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# Mathematical Thinking: Problem-Solving and Proofs (2nd Edition)

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4.5. There is such a bijection if and only if A contains at least two elements. If A contains only one element or is empty, then the only function from A to A is the identity function. If A does contain at least two elements, then let x, y A, x 6 = y . Define f : A A by f ( x ) = y, f ( y ) = x , and f ( z ) = z for x 6 = z 6 = y . Clearly f is not the identity on A since f ( x ) = y 6 = x . Intuitively it is obvious that f is a bi- jection. We prove it by noticing that f f = id A , and hence f = f - 1 and f is bijective. 4.7. ( A ) A is not injective since A (0 , 0) = 0 = A (1 , - 1). A is surjective since A (0 , y ) = 0 + y = y for any y R . ( M ) M is not injective since M (0 , 1) = 0 = A (1 , 0). M is surjective since M (1 , y ) = 1 · y = y for any y R . ( D ) D is not injective since M (0 , 1) = 1 = A (1 , 0). D is also not surjective since M ( x, y ) 0 for all x, y R . 4.9. The statement is true. A full proof involves four cases, but they are very similar. Therefore we prove only one case: f is nondecreasing and g is nonincreasing. Let x < y . Then f ( x ) f ( y ), so ( g f )( x ) = g ( f ( x )) g ( f ( y )) = ( g f )( y
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Unformatted text preview: g ◦ f is nonincreasing and therefore monotone. 4.10. By a previous exercise, the equations f ( x ) = y and g ( x ) = y each have unique solutions for every y ∈ R . This says precisely that f and g are bijective. Now g ( f ( x )) = c ( ax + b ) + d = acx + bc + d and f ( g ( x )) = a ( cx + d ) + b = acx + ad + b , so g ◦ f-f ◦ g is a constant function and hence is neither injective nor surjective. 4.11. Multiplication by 2 is both surjective and injective on R since 2( x/ 2) = x and if 2 x = 2 y , then x = y . Thus it is a bijection on R . It is not surjective on Z , since the image of any element of Z under multiplication by 2 is even. Therefore multiplication by 2 is not bijective on Z . (Note that it remains injective on Z since Z ⊂ R .) 1...
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