Show that the function `f: R_0->R_0` , defined as `f(x)=1/x` , is one-one onto, where `R_0` is the set of all non-zero real numbers. Is the result true, if the domain `R_0` is replaced by `N` with co-domain being same as `R_0` ?

Explanation

`f: R_(**) to R_(**)`
and `" " f(x) = (1)/(x) AA x in R_(**)`
Let `x, y in R_(**) and f(x) = f(y) `
`rArr " " (1)/(x) = (1)/(y) rArr x =y`
`therefore f` is one-one.
For each `y in R_(**)`, ` " "x = (1)/(y) in R_(**)`
such that `" " f(x) = f((1)/(y)) = (1)/((1)/(y)) = y `
`therefore f `is onto.
Therefore, function `f` is one-one onto function.
Again, let `" " g: N to R_(**)`
then, `" " g(x) = (1)/(x) AA x in N`
Let `x, y in N and g(x) = g(y)`
`rArr (1)/(x) = (1)/(y) rArr x =y`
`therefore g ` is one-one.
`because ` For 12`inR_(**)` there does not exist `x in N ` such that
`g(x) = (1)/(12)`
`therefore g` is not onto.
Therefore, g is one-one but not onto.