Question

Question #a6851

Answer 1

`f(n)` is not bijective

Explanation:

We have:

`f : NN -> NN`
`f(n) = { ( (n+1)/2, "n is odd"), (n/2, "n is even") :}`

Lets tabulate the value of `f(n)` for the first few Natural numbers to establish the pattern of how `f(n)` works.

`{: ( 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)` maps each consecutive pairs of natural numbers in the range set to a single number in the domain set, and therefore we concluse that:

`f(n)` is not bijective

Answer 2

No it is not a bijection.

Explanation:

If `k` is an odd integer, then `k+1` is even,

and `f(k) = (k+1)/2 = f(k+1)`

Therefore, `f` is not injective.

Bonus

`f` is surjective.

For any integer `m`,

`2m` is an even integer and `f(2m) = m`