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Home  >>  CBSE XII  >>  Math  >>  Relations and Functions
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Show that the function \(f : R_* \rightarrow R_* \) 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 \(R_*\)?

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Toolbox:
  • A function $f: X \rightarrow Y$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one function.
  • A function$ f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
Given $f:R_* \rightarrow R_*$ defined by $f(x)=\Large \frac{1}{x}$ where $R_*$ is a set of nonzero real numbers:
Let $x$ and $y$ be two elements in $R_*$.
Step1: Injective or One-One function:
For a one-one function, $f(x) = f(y)$
$ \Rightarrow \Large \frac{1}{x}=\frac{1}{y}$$ \Rightarrow x = y.$
Therefore $f:R_* \rightarrow R_*$ defined by $f(x)=\Large \frac{1}{x}$ is one-one.
For an on-to function, for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
$ \Rightarrow$ For every $y \in R_*$ there must exist $ x=\large \frac{1}{y}$$ \in R_*$ such that $ f(x)=\frac{1}{\Large ( \frac{1}{y})}$$=y$.
Therefore $f:R_* \rightarrow R_*$ defined by $f(x)=\Large \frac{1}{x}$ is onto..
Step 2: Surjective or On-to function:
Now, let us consider a function $f_2:N \rightarrow R_*$ defined by $f_2(x)=\Large \frac{1}{x}$:
For a one-one function, $f_2(x_2) = f_2(y_2)$
$ \Rightarrow \Large \frac{1}{x_2}=\frac{1}{y_2}$$ \Rightarrow x_2 = y_2.$
Therefore, $f_2:N \rightarrow R_*$ defined by $f_2(x)=\Large \frac{1}{x}$ is one-one.
$ \Rightarrow$ For and onto function, for every $y_2 \in R_*$ there must exist $ x_2=\large \frac{1}{y_2}$$ \in R_*$ such that $ f_2(x)=\frac{1}{\Large ( \frac{1}{y_2})}$$=y_2$.
However, we see that for $y_2 = 1.5 \in R_*$, there is no $x_2$ in $N$ such that $f_2(x) = \Large \frac{1}{1.5}$.
Solution: Therefore  the function $f_2$ is not onto.

 

answered Mar 13, 2013 by balaji.thirumalai
edited Mar 19, 2013 by meena.p
 

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