# so2 - Math 55 - Homework 2 Solutions Thursday, June 30,...

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Math 55 - Homework 2 Solutions Thursday, June 30, 2011 1. From section 1.7, do problem 9. 8 and 9 are consective, 9 is a perfect square, and 8 is a perfect cube. (0 and ± 1 also works.) 2. From section 2.1, do problem 38. a) If S is in S , then by the deﬁnition of S , S / S . Contradiction. b) If S is not in S , that is S / S , then S meets the requirements to be in S . So S S . Contradiction. 3. From section 2.2, do problems 45. a) Notice that if i < j then A i A j and A i A j = A j . So n [ i =1 A i = A 1 A 2 ... A n = A n b) Notice that if i < j then A i A j and A i A j = A i . So n \ i =1 A i = A 1 A 2 ... A n = A 1 4. From section 2.3, do problems 30 and 31. 30) Yes. In fact, we don’t even need f to be one-to-one, just f g . Consider some x and y so that g ( x ) = g ( y ) . Now ( f g )( x ) = f ( g ( x )) = f ( g ( y )) = ( f g )( y ) . Since f g is one-to-one and ( f g )( x ) = ( f g )( y ) , we know x must equal y . Which is what we needed to show to get that g is one-to-one. 31) No. Let f : { 1 , 2 } → { 1 } deﬁned by f (1) = f (2) = 1 and g : { 1 } → { 1 , 2 } deﬁned by g (1) = 1 . Now f is onto because 1 = f (1) , and f g is onto because 1 = ( f g )(1) . But g is not onto because it doesn’t hit 2 . 5. You are given a piece of chocolate

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## This note was uploaded on 03/14/2012 for the course MATH 55 taught by Professor Strain during the Summer '08 term at University of California, Berkeley.

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so2 - Math 55 - Homework 2 Solutions Thursday, June 30,...

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