Question

Let A = R − {−2}, B = R − {1} and f : A → B be given by the rule f(x) = x + 1 x + 2 . (a) Prove f is one-to-one. (b) Prove f is onto B. (Comment: don’t forget that if given b ∈ B, you construct a such that f(a) = b, you must also show a ∈ A.)

Answer 1

Step-by-step explanation:

In order to prove that f is one-to-one and surjective, we can prove directly that f is bijective by findng
`f^(-1)` .

`f(x) = (x+1)/(x+2)`

To calculate the inverse of f, first we swap the variables x and y. Then we write y in values of x.

`x = (y+1)/(y+2)`

`x/(y+2) = y+1`

`xy +2x = y+1`

On this step it is important to separate what terms contain y and what terms do not. After that we can separate y as a factor

`2x = y+1 -xy`

`2x-1 = y-xy`

`2x-1 = y(1-x)`

`y = (2x-1)/(1-x)`

Thus,
`f^(-1)(x) = (2x-1)/(1-x)` . Note that
`f^(-1)` is well defined en every element of B, because it can be defined on any element different from 1, where the divider is zero.

An element of the form
`(2x-1)/(1-x)` , for some x, is an element of A because it is different from -2. Lets suppose that there exist an x such as
`(2x-1)/(1-x) = -2` , thus 2x-1 = -2(1-x) = -2+2x, from where we conclude that 0 = -1, that is a contradiction. As a result,
`(2x-1)/(1-x)` cant be equal to -2.

I hope this works for you!