1873_solutions

# From exercise 3 that card p s card p s u w 2 n 2 n

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Unformatted text preview: if we define  j for each j function from 1 to 1, 2, , m whose value at 1 is j then  is a one-one function from 1, 2, , m onto 1 . Therefore card 1  m  m 1 . Now suppose that n is any positive integer for which the assertion p n is true. In order to prove that card n1  m n1 we shall show that 1, 2, m ß n1 n and for this purpose we shall define a function : n 1, 2, , m n 1 as follows: For each member f, j of the set n 1, 2, , m we define  f, j k  1, 2, , n fk if k j if k  n  1 . The proof will be complete when we have seen that  is one-one and is onto the set n1 . 1, 2, , m and that To see that  is one-one we suppose that f 1 , j 1 and f 2 , j 2 belong to n 1, 2, , n we have  f 1 , j 1   f 2 , j 2 . For each k f1 k   f1, j1 k   f2, j2 k  f2 k and so f 1  f 2 . We see also that j1   f1, j1 n  1   f2, j2 n  1  j2. Finally, to see that  is onto the set n1 , suppose that g n1 and define j  g n  1 and define f k  g k for every k 1, 2, , n . We see at once that g   f, j . 56 11. a. Prove that if for each positive integer n we define F n to be the set of all one-one functions from 1, 2, , n onto 1, 2, , n then for each n we have F n 1 ß F n 1, 2, , n, n  1 . Solution: To motivate the solution of this exercise we should consider that a one-one function from the set 1, 2, , n onto itself is a way or arranging the members of the set 1, 2, , n in a sequence x 1 , , x n . For each such arrangement we can obtain an arrangement of the members of the larger set 1, 2, , n, n  1 by placing the number n  1 in any of the n  1 positions shown n  1, x 1 , x 2 , , x n x 1 , n  1, x 2 , , x n  x 1 , x 2 , , x n , n  1 Thus the number of ways of arranging the members of 1, 2, , n, n  1 should be n  1 times the number of ways of arranging the members of 1, 2, , n in a sequence. Now we begin: We define a function  from F n 1, 2, , n, n  1 member f, m of the set F n 1, 2, , n, n  1 we define fj if j  m n1 if j  m . fj  f, m j  F n1 as follows: Given any if j  m 1 Each such function  f, m is a one-one function from 1, 2, , n, n  1 onto 1, 2, , n, n  1 . To see that the function  is one-one, suppose that f, m and g, k belong to the set 1, 2, , n, n  1 and that  f, m   g, k . Since the function  g, k is one-one and since Fn  g, k m   f, m m  n  1   g, k k we deduce that k  m. Given any j  m we have f j   f, m j   g, m j  g j and given any j m we have f j   f, m j  1   g, m j  1  g j and we conclude that f  g. Therefore the function  is one-one. Finally, to show that  is onto the set F n1 we assume that h F n1 . In other words, h is a one-one function from the set 1, 2, , n, n  1 onto 1, 2, , n, n  1 . We see easily that if m  h 1 n  1 and if we define f : 1, 2, , n 1, 2, , n by the equation hj then f if j  m h j1 fj  if j m F n and h   f, m . b. Prove that for each positive integer n, the number of ways of ordering the numbers in the set 1, ...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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