Let a1 be the number of hours for which the child whatched TV on the first day, a2 the number of hours on the first and second day, a3 the number of hours on the first, second and third day, and so on. The sequence of a1,a2,...,a49 is a strictly incresaing sequence since at least one hour the child spend on watching TV. Moreover, a1 >= 1, and since at most 11 hours are spent during one week, a49 <= 11 X 7 = 77. Hence we have:
The sequence a1+20, a2+20, ... , a49+20 is also a strictly increasing sequence:
Thus ecah of the 98 numbers
is an integer between 1 and 97. It follows that two of them are equal. Since no two of numbers a1, a2, .., a49 are equal abd no two of a1+20, a2+20, ... , a49+20 are equal, there must be an i and j such that ai = aj+20. Therefore on days j+1, j+2, ... , i, the child watched TV for exactly 20 hours.
[13]
From Ramsey theorem, we can conclude there must be at least on triangle whost all edges are colored by red or blue. Let us name the 6 points as A, B, C, D, E, F. Without lossing generality, we assume we find one red triangle ABC.
Now let's start from point D to find a second triangle. If edge DA, DB, DC are not of same color, by the same method as used in Ramsey theorem proof, we must find a triangle which is different from ABC and all of its edges are same colored(In this case, second triangle is found). In case edges DA, DB, DC are all red, triangle DAB are the second triangle whost edges are all of same color.
So in the other words, in case edges DA, DB, DC are not all blue, we can find a different triangle from ABC whose edges are all of same color. It is the same for point E, F.
In the case DA, DB, DC, EA, EB, EC, FA, FB, FC are all blue, suppose one of edges DE, EF, FD is blue. Without lossing generality, we assume DE is blue, then triangle DEA is the second triangle we want. If none of edges DE, EF, FD are blue, then triangle DEF is the second triangle we want. End.
[19]
(a) For equilateral traingel ABC, let point D, E, F be the midpoint of each edge. Then we divide the triangle ABC into four equilateral triangles according to figure 1.a. Then in each small triangle, the distance between any two inside points is at most 1/2. So if 5 points are chosen within the four triangles, then two of them must fall into same triangle, whose distance apart is at most 1/2.
(b)Let us divide equilateral traingel ABC to smaller equilateral triangles as shown in figure 1.b, in which each samll equilateral triangle's side length is 1/3. So any two points in same small triangle are apart from the other at most 1/3. If 10 points are chose from triangle ABC, then at least two of points fall into same small triangle, so their distance apart is at most 1/3.
(c) If the triangle ABC are divided into smaller equilateral triangles, in which each triangle's side length is 1/n, then there are n^2 triangles. So if n^2+1 points are chosen with an equilateral triangle of side length 1, there are at least points whose distance apart is at most 1/n. So m = n^2+1.
[22]
Let's name the left Ramsey number as R(k+1), the right one as R(k); So we need to prove R(k+1) <= (k+1) (R(k)-1) +2.
i) Let P = R(k+1). Suppose all edges between any two of P points are colored by k+1 colors. Select a random point q from these P points. Point q has P-1 connected neighbors. Then from pigeonhole pricinple, at least (P-1)/(k+1) edges are of same color. Without loss of generality, let's see (P-1)/(k+1) edges has the same color c0. In fact, (P-1)/(k+1) = R(k). So if one edge conecting two points of those R(k) points, we have a triangle whose edges are of same color c0. Otherwise, no edges among theose R(k) points has color c0. Then it means all edges between any two of R(k) points are colored by k colors. It is known that at least one triangle is of same color. So it is proven that if the edges connecting a set of P points are colored with k+1 colors, then at least one triangle is of same color.
ii)
R(k+1) <= (k+1)(R(k)-1) +2
<= (k+1) R(k) -k+1
<= (k+1) R(k)
<= (k+1)k R(k-1)
<= (k+1)k...4R(3) < 3(k+1)!
[27]
There are 2^n subsets of {1,2,..,n}. Let's group them into pairs. Each pair is one subset and its complement. It is a one-to-one mapping. So there are 2^(n-1) pairs. For each pair, there is no common elements in the two subsets. If a collectiong has more tham 2^(n-1) subsets, then at least one pair is in the collection. Then there is some subsets that do not have common elements. So there are at most 2^(n-1) subsets in the collection to gurantee each pair of subsets has at least one element in common.