Question

Show that the function f: R_* \(\to\) R_* defined by f (x)= \(\frac {1} {x}\) 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_* ?  

Answer 1

It is given that f : R_* \(\to\) R_* is defined by  f (x)= \(\frac {1} {x}\) .
One-one: f (x)=f (y)
\(\Rightarrow\)  \(\frac {1} {x}\) = \(\frac {1} {y}\)
\(\Rightarrow \) x=y.
∴ f is one-one.
Onto: 
It is clear that for y∈ R_* , there exists x = \(\frac {1} {y}\)  ∈ R_*  (exists as y ≠ 0 ) such that f (x)= \(\dfrac {1} { \frac {1} {y}}\)=y. 
∴ f is onto.
Thus, the given function (f) is one-one and onto.
Now, consider function g: N \(\to\) R_* defined by 
g (x)= \(\frac {1} {x}\)
we have g (x₁)=g (x₂) \(\Rightarrow\) \(\frac {1} {x_1}\)= \(\frac {1} {x_2}\)\(\Rightarrow\) x₁ = x₂ 
∴ 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) = \(\frac {1} {1.2}\)

Hence, function g is one-one but not onto.