how that the function f : R* → R* defined by
is one-one and onto, where R* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R*?
f : R* → R*
Let x1, x2 ∈ R* such that f (x1) = f (x2)
∴ f is one-one Again, given y ∈ R*, there exists such that
∴ f is onto.
If f : N → R0. then function f is one-one, but every real number in codomain may not have a pre-image in N.
∴ f is not onto.