Answer:
(a) It is a permutation of Z
(b) It is not a permutation of Z
Step-by-step explanation:
Recall that a permutation of a set S, is just a function g:S--->S which is both injective and surjective. That is to say, g is bijection from S onto S.
(a)
Let's show that g is injective, to do this, we need to show
[tex]g(a)\neq g(b)\Rightarrow a \neq b[/tex]
But
[tex]g(a)\neq g(b)\Rightarrow -a+2\neq -b+2 \Rightarrow -a\neq -b \Rightarrow a\neq b[/tex]
So g is injective.
To prove g is surjective, let's take any integer b and show that there exists an integer a such that g(a) = b.
If we take the integer a = 2-b, then
g(a) = g(2-b) = -(2-b)+2 = -2 +b +2 = b.
So g is also surjective, and thus, a permutation of the integers.
(b)
Although g is injective, g fails to be surjective. To show this, let's show that there is no integer a such that g(a) = 2.
[tex]g(a)=2\Rightarrow a^3=2\Rightarrow a=\sqrt[3]{2}[/tex]
but a is not an integer, and g is not surjective. So it is not a permutation.