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# functions_1 - (upon expanding ⇒ x-xy =-y ⇒ x(1-y =-y...

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Analysis 1: A function example Example Let f : R - { 1 } → R - { 1 } ,f ( x ) = x/ ( x - 1). i) Show that f is an injection. ii) Show that f is a surjection. Solution. i) Suppose x 1 ,x 2 R - { 1 } , the domain of f . We need to show ( f ( x 1 ) = f ( x 2 )) ( x 1 = x 2 ) . So assume that f ( x 1 ) = f ( x 2 ) . From the deﬁnition of the function, we then have that f ( x 1 ) = f ( x 2 ) x 1 ( x 1 - 1) = x 2 ( x 2 - 1) x 1 ( x 2 - 1) = x 2 ( x 1 - 1) (by cross-multiplying) x 1 x 2 - x 1 = x 2 x 1 - x 2 (upon expanding) x 1 = x 2 (after cancellation and multiplication by -1) Therefore, f is an injection. ii) Let y R - { 1 } , the co-domain of f . We need to show that ( x R - { 1 } )[ f ( x ) = y ] . [We ﬁrst attempt to solve the equation f ( x ) = y . This could be considered as ‘working’. We have f ( x ) = y x ( x - 1) = y x = ( x - 1) y (upon multiplying by ( x - 1); note that x - 1 6 = 0) x = xy - y

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Unformatted text preview: (upon expanding) ⇒ x-xy =-y ⇒ x (1-y ) =-y ⇒ x =-y 1-y = y y-1 Now we return to the proof. ] Deﬁne x = y y-1 . Then x ∈ R as y ∈ R- { 1 } . Also, x 6 = 1. [ To check this: Assume for a contradiction that x = 1 . Then x = y y-1 = 1 ⇒ y = y-1 ⇒ 0 =-1 , a contradiction. ] Thus x ∈ R- { 1 } , the domain of f . We then have that f ( x ) = f ( y y-1 ) = y y-1 y y-1-1 = y y-( y-1) = y 1 = y. Thus, for each y in the co-domain, we can ﬁnd an x in the domain such that f ( x ) = y . Therefore, f is a surjection. 2...
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functions_1 - (upon expanding ⇒ x-xy =-y ⇒ x(1-y =-y...

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