let f:n-n defined by f(x)= n+1/2, if n is odd

n/2, if n i even

state whether f is bijective ?

i know how to prove that f is 1-1 . i need help for if f is onto or not ..

**Q4.**

*f*: N ® N is defined by

For *n* = 1, we have

For *n* = 2, we have

Thus, *f *(1) = *f* (2) for 1 ≠ 2.

So, *f* is not one-one as two distinct elements in the domain have the same image under function *f*.

Suppose *n* be an arbitrary element of N.

If *n* is odd natural number, then 2*n* –1 is also an odd natural number.

If *n* is even natural number, then 2*n* is also an even natural number.

Thus, for every *n* in the codomain of *f*, there exists its pre image in the domain N. Hence, the range of *f *is equal to the codomain of *f*. So, *f *is onto.

The function is not one-one but onto. Hence, *f* is not bijective.

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