dimension of x is equal to the dimension of y if x and y are birational. We know that because we have seen that the dimension of a variety is equal to the dimension of any open subset. And from this it follows. So on the other hand, we know that x is birational to y if and only if the function field, the field of rational functions on x is isomorphic as a k-algebra to the function field of y. So therefore, it must be possible to express the dimension of x in terms of the function field. So thus, time x must be computable or determined by kx. And in fact, the answer is that it is equal to the transcendence degree of kx over k. We will see that the dimension of x is the transcendence degree of kx over k. Now, I don't expect that you know what that is, so I have to introduce it first. So maybe you know what it is, but I have to tell you what the transcendence degree is and then to show you that this will be the case. So I want to introduce this transcendence degree. So this is, I have to say a bit about field extensions. So if say k, let's let k over k be a field extension. So that just means that large k and small k are both fields and small k is a subring of the big k. And so if say a1 to an are some elements of k, we can look at the field, the extension of k generated by these elements. So denote k a1 to an with round brackets, the smallest subfield of k of the large k, which contains the small k and these elements. It's just what if you want its intersection over all these, but it's just can be obtained by taking, OK, we can look at this. So this is the field extension of k generated by these elements. And so if there are, finally, many such elements such that k is equal to this thing, we say that k over k is a finitely generated field extension. So if a1 to an k such that k is just equal to this, then we say the field extension is finitely generated. For instance, complex numbers over q is not finitely generated field extension. OK, so let's see. And now we want to come to the transcendence degree. We only do this for finitely generated field extension. So let's assume we have such a thing. So let k over k be a finitely generated field extension. So it means it's obtained from the smaller field by adding finitely many elements. And now we take some elements. So elements b1, b1 to bn, and k are called algebraically independent. Well, if you cannot get them, b1 to bn cannot be a 0 of a polynomial with coefficients, so over k, of course, more k, of a polynomial with coefficients in small k. So if there is no polynomial, say f in n variables, except for the 0 polynomial, because that always would work, such that if I take f and I replace the variables by these elements, you get 0. So there's no polynomial in several variables that vanishes on them. So this is the generalization of the concept of transcendent, because you say an element b in k would be called algebraic if there is a non-zero polynomial, such that f of b is equal to 0. And you call it transcendence if it's not algebraic. So you call an element b in k transcendent if there's no polynomial f in kx, such that f of b is equal to 0. And now this is the generalization for more elements, called them algebraically independent. And for the intuition, it's quite good to think of algebraically independent behaving very similar to linearly independent and vector spaces. I mean, so the formal properties are similar. Obviously, it's a different concept. But there are some, in particular, so let me, we will be able to talk about something which you call a transcendence basis. So just to say it again, if b in k is algebraically independent over k, if it's just one element, if and only if b is transcendent. So there's no polynomial which vanishes on it. So if b1 to bn are algebraically independent, transcendent. You know the word transcendent? Oh. Well, OK, so I'm not so sure about the spelling. No, I don't think so. I would write it like that. But it might, you know, you have to, you can check it. I would write a z. But it might be that I'm confused by the fact that I'm German. I cannot guarantee. Anyway, it's a, you know, there's no, at least there are no two different concepts where once it's with z and once with z. OK, so I will, OK, I'm willing to believe you. OK, OK. So if these elements b1 to bn, these are algebraically independent, then it's easy to see that if I take this field k, so algebraically independent over small k, if I take the field generated by them, that this is isomorphic to the field of rational functions in n variables. So whose elements would just be quotients of polynomials in the xi. I mean, basically, for some of the definition, it would be isomorphic to this. So, and then finally, a maximal set of algebraically independent elements over small k would be called the transcendent spaces. OK, so this is this definition. And now I get to do this theorem, which is basically a statement that the transcendent spaces will always exist, which is analogous to the statement that every vector space is a basis. And also that if you have a set of generators for the field extension, you can choose the transcendent spaces from the set of generators, which is also a fact that you have for field extensions, for vector spaces. If you have a set of elements which generate the vector space, then you can take a subset of it, which will be a basis. So as you see, now I have made some compromise. So this power, I will do this without proof. Although it's not particularly difficult, but it takes some effort. It's not what we are talking about. So let k over k be a finitely generated field extension. So then first, there is a transcendent spaces. So there exists a transcendent spaces of k over k. And I can choose it as a subset of the set of generating elements. It's supposed to be set to the end. OK, that's now the main thing. Then the second one is that every transcendent spaces has the same number of elements. So that every base, analogous to the state and world vector space, that every basis has the same number of vector space, which is called dimension of the vector space. So every transcendent spaces of k has the same number of elements. And this number is called the transcendence degree. As I said, the last one is that if you have a transcendence basis for field extension, the rest is a finite algebraic extension. So let b1 to br be a transcendence basis of k over k. Remember that k over k is a finite, finite degenerated field extension. Then if I take k over this thing, this is a finite algebraic extension. OK, so every other element will be algebraic over this. And so I mean, it's just that you have a finite degenerated algebraic extension, so it's a finite algebraic extension. OK, so this is this thing. I mean, I said I will not prove it, so I don't have to worry about the proof. But it's not very difficult. But as I said, it's not what we are concerned with. So now we want to use this. So we want to now start relating this to our thing of dimension. So we want to say that, as I said, the dimension of x is equal to the transcendence degree of the function field of x over k. So let's first see. So let now x be an algebraic variety over k. So then x contains an open affine subset, which I call v. And as the function field for x and open subset is the same, we have that kx is equal to kv. So we can assume that v is a closed sub-variety of some n by applying its isomorphic to closed sub-variety, so we can find isomorphism. So we want to describe kx. So kx is equal to kv, which is this. So let x1 to xn be the coordinates on a n, and say y1 to yn be the coordinate functions. So just the restrictions of the coordinates to v. Well, then we know that kx is equal to kv is equal to the quotient field of av. And so that's the, and av will be the extension of the ring k by the elements y1 to, so by this I mean that I'm not saying it's a polynomial ring generated by these, you just have k and you add to it all polynomials in y1 to yn. They don't have to be all different from each other, like we had also before. And so this will then be k y1 to yn. So kx, so we see that this is indeed a finitely generated field extension of k. So now let say the transcendence degree of kx over k be the transcendence degree of this thing. And now the claim is that, so what one can easily see, so if x is equal to an, or x is equal to pn, then we know that kx is equal to kx1 to xn. So the field of rational functions in these, the field of rational functions in these variables. And so the transcendence degree is precisely n. So which happens to be the same as the dimension of an or pn. And the claim is that this always holds. So we want to claim the following theorem. Let x be a variety, then the dimension of x is equal to the transcendence degree of kx over k. And in many books you find that also as a dimension, as a definition of the dimension. For some things, this definition of the dimension is easier for other things, our definition is easier. So anyway, we have chosen. Now we want to see, at least to sketch a proof, why this is true, why these two definitions of dimension are equal. So we will prove the next time. So we'll show another theorem. Every variety x is birational to a hypersurface in a fine space. So in, say, a to the dimension of x plus 1, OK? So it's actually somewhat surprising. So it shows that this equivalence relation by rationality actually identifies many things because obviously there's a big difference between hypersurfaces and varieties of higher core dimension. But every variety has an open dense subset, which is isomorphic to an open dense subset in a hypersurface in a n plus 1. So that's a surprising fact, but that's how it is. And now we will prove that the next time, or if you're very fast even today, but I don't think so. And now we use this to prove this theorem. So this is next time. So now comes the proof of theorem 1. So if this is the case, we can assume that our variety is a hypersurface. So by this previous theorem, we can assume x in a n, x is 0 for f in a n, is a hypersurface. So where f in k x 1 to x n is an irreducible polynomial, OK? So we know, therefore, that the dimension of x is equal to n minus 1. Because we know the dimension of hypersurface is 1 less than the dimension of the ambient space. So now we want to show that also the transcendence degree of the function field of x is n minus 1. So to show k x over k is equal to n minus 1. OK, so we take, just to check, let x 1, x n be the coordinates on a n and say y 1 to y n in a x, the coordinate functions. We have just seen that in that case, we can say that k x is the field generated by the field extension of small k generated by these coordinate functions. And now, first we want to see that these n elements cannot be algebraically independent, which then implies that this thing is smaller equal to n minus 1. And that's very simple. If we take f of y 1 to y n, this is just the class of f in the ideal of x in a x, which is, after all, k x 1, x n divided by f. So in other words, this is equal to 0. Because obviously, the class of f is 0. And so it does y 1 to y n are algebraically dependent. And so we know that the transcendence basis can be taken as a subset of these elements. And we cannot take all of them. So we can take at most n minus 1. So it follows that the transcendence degree of k x over k is smaller equal to n minus 1. But we want to prove it's equal to n minus 1, so we are not done. So we have to show the opposite inequality to show equality. We can assume that the last variable x n occurs in f. I mean, we have a polynomial in the variable x 1 to x n. It will depend on some. It might not necessarily have to depend on all the variables, but it will depend on some. It's not a constant polynomial. That then you wouldn't have a hyper surface. So we could, in the worst case, remember the coordinates in such a way that the last one actually occurs in the polynomial. So that's kind of trivial. And then we want to show that the previous ones, y 1 to y n minus 1 are algebraically independent. Want to show y 1 to y n minus 1 are algebraically independent. So then that means that the transcendence degree is at least n minus 1, and so they're equal. And this is quite simple. So assume they are algebraically independent. So otherwise, we would have a polynomial in these variables which vanishes on them. So there exists an element g in k x 1 to x n minus 1 with g of y 1, so a non-zero element, with g applied to these elements is equal to 0. But these are just the classes of the xi in ax. So this means, in other words, this just means that if you take the polynomial and divide by f, you get 0. I mean, you take the polynomial and take its class in the polynomials divided by f. In the quotient ring, you get 0. So in other words, we have that g is in the ideal generated by f. But this is not possible because if you take the ideal generated by f, you take f and multiply it by any polynomial. You know that f itself depends on x n. So every polynomial, if you multiply f by any non-zero polynomial, you get a polynomial which still depends on x n. So the only way how a polynomial which depends on less variables would lie in the ideal is if it's 0. So thus, we have a contradiction to the choice. And so we find that, therefore, in fact, these elements are algebraic, independent, n to the 10 sense degree is equal to n minus 1. So thus, OK, so maybe I will stop here. So next time, we will have to prove this other result, which actually is not so difficult, but looks in the moment rather surprising that every variety is vibrational to a hyper surface. OK, so that's it for today. We meet again on Wednesday, I think.


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So let's start lesson one y'all there The following says warm up the following sample probability questions were asked in junior high math. I need the chatter to stop Hey, we're starting that corner. Thank you. I Don't know. What did I say page 423? I think I said or 424 something like that Okay A fair die is rolled by the way English vocabulary to dice one die I'll often say one dice because it's kind of become an idiom that's part of our language But if you're reading a book and they talk about a die, that is the singular of dice We'll be grammatically correct in our typing or on the exams a fair die is rolled What's the probability of rolling a one now? Intuitively, you know the answer to this in fact basic probability We have a reasonably good intuition on but anything beyond really basic. We're lousy on Hence Vegas hence Horoscopes hence Gamma all the super all the patterns we try and find we're really lousy at it. So basics one out of six a Circle is divided into four equal sectors labeled clubs diamonds hearts and spades When it's spun, what's the probability it lands on hearts? one out of four two coins are thrown and The number of heads is counted What's the probability of getting two heads when you throw two coins? It's a bit of a trickier question. Let's find out It seems to me that you could get heads on the first heads on the second Tails on the first heads on the second tails on the first Tails on the second am I missing anything or me? I'm missing one heads on the first Tails on the second have I missed any possibilities? By the way, can you see the possibility for permutations and combinations to start showing up if I treat this a little later? Yes, that's why we always do combinations before we do probabilities Yeah, I think it's one out of four looks like and one of the first things We're gonna start to ask ourselves is there a way to get this without counting But a short rule we have in probability is this if you can count it you can calculate it when in doubt Write them out count them up Now that works great for small questions. It doesn't work great for picking five cards or anything like that too many possibilities Terminology probability theory deals with the mathematics of chance or Prediction we're going to use the following terminology a trial is any operation whose outcome cannot be predicted with certainty For example flipping a coin Rolling a dice Dropping a ball is not a trial because I know it's gonna hit the ground What are the odds of the ball hitting the ground? 100% long as there's no desk or something in the way But any event where there is some uncertainty about more than one outcome called a trial an Experiment consists of actually doing one or more trials We're gonna use those fairly interchangeably though an Outcome is the result of carrying out an experiment flip a coin get a heads the outcome is heads Roll three dice the outcome is a six a six and a four the sample space Often used with a capital letter s is the set of all possible outcomes Here is the sample space for flipping two coins there are no other outcomes beyond those four and event is A subset of the sample space for example this question was asking about the event two heads from this sample space heads an even number could be an event if you were rolling a dice and Here's our first official definition of probability This one's pretty intuitive. I'll you know highlight it But I think most of you naturally understood this when you guys gave me the one out of six answer If an experiment has a set of equally likely possible outcomes Then the probability of a particular event a is given by the formula We're going to use some notation gen lots of shorthand instead of writing the word probability You know what letter we're going to use a capital letter what? P and then we're going to do an open bracket and the event is going to go in the brackets in this case the event a and It's the number of outcomes favorable to a Divided by the total number of possible outcomes probability of heads one of a two Probability of getting a five when you roll a die one out of six. So this is going to be our notation Pretty good pretty short What's the probability of something being impossible if something's impossible? What's its probability? if it's impossible we would say P of x is zero for example, what's the probability of rolling a nine on a six-sided standard die? You get a nine zero Okay, that doesn't sound very profound. It's more profound than you think oh and if something is guaranteed to occur We say that it has a probability of one 100% if you want to think about it as a percentage for example, what's the probability of rolling a natural number less than seven? What are the natural numbers less than seven six five four three two and one? What's the probability of rolling a one or two or a three or a four or five or a six on a die? Guaranteed what's the probability of rolling a heads or tails? guaranteed Okay So guaranteed mathematically one that gives us our first built-in error check gives us our first built-in error check Rio for any event a The odds of it occurring have to be between zero and one if you get an answer bigger than one you've messed up You've done something wrong on your calculator. You've forgotten brackets. You've done something if You get a negative answer you messed up All probabilities are either impossible or guaranteed or anything in between Often we're interested in the odds of something not occurring And we have an abbreviation or a shorthand for the word not if we draw a Horizontal bar above something we read that as not a when I write this this is actually the probability of not a occurring and We call it The complement of a oh and there's a handy dandy little equation The probability of not a or not a is equal to one minus the probability of a That's on your formula sheet, but it's worth having in your brain Let me give you an example Supposing they tell you the probability that it's gonna rain tomorrow is one out of three What's the probability that it's not gonna rain tomorrow then? Two out of three That's gonna become handy because Dylan you're gonna find for a lot of questions It's way easier to find the complement than it is to actually find the answer to the question that you're looking for But then just go one minus the complement. You're done. We're gonna use that all the time So this formula here is on your formula sheet But Jen I'm gonna say you'll memorize it naturally because it's kind of intuitive for each of the examples It says state the span sample space Which outcomes are favorable? Whether the outcomes are equally likely and the probability of the event So a fair die is rolled. What's the sample space for a fair die and? Traditionally we use the squiggly brackets because it's a set what are the outcomes for a fair die one comma two comma three comma four comma five comma six Now the event it asks is what's the probability of rolling a one? Okay, I think that's and I'm just gonna circle it that event right there What is the probability of rolling a one we all know this one? If it's a fair die, what is the probability? 106 oh Are all these outcomes? I should have answered part three are all these outcomes equally likely is it this? Yeah, that's gonna be a key right now right now. We can't handle Non-equally likely outcomes. We will in a couple of days. We'll add a clever trick a Circular spinner is divided into four equal second sectors. We did this one already We said the answer was one out of four Two coins are thrown and the number of heads is counted. What's the probability of obtaining two heads? We did this one already as well. We listed our sample space. We'll do this one over here though We said the sample space was Now it depends how you list the sample space if you list the sample space as heads heads tails tails heads tails You can get two heads one head or no heads Are all of these equally likely the answer is no Because you can get that two different ways And so we would either have to list it the way that I did earlier by calling it heads one heads two tails One tails two heads one tails two tails one heads two and then they're all equally likely or Preview of coming attractions So not equally Likely in fact if I was doing this question. I wouldn't list my sample space this way I would list my sample space as The way I did before heads one heads two heads one tails two tails one heads two and What have I missed? Tails one Tails two and there I think those are all equally likely Because each of them will only occur once and then I could get the answer of One out of four not one out of three probability of heads heads Is one out of four So in part B says why can't you use the number of outcomes divided by the total number of Outcomes you can't for the blue set here because they're not equally likely next page compound events Events formed from repeated trials or from a combination of simple events are called compound events and Often a table a chart or a tree diagram is Useful in determining the sample space the two we're going to use the most are a tree diagram and a chart In fact, we'll use a tree diagram Nick 90% of the time We'll use a chart a small amount of the time and you'll figure out pretty quickly when we use want when we use what Says this consider an experiment of rolling three equally spaced So I've rolling an angle try that again consider an experiment of rolling an equally spaced Triangular spinner number one to three. There's a three somewhere and tossing two coins Says draw a tree diagram to show all of the outcomes So the first thing we can do is we can roll the spinner. There is Three possible outcomes a one or a two or a three then You can flip two coins What are all the possible outcomes when you flip two coins? heads heads heads tails Tails tails heads heads tails tails tails heads heads tails tails tails Now this is using the non equally likely outcomes approach. I don't like that I want instead of three options here. We said technically in the sample space How many sample how many elements are there in the sample space? For do you know your rights? I'm gonna go like this heads one heads two one two one two One two one two one two one two one two And then I'm gonna add one more branch to each one of these if I can. I'm gonna say, okay Here is which one am I missing? tails one heads two Tails one heads two Tails one heads two Tails one heads two Now why is that so handy? Oh, it says list the sample space at the end of the tree I don't want to list them all Instead, I can show them to you Jan one sample space is one tails heads Another one is one heads heads another outcome is one heads tails Another outcome is one tails tails You can walk down any branch and if you walk down every single branch systematically You'll have listed the sample space, but they're contained in the tree I like that better because it's way less writing Dina, can you see one possible outcome is a three tails tails as well as a three tails heads a three heads tails and a three heads heads They're all there How many outcomes are there count the number of branches? Brendan is there a way that we could have used the fundamental counting principle to figure that out? Well, it would have been yeah three numbers times four outcomes the tricky part would have been ahead of time spotting There was four outcomes. You probably had to make a list Are all the outcomes equally likely the way I've drawn it here? Yes So now I can answer these questions for part D the probability of getting a three comma H H is How many outcomes were there grand total? 12 How many outcomes are favorable? How many branches end in a go with a three heads heads? Just one what about the probability of getting a prime number and Exactly one tail Which of these numbers are prime One is not prime by definition primes are any number greater than one There's two prime Yeah, two and three are both prime numbers because they only go into the only divisors They have are themselves and one one is not considered prime. So Here here and exactly one tail I find what helps me a lot if I've done a tree is to simply change colors or not and put checkmarks under Satisfactory outcomes. Here's a two and exactly one tail. Here's a two and exactly one tail Here's a prime and exactly one tail. Here's a prime and exactly one tail by doing that It's very easy for me to spot the number of favorable outcomes Oh and by writing them below the tree Madison if I need to use this tree again I just pull up my eraser erase those checkmarks and start over or if it's pen scribble it out and start over But I would think well, how many favorable outcomes are there? four out of I know that's one third Doug we're not going to reduce probabilities Because 99% of the time you're gonna want to common denominator anyways now They will reduce them in their final answer. So I will trust that you still remember your fraction button math enter enter Now this one, you know is one third but for the more complicated ones like the card questions that are over a question in the thousands If they want me to reduce the fraction that's gonna I'm just gonna go like this math enter enter and that's gonna write it as a lowest term tree When will I use a chart? Any time I'm dealing with two dice You know why how many outcomes are there using the fundamental accounting principle with two dice How many branches would my tree have if I use a tree then? 36 no Trees are for smaller samples, but they're better in a lot of ways I use a chart for a dice and in fact what I would do here We have a red die and a blue die. They've started this chart for me But often I'll just freehand it it would I would write down red blue I would write numbers one through six one through six and then I would just add a line like that and like that This is one one one two One three fill in one four one five one six It takes about ten seconds to do a dice chart, but it's so handy. It's well worth the while two one two two two three two four two five two six three one three two three three Five three four is already there 36 says show all the possible outcomes in the array done How many points are in the sample space? What did we say by using the fundamental accounting principle? 36 says List the event the same number appears on both dice as a subset of The sample space by the way, what do we call it when we're playing a game and we get the same number? You roll it in Monopoly Huh doubles do we not still call it doubles? No, that's only for two ones. That's only for a subset of the subset I agree with you Brandon that that's one possible outcome, but if we're talking the same number that's not snake eyes That's a Brandon Brett caught myself now There's a list of all the doubles. Of course. I ran out of room Okay By the way, if they don't give you this chart and I will ask you a dice question on your test. I Just go like this Freehand one two three four five six one two three four five six One one one two one three you can whip you can write it very fast so What's the probability? That the same number appears on both die or as we like to say when we play Monopoly Brett, what's the probability of doubles? How many outcomes are there grand total? 36 how many favorable outcomes are there? Six and I know that's one out of six and lowest terms who cares What's the probability that a different number appears on each die or a shorter way to write that is what's the probability of? not doubles Don't count use the compliment If six out of 36 is the odds of getting doubles, what are the odds of not getting doubles? 30 out of 36 because they are opposites to each other. There's no overlaps. It's the compliment That's where we use the compliment. It would be way easier if they said what are the odds of not getting doubles I would find the odds of getting doubles because I know there's only six of them count that and then subtract Part of your homework today, but we're gonna move on to lesson two as well number one three four So I think I skipped two. Yes So one skip two three four skip five six is nice eight and 11 Having said that let's go to lesson two Here is where the fundamental meat and potatoes kind of begin here more terminology and we're gonna talk about We're gonna talk about the events a or B and the events a and B We got a bit of a problem here because the word or in math means a little bit different than it does in English so in Mathematics the event a or B is Said to occur if a occurs or B occurs or both occur and That's the difference between math and the English language if I say to Doug Doug I have two chocolate bars. You can have a Snickers bar or a Mars bar. We understand in English Both is not an option in math. It is so in math or means one or the other or both and What oh the event a and B occurs if both events occur simultaneously at the same time And one of the better ways one of the easy ways to visualize this is through something called a Venn diagram So it says this Consider the experiment of noting it of rolling a die and noting the result Let the event a be an even number is thrown Let the event be be an odd number is thrown put the outcomes into a Venn diagram So our outcomes for rolling a die are the numbers one through six which ones would end up in event a If event a is an even number is thrown which ones are here two four six which ones are here one Three five Combine those two together and that is the event rolling a die So it says list the outcomes for the event a can you see them from the Venn diagram? two or Six the event be one Three or five What about the event a or B? So that means one or the other or both? Any numbers that appear in one or the other or both? I think Madison That's all of them now the event a and B means which ones are happening at the same time Are there any numbers that are odd and even at the same time? Empty set zero with a line through it. Don't just put a zero because zero is a number zero with a line through it, okay? C says let N of a I think we're gonna skip straight to D Looking at this Venn diagram. What's the probability of event a I need something out of something? three out of Six I know it's one of two. We're not gonna reduce What's the probability of event B? Also three out of six. What's the probability of events? Oh a or B? Six out of six which is one Right or means one or the other or both odd or even or both there is no both in this case But we're gonna talk about what that means. Oh, what's the probability of a and B? zero Out of six zero Here's what I want you to notice. So here's some more terminology lots of terminology today This one we're gonna put a bit of a star around because this one they will ask you on the other ones You'll learn them naturally this one kids get mixed up.  Yes Okay And you can easily tell an event diagram if they're mutually exclusive because there's no physical overlap Says verify the following rule the probability of a or B equals the probability of a plus the probability of B I think I said to you guys last unit or means add. It does work next page Once again, we're going to roll a die and note the result The event a is an even number is thrown The event B is a multiple of three is thrown Okay Where is what's going to go in a? to a for Can I put the six right there? Why can't I put the six right there? It's also a multiple of three. I need to put it right there because it's both What else is a multiple of three? That's going to go right there because it's not even but it is a multiple of three. What number am I missing? What numbers am I missing? They go Outside there's my whole set so it says list the outcomes for the event a to for or six List the outcomes for the event B to Sorry B. Mr. Dewick look at the right diagram three and six The event a or B two three Four or six one or the other or both you don't write the six twice though even though it appears in both What's the event a and B? six Once again, we're going to skip C. We're going to see if we can go right down to D. So What's the probability of a occurring? three out of six What's the probability of B occurring? two out of six What's the probability of a or B occurring? four out of six What's the probability of a and B occurring? Here's the question Can you use these numbers to get an equation and the shorter answer is yeah, if you want to find the answer to or It's the first one plus the second one minus The overlap See it five minus one Four and that works from the previous page as well if I wanted to find this answer here or It's the first one plus the second one minus the overlap But what was the overlap in this earlier example because they're a mutually exclusive The overlap was zero and this is going to give us our first official probability formula equation So because these have common outcomes, they are not mutually exclusive and This gives us our or equation right here the probability of a or B is the probability of a Plus the probability of B minus the overlap oh and if the overlap is zero They're mutually exclusive and You can use that because the overlap is zero But I really don't do a big song and dance about this one because Jen It's a special sub case of the general equation right there Jen how many terms are there in this equation count them? For that means if you know three you know one if you know three and over the fourth So usually they want you to find that sometimes they'll give you this and this and say find the overlap or Sometimes they'll give you that and that and or that that and that and say find the probability of be any any three gives me four so Here are our key ideas so far Events are said to be mutually exclusive if no overlap in the Venn diagram and Or means add Except we have to subtract any overlap because we'll have counted it twice. I'm gonna put a box around this one Remembering that the upper equation is a special sub case In other words if this is true, then you know, they're mutually exclusive Lots of terminology, but hopefully Doug you're seeing the math so far not too tough. We're counting and I don't even think we need to pull off our shoes yet. We've been counting up to ten at the most I think oh no, we did have a 36 didn't my bad. So example one State whether turn the page if you haven't already state whether the events a and b are mutually exclusive or not a Lot of this you can do with straight common sense what you're asking is is it possible for something to be both of these at the same time? Can you be a face card and a club? What's a face card? Jack Queen or King is it possible to be a club and the Jack Queen King? So these are Not mutually exclusive What about this next one? To dice or thrown is our event our experiment Event a is the dice both show the same value Brett what we call doubles Event B is the total score is 11 Can you have a score of 11 and be showing doubles? example to Which of these events are mutually exclusive? Which of these events can't happen at the same time? Adam Event what an event what? Okay event a and c you can't be a face card and you can't be a jack-queen king and an ace at the same time You're one or the other B and D. Can you be a club and a red card any others? I don't think so So this is Megan the common sense approach to mutual exclusivity and that's if you know enough about the event that you just know There's also a mathematical approach where you don't know what the event is. You just know the probabilities First of all from a Venn diagram here says which ones are mutually exclusive a is mutually exclusive from everything B and D are mutually exclusive Shit That like a and c but a and b and a and a is from everything what I want to do is go to example for here's a mathematical one It says use the following information to determine whether the events a and b are mutually exclusive And they've given me the probability of a the probability of B and or I'm gonna write down our equation We said this the probability of a or B is equal to the probability of a Plus the probability of B minus The overlap a and b this is on your formula sheet You don't need to memorize it. Although many of you will because you're lazy When will this be mutually exclusive? This will be mutually exclusive if the and term ends up working out to what? Zero so let's plug in everything else What's a or B? Seven over twelve one over four Plus one over three is that zero I'm not gonna go fancy.  Does work out to zero? Therefore P of a and b equals zero so mutually Exclusive no overlap and you'll notice I was able to do that deal even though I have no idea what events those are No clue Could be betting on something. I don't know could be the odds of your car starting I don't know a Venn diagram is very very useful. In fact, I would say this anytime they give me two outcomes and Percentages I go Venn diagram Example five a grade nine class was surveyed to find out whether they did math homework or English homework last night so Looking at this and assuming we're not talking about this class What percent did math homework and don't say 47 because 47 is wrong? What percent did math homework not 47? Evan what or said what the entire circle with math in it adds to what? Do the math please I'm wanting more specific than 63% what percent did math and English homework? See the and What's the and 16? What percent did math or English homework? now From the formula that means One plus the other minus the overlap, but from the quick Venn diagram I think it's that plus that plus that it's anything that appears in Those circles, so I'm just quickly going to go 47 plus 16 plus 25. What's the correct answer? 88 can you tell me another way I could have gotten the answer without adding 100% 1 as a decimal minus 12 percent point one two as a decimal Now if I had used the formula by the way to get or probability of write this down, please M or E it would be probability of E of M Plus the probability of E minus the overlap and it would have been Right does that match this equation up here? Yep, except instead of using A and B. I'm using M and E So M or E first one second one overlap and What did we say the first one was? 63 plus What percent did English homework not 25? 41 Minus the overlap does that also give you 88 check me does it also give you 88? Yes. No. Yes Yeah, which way is fastest compliment? Which way second fastest? But for what it's worth the formula would also have gotten me there if they'd given me that Two more with them a single card is drawn from a standard deck of 52 cards Use formulas to determine that the probability that a Nine of diamonds or a heart is drawn Do you know what word is that right there in the middle? So at the top of the page, let's write down the probability of a or B is equal to the probability of a Plus the probability of B Minus the overlap There's our generic template Instead of a Amy can I use? 9d for nine diamonds and Instead of B what would be a clever letter to use? H for heart Okay, so this is wanting to find the probability of 9d or H and that's going to be the probability of The first one plus the probability of the second one minus the probability of Both how many nine of diamonds are there in the deck? one out of How many cards are there in the deck? 52 52 52 Plus How many hearts are there in the deck? 13 divided by how many cards are there in the deck? 52 Minus How many cards are nine of diamonds and hearts at the same time? I think that's a trick question I think they're mutually exclusive Zero and This is why I said don't reduce fractions because technically this is one quarter, but if I had reduced this Don't they have a common denominator already built in if I don't reduce In fact, what is 1 over 52 plus 13 over 52 in your head with no calculator? 14 out of 52 good enough now let's compare that with B Instead of a nine of diamonds or a heart Just a nine or a heart Okay, that's going to be the probability of nine or H is going to be the probability of nine Plus the probability of H minus the probability of nine and H Right. I'm just filling in that template formula. I rarely use a and B I usually try and pick letters that makes or symbols that make sense to me. All right Evan What's the probability of a nine? four out of 52 plus 13 out of 52 minus How many nines are hearts at the same time one out of 52 Four plus 13 minus one 17 take away one 16 out of 52 So here's a question If you had to bet which of these two would be a better bet Yeah, B. They're both lousy by the way because they're not above 50 percent. They're both terrible bets So I guess I should really say instead of asking which of these would be the better bet Which of these would be the less terrible bet B? The odds are slightly better barely but slightly and I don't think you would see that intuitively just by glancing at it Example seven two hundred people with neurology symptoms Which includes headaches and backaches back aches participate in a study to evaluate a pain relief medicine 60 people experienced headache relief 126 experienced backache relief 36 experienced both How many categories are there in this question? There's two headache and backache This is where a Venn diagram not a formula shines I used a formula with the previous one because in cards there's all sorts of catty black red face Forget it here. What I'm going to do is I'm going to draw a Venn diagram like this with some overlap and whenever I'm doing a Venn diagram I want to try and find the overlap first if at all possible How many people are in the overlap category? How many people both? 36 I'll put a little 36 right there. How many people got relief from headaches? 60 now a lot. Oh, I'm going to call this one headaches and I'll call this one back aches in my high-tech numbering system And a lot of people Doug want to put a 60 right there. That's wrong Why is that wrong? How many people do you have in your headaches by putting a 60 right there? Not 60 How many do you have? 96 so I don't put a 60 right there What number is going to go right there? Gotta be 24 Can't be 60. Oh Kelvin what do those two together add up to? 60 there's your 60 that got headache relief 36 got both only 24 it got their headache, but it didn't get their back ache. Oh 126 people got backache relief. I'm not gonna put this is wrong.  Do the math? How will you calculate that? 200 minus 24 minus 36 minus 90 how many people did this drug not work for and This is some of the basic math behind clinical drug trials and things like that How many? anyone 50 5 0 or 1 5 5 0 Okay Now I've got my bent diagram set up now. I can answer the questions What's the probability that you get at least one of the two symptoms now at least one of means? headache or Backache because remember we said Or means one or the other or both now we could do this with a formula But isn't it gonna be? 26 plus 36 plus 90 or Better yet, isn't it gonna be 200 minus 50? I? think the probability That someone got relief from at least one symptom is that 75% I heard somebody say yeah, which is for a medication pretty good What are the odds that it didn't work for you now? How would I write this? I would write this as Not a or B. That's the abbreviation for neither and it's 50 out of 200 so we've thrown a lot at you terminology and Then diagrams and the or equation. I think you folks are getting the well Okay, I guess number one. It's gonna skip it, but no for skip five C So five a and b six seven eight nine ten Any election? Yeah, I'm gonna go 11 12 There's not gonna be as much homework next class. It's a lot I threw at you. I know but we're trying to Get the sprint going for the homestretch and you do have about Almost 15 12 well, yeah 15 almost minutes to work


For me, since I'm slightly from an adjacent field, I think every single point my panelists made were a takeaway. But the good thing to note is performance marketing, which was our panel on, is alive and kicking and it's reassuring to note that all the marketing dollars are getting accounted for. I think what mobile marketers should keep in trend is most of the penetration is now rural, digital, tier 2, tier 3 India and in regional languages and we need to respect that, create the content in that and speak and a lot of video waste content, right? So those are the trends. Every marketer is on it, but it's going to go far higher. Even in BFSI, when you add video or animate the existing pictures, right, you get significant, just animating the pictures, it's not even real video, you get significantly higher hits on that ad. So I would suggest that your video is the other thing, apart from regional languages.