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Unformatted text preview: MATH 135: SOLUTIONS TO HOMEWORK SET #3 1. (page 133, # 2) Show that in Fig. 32 we have J ( m,n ) = [1 + 2 + + ( m + n )] + m = 1 2 [( m + n ) 2 + 3 m + n ] Proof. We demonstrate that 0 = J (0 , 0) and the J ( m,n ) + 1 = J ( m + 1 ,n 1) if n > 0 and that J ( m,n ) + 1 = J (0 ,m + 1) if n = 0. Certainly, J (0 , 0) = 1 2 [(0 + 0) 2 + 3(0) + 0] = 0. We compute, J ( m,n ) + 1 = 1 2 [( m + n ) 2 + 3 m + n ] + 1 = 1 2 [( m + n ) 2 + 3 m + n + 2] = 1 2 [([ m + 1] + [ n 1]) 2 + 3( m + 1) + ( n 1)] = J ( m + 1 ,n 1) So, if n > 0, this gives the desired result. We compute, J ( m, 0) + 1 = 1 2 [( m + 0) 2 + 3 m + 0] + 1 = 1 2 [ m 2 + 3 m + 2] = 1 2 [( m + 1) 2 + ( m + 1)] = 1 2 [(0 + [ m + 1]) 2 + 3(0) + ( m + 1)] = J (0 ,m + 1) 2. (page 133, # 3) Find a onetoone correspondence between the open unit interval (0 , 1) and R that takes rationals to rationals and irrationals to irrationals. f ( x ) =  1 4 x if 0 < x 1 4 1 4 x if 1 4 < x < 1 2 if x = 1 2 4 x 2 if 1 2 < x 3 4 1 4 x 3 if 3 4 < x < 1 3. (page 138, # 6) Let be a nonzero cardinal. Show that there is no set K for which  X  = X K ....
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This note was uploaded on 10/13/2009 for the course MATH 135 taught by Professor Krueger,j during the Spring '08 term at University of California, Berkeley.
 Spring '08
 Krueger,J
 Math

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