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)# is not bijective
No it is not a bijection.
Explanation:
If
and
Therefore,
Bonus
For any integer