It is given that f :R→ {x ∈ R : −1 < x <1} defined by f( x) = x/(1 + |x|), x ∈ R
Suppose, f(x) = f(y), where x, y ∈ R ⇒ x/(1 + |x|) = y/(1 + |y|)
It can be observed that if x is positive and y is negative, then we have
x/(1 + x) = y/(1 - y) ⇒ 2xy = x - y
Since, x is positive and y is negative, then x > y ⇒ x − y > 0
But, 2xy is negative. Then, 2xy ≠ x − y.
Thus, the case of x being positive and y being negative can be ruled out.
Under a similar argument, x being negative and y being positive can also be ruled out. Therefore, x and y have to be either positive or negative.
When x and y are both positive, we have f(x) = f(x) ⇒ x/(1 + x) = y/(1 + y)
⇒ x + xy = y + xy ⇒ x = y
When x and y are both negative, we have f(x) = f(y) ⇒ x/(1 - x) = y/(1 - y) ⇒ x - xy = y - xy ⇒ x = y
Therefore, f is one-one. Now, let y ∈ R such that −1 < y < 1.
If y is negative, then there exists x = y/(1 + y) ∈ R such that
If y is positive, then there exists x = y/(1 - y) ∈ R such that
Therefore, f is onto. Hence, f is one-one and onto.
