Question

Suppose $ {f_1} $ and $ {f_2} $ are non-zero one-one functions from $ R $ to $ R $ . Is $ \dfrac{{{f_1}}}{{{f_2}}} $ necessarily one-one? Justify your answer. Here $ \dfrac{{{f_1}}}{{{f_2}}}:R \to R $ is given by $ \left( {\dfrac{{{f_1}}}{{{f_2}}}} \right)\left( x \right) = \dfrac{{{f_1}\left( x \right)}}{{{f_2}\left( x \right)}} $ for all $ x \in R $ .

Answer 1

Hint: To solve the given problem, we will consider two non-zero one-one functions $ {f_1} $ and $ {f_2} $ from $ R $ to $ R $ and we will prove that $ \dfrac{{{f_1}}}{{{f_2}}} $ is not necessarily one-one. That is, we will prove this by taking counter examples.

Complete step-by-step answer:
Let us consider the function $ {f_1}:R \to R $ given by $ {f_1}\left( x \right) = x $ for $ \forall x \in R $ . First we will prove that $ {f_1}:R \to R $ is a one-one function. For this, let us consider two elements $ a,b \in R $ . We have $ {f_1}\left( x \right) = x $ . Therefore, we can write $ {f_1}\left( a \right) = a $ and $ {f_1}\left( b \right) = b $ . Now consider $ {f_1}\left( a \right) = {f_1}\left( b \right) $ . Therefore, we get $ a = b $ . So, we have $ {f_1}\left( a \right) = {f_1}\left( b \right) \Rightarrow a = b $ . So, we can say that $ {f_1} $ is a one-one function.
Let us consider the function $ {f_2}:R \to R $ given by $ {f_2}\left( x \right) = {x^3} $ for $ \forall x \in R $ . First we will prove that $ {f_2}:R \to R $ is a one-one function. For this, let us consider two elements $ a,b \in R $ . We have $ {f_2}\left( x \right) = {x^3} $ . Therefore, we can write $ {f_2}\left( a \right) = {a^3} $ and $ {f_2}\left( b \right) = {b^3} $ . Now consider $ {f_2}\left( a \right) = {f_2}\left( b \right) $ . Therefore, we get $ {a^3} = {b^3} \Rightarrow a = b $ . So we have $ {f_2}\left( a \right) = {f_2}\left( b \right) \Rightarrow a = b $ . So, we can say that $ {f_2} $ is a one-one function.
Now we will find $ \dfrac{{{f_1}}}{{{f_2}}}:R \to R $ . In this problem, it is given that $ \left( {\dfrac{{{f_1}}}{{{f_2}}}} \right)\left( x \right) = \dfrac{{{f_1}\left( x \right)}}{{{f_2}\left( x \right)}} $ for all $ x \in R $ .
Therefore, we can write
 $ \left( {\dfrac{{{f_1}}}{{{f_2}}}} \right)\left( x \right) = \dfrac{x}{{{x^3}}} $
 $ \left( {\dfrac{{{f_1}}}{{{f_2}}}} \right)\left( x \right) = \dfrac{1}{{{x^2}}} $
Now we will check if the function $ \dfrac{{{f_1}}}{{{f_2}}} $ is one-one or not. Let us take $ x = 1 $ and $ y = - 1 $ .
Now $ \left( {\dfrac{{{f_1}}}{{{f_2}}}} \right)\left( x \right) = \left( {\dfrac{{{f_1}}}{{{f_2}}}} \right)\left( 1 \right) = \dfrac{1}{1} = 1 $ and $ \left( {\dfrac{{{f_1}}}{{{f_2}}}} \right)\left( y \right) = \left( {\dfrac{{{f_1}}}{{{f_2}}}} \right)\left( { - 1} \right) = \dfrac{1}{{{{\left( { - 1} \right)}^2}}} = 1 $
Here we can see that $ x \ne y $ but $ \left( {\dfrac{{{f_1}}}{{{f_2}}}} \right)\left( x \right) = \left( {\dfrac{{{f_1}}}{{{f_2}}}} \right)\left( y \right) $ . So, we can say that $ \dfrac{{{f_1}}}{{{f_2}}} $ is not a one-one function. Hence, if $ {f_1} $ and $ {f_2} $ are one-one functions from $ R $ to $ R $ then $ \dfrac{{{f_1}}}{{{f_2}}} $ is not necessarily one-one function.

Note: A function $ f:A \to B $ is said to be one-one function if for every $ x,y \in A $ , $ x \ne y \Rightarrow f\left( x \right) \ne f\left( y \right) $ or $ f\left( x \right) = f\left( y \right) \Rightarrow x = y $ . A function $ f\left( x \right) $ is one-one if and only if $ {f^{ - 1}}\left( x \right) $ exists. If $ f $ and $ g $ are one-one functions then we can say that $ f + g $ , $ fg $ are always one-one functions. Note that here the function $ fg $ is given by $ \left( {fg} \right)\left( x \right) = f\left( x \right) \cdot g\left( x \right) $ .