Show that the function f: R* → R* defined by `f(x) = 1/x` 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?
It is given that f: R* → R* is defined by `f(x) = 1/x`
f(x) = f(y)
`=> 1/x = 1/y`
=> x = y
∴f is one-one.
It is clear that for y∈ R*, there exists ` x= 1/y in R ("Exists as y" != 0)`such that
`f(x) = 1/((1/y)) = y`
∴f is onto.
Thus, the given function (f) is one-one and onto.
Now, consider function g: N → R*defined by
`g(x) = 1/x`
`g(x_1) = g(x_2) => 1/x_1 = 1/x_2 => x_1 = x_2`
∴g is one-one.
Further, it is clear that g is not onto as for 1.2 ∈R* there does not exit any x in N such that g(x) = `1/1.2`
Hence, function g is one-one but not onto.