Show that the function `f: R->{x in R :-1ltxlt1}` defined by `f(x)=x/(1+|x|),x in R` is one-one and onto function.

Explanation

Let x, `y in R` and ,
f(x)=f(y)
`rArr x/(1+|x|)=y/(1+|y|)`
If x is positve and y is negative then `x gt y `
`rArr x-y gt 0 and 2xy lt 0 `
`therefore x/(1+x)=y/(1-y)`
`rArr y+ xy =x -xy`
`rArr 2xy = x-y` which is impossible then `f(x)=f(y)rArr (x)/(1+x)=(y)/(1+y)`
`rArr x+y`
If x and y both are negative then `f(x)=f(y)rArr x/(1-x)=y/(1-y)`
`rArr x-xy = y-xy`
`rArr x=y`
Therefore , f is one - one Let y `in ` R be such that `-1 lt y lt 1`
If y is negative then `x=y/(1+y) in R` is such that
`f(x)=f((y)/(1+y))=((y)/(1+y))/(1+|(y)/(1+y)|)=((y)/(1+y))/(1-y/(1+y))=y`
If y is positive then `x=y/(1-y) in R` is such that `f(x)=f(y/(1-y))=((y)/(1-y))/(1+((y)/(1-y)))=((y)/(1-y))/(1-y/(1+y))=y`
`therefore` f is onto
Therefore f is one - one onto .