hw1solutions - Chapter 0: Sets and Functions (PSU Notes) 1....

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1. A = { 1 , 3 , 5 , 7 , 9 , 11 } ,B = { 3 , 6 , 9 , 12 , 15 , 18 } ,A B = { 3 , 9 } ,A - B = { 1 , 5 , 7 , 11 } . 2. (1) A B C = { 4 } . (2) A - B = { 1 , 3 } . (3) ( A B ) C = { 5 , 7 , 9 , 10 } . (4) A C B C C C = { 1 , 2 , 3 , 5 , 6 , 7 , 8 , 9 , 10 } . It is easy to see that { 4 } C = { 1 , 2 , 3 , 5 , 6 , 7 , 8 , 9 , 10 } . 4. 6. Solve ( y - 2 = x - 1 2 x + 1 = y + 2 ( y = x + 1 2 x + 1 = y + 2 2 x +1 = x +1+2 x = 2 and y = 3. 8. R = { (1 , 3) , (1 , 5) , (2 , 3) , (2 , 5) , (3 , 5) , (4 , 5) } , so dom R = { 1 , 2 , 3 , 4 } and range R = { 3 , 5 } . 10. (a) Reflexive: (1 , 1) , (2 , 2) , (3 , 3) , (4 , 4) , (5 , 5) R . (b) Symmetric: (1 , 2) R (2 , 1) R , (3 , 4) R (4 , 3) R . No other pairs. (c) Transitive: (1 , 2) , (2 , 1) R (1 , 1) R ; (3 , 4) , (4 , 3) R (3 , 3) R ; (2 , 1) , (1 , 2) R (2 , 2) R ; (4 , 3) , (3 , 4) R (4 , 4) R . No other pairs. Then 1 2 , 3 4 , 5 is alone, i.e. E = { 1 , 2 } ∪ { 3 , 4 } ∪ { 5 } . 12. g ( - 2) = - 7 ,g ( - 1) = 0 ,g (0) = 1 ,g (1) = 2 ,g (2) = 9. We can verify that for all x 1 ,x 2 X , x 1 6 = x 2 will imply g ( x 1 ) 6 = g ( x 2 ), so g is one-to-one, and im( g ) = {- 7 , 0 , 1 , 2 , 9 } . By solving y = x 3 + 1, we have y - 1 = x 3 , i.e. x = 3 y - 1. Then g - 1 ( x ) = 3 x - 1 (if we assume that the independent variable is always x ). 14. Since f ( a ) = b,f ( b ) = d,f ( c ) = a , and f ( d ) = c , f maps different elements of A to different elements in A , so f is one-to-one. Meanwhile, im( f ) = { b,d,a,c } = A , so f is onto. Its inverse function is: f - 1 ( b ) = a,f - 1 ( d ) = b,f - 1 ( a ) = c , and f - 1 ( c ) = d , i.e. f
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hw1solutions - Chapter 0: Sets and Functions (PSU Notes) 1....

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