Question #a6851

2 Answers
Feb 16, 2017

Answer:

#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

Feb 16, 2017

Answer:

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#