For each function below, determine whether or not the function is injective and whether| or not the function is surjective. be sure to justify your answers. (a) f : n -> n given by f(n) =n+ 2
(b) f p({1,2, 3}) -» n given by f(a) = |a| (note: p(s) denotes the power set of a set s.)

a) injective but not surjective. b) neither injective nor surjective.

Step-by-step explanation:

A function is injective if there aren't repeated images. To check if a function is injective we are going to suppose that for two values in the domain the image is equal, then we need to find that the two values are equal.

a) f : N → N with f(n) = n+2.

suppose that for n and m natural numbers, f(n)=f(m). Then

n+2 = m+2

n+2-2 = m

n = m.

Then, f(n) is injective.

Now, a function is surjective if every term m in the codomain there exists a pre-image of that element, that is to say, there exists an n such that f(n) = m. That is, the range of the function is equal to the codomain.

In this case, f(n) is not surjective. For example, if we would have that

n+2 = 1

n = 1-2

n = -1

but -1 is not a natural number, then for m=1 we don't have a pre image in f.

b) f: P({1, 2, 3}) → N with f(A) = |A|  (amount of elements in the subset A).

Now, for {1,2,3} we can have subsets of 0, 1, 2 or 3 elements. Then, the range of the function f is {1, 2, 3} (0 is not included because 0 is not natural). Then, the range is not all the natural numbers and therefore the function is not surjective.

Now, let's check if f is injective. Let {1} and {2} subsets of {1, 2, 3}. Then

f({1}) = |{1}| = 1.

f ({2}) = |{2}| = 1.

We have two different subsets with the same image, then f is not injective.

its d. the first coordintaes are 10,32

step-by-step explanation:

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)