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Question
A function f from the set of natural numbers to integers defined by
`{([n-1]/2," when n is odd" is ),(-n/2,when n is even ) :}`
Options
neither one-one nor onto
one-one but not onto
onto but not one-one
one-one and onto both
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Solution
one-one and onto both
Injectivity:
Let x and y be any two elements in the domain (N).
\[\text{Case}-1: \text{Bothxandyare even}.\]
\[\text{Let}f\left( x \right) = f\left( y \right)\]
\[ \Rightarrow \frac{- x}{2} = \frac{- y}{2}\]
\[ \Rightarrow - x = - y\]
\[ \Rightarrow x = y\]
\[\text{Case}-2: \text{Bothxandyare odd}.\]
\[Letf\left( x \right) = f\left( y \right)\]
\[ \Rightarrow \frac{x - 1}{2} = \frac{y - 1}{2}\]
\[ \Rightarrow x - 1 = y - 1\]
\[ \Rightarrow x = y\]
\[Case-3:\text{Let x be even andybe odd}.\]
\[\text{Then},f\left( x \right) = \frac{- x}{2}\text{and}f\left( y \right) = \frac{y - 1}{2}\]
\[\text{Then, clearly}\]
\[x \neq y \]
\[ \Rightarrow f\left( x \right) \neq f\left( y \right)\]
\[\text{From all the cases,f is one-one}.\]
Surjectivity:
\[\text{Co-domain of f} = Z = \left\{ . . . , - 3, - 2, - 1, 0, 1, 2, 3, . . . . \right\}\]
\[\text{Range of f } = \left\{ . . . , \frac{- 3 - 1}{2}, \frac{- \left( - 2 \right)}{2}, \frac{- 1 - 1}{2}, \frac{0}{2}, \frac{1 - 1}{2}, \frac{- 2}{2}, \frac{3 - 1}{2}, . . . \right\}\]
\[ \Rightarrow \text{Range of f} = \left\{ . . . , - 2, 1, - 1, 0, 0, - 1, 1, . . . \right\}\]
\[ \Rightarrow \text{Range of f} = \left\{ . . . , - 2, - 1, 0, 1, 2, . . . . \right\}\]
\[ \Rightarrow \text{Co-domain of f} = \text{Range of f}\]
⇒ f is onto.
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