Question

Prove that the inverse of one-one onto mapping is unique

Answer 1

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Answer 2

Answer:

Hence proved that the inverse of one-one onto mapping function is unique

To prove:

An unique value is obtained when we perform the inverse of one-one onto mapping function.

Solution:

Theorem:  

If $\mathrm { f } : \mathrm { x } \rightarrow \mathrm { y }$ is a one- one and onto then, $f^{-1}$ is a one-to-one and onto function.

Let, $\mathrm { f } : \mathrm { R } \rightarrow \mathrm { R }$ be given by

$F(x) =7x-1=y$

Now swap x and y to solve y

$\begin{array} { c } { 7 y - 1 = x } \\\\ { 7 y = x + 1 } \\\\ { y = \frac { x + 1 } { 7 } } \end{array}$

Thus

$\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \frac { \mathrm { x } + 1 } { 7 }$

Hence proved that the inverse of one-one onto mapping function is unique