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 allnon-zero real numbers. Is the result true, if the domain `R_0`is replaced by `N`with co-domain beingsame as `R_0`?

Show that the function `f: R_0->R_0`, defined as `f(x)=1/x`, is one-on

1.	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 allnon-zero real numbers. Is the result true, if the domain `R_0`is replaced by `N`with co-domain beingsame as `R_0`?
Answer» `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.