Mathematics: 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.)

Let a = r − {−2}, b = r − {1} and f : a → b be given

Ответ:

kaffolter25

09.01.2022

Mathematics

Explanation:

In order to prove that f is one-to-one and surjective, we can prove directly that f is bijective by findng $f^{-1}$ .

$f(x) = \frac{x+1}{x+2}$

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

$x = \frac{y+1}{y+2}$

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)

$y = \frac{2x-1}{1-x}$

Thus, $f^{-1}(x) = \frac{2x-1}{1-x}$ . 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 $\frac{2x-1}{1-x}$ , for some x, is an element of A because it is different from -2. Lets suppose that there exist an x such as $\frac{2x-1}{1-x} = -2$ , thus 2x-1 = -2(1-x) = -2+2x, from where we conclude that 0 = -1, that is a contradiction. As a result, $\frac{2x-1}{1-x}$ cant be equal to -2.