# Question #a6851

##### 2 Answers

#### Explanation:

We have:

#f : NN -> NN#

#f(n) = { ( (n+1)/2, "n is odd"), (n/2, "n is even") :}#

Lets tabulate the value of

# {: ( n, "even/odd", "mapping", f(n)), ( 1, "odd", (1+1)/2, 1 ), ( 2, "even", 2/2, 1 ), ( 3, "odd", (3+1)/2, 2 ), ( 4, "even", 4/2, 2 ), ( 5, "odd", (5+1)/2, 3 ), ( 6, "even", 6/2, 3 ) :} #

And so the pattern is quite clear.

The definition of a bijective function (or one-to-one function) is that each element of the domain set is paired with exactly one element of the range set and vice versa.

We can see that the function

#f(n)# isnotbijective

No it is not a bijection.

#### Explanation:

If

and

Therefore,

**Bonus**

For any integer