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Let f" ": N->N be defined by f(n)={(n...

Let `f" ": N->N` be defined by `f(n)={(n+1)/2,""if""""n""""i s""""od d n/2,""if""""n""""i s""e v e n` for all `n in N` . State whether the function f is bijective. Justify your answer.

Text Solution

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In `f : N to N `
`f(x) = {{:((n+1)/(2)"," if n "is odd"),((n)/(2) ","" " if n "is even"):}`
`because f(1) = (1+1)/(2) = 1 and f(2) = (2)/(2) = 1 `
`therefore f(1) = f(2)`
but ` 1 ne 2 `
` therefore f` is not one-one.
Let `n in N `
First Case : When `n = 2r + 1, r in N `
then `4r + 1 in N `
and `" " f(4r + 1) = (4r+ 1 + 1)/(2) = 2r + 1`
`therefore f` is onto.
Second Case : When ` n= 2r`
then `" " 4r in N `
and `" " f( 4r ) = (4r)/(2) = 2r`
`therefore f` is onto.
Therefore, `f` is not one-one but onto.
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