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Unformatted text preview: Srping 2007 Math 510 HW2 Solution Section 2.4. 16. ( 3 2 ) Prove that in a group of n > 1 people there are two who have the same number of acquaintances in the group. (it is assumed that no is acquainted with himself or herself.) Proof. Let a 1 , a 2 , . . . , a n be the number of people acquainted with the first, second, ... , and the last person respectively. Then 0 ≤ a i ≤ n- 1 for all n . If these n numbers are all distinct, then they have to take each and all the values 0 , 1 , 2 , . . . , n- 1 by Pigeonhole principle. Thus a i = 0 and a j = n- 1 for some i and j . Since n > 1, then i 6 = j . But this is impossible since the j-th person is acquainted with all other n- 1 people in the group (everyone except himself), in particular the i-th person. But the i-th person is acquainted with no one, in particular the j-th person. Since we assumed that acquaintance is a mutual acquaintance. This is a contradiction and is impossible to both numbers 0 and n- 1 be taken as the values of a 1 , . . . , a n . 18. Prove that of any five points chosen within a square of side length 2, there are two whose distance apart is at most √ 2. Proof. We divide the square with side 2 into four squares with side 1 by bisecting sides. Each of the five points has to be in one of the four small squares. By Pigeonhole principle, there must be a small square containing at least two of the five points. The diameter of the square (i.e., the largest possible distance between two points in the square) is the length diagonal, which is √ 2. Thus the distance between those two points which lies in the same small square is at most √ 2....
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- Spring '10
- Playing card, Line segment, red line, line segments