Let o be the set of even integers. define f : z ? o by f(n) = 2n. prove that f is a bijection in the following two different ways.
(a) prove that f is both injective and surjective.
(b) find an inverse g : o ? z of f. (show that g is an inverse of f.)

Check.

Step-by-step explanation:

a. Let such that f(n) = 2n.

Injective:

Let n, m such that f(n)=f(m). Then,

2n = 2m and dividing both sides by 2 we obtain that n=m. Then f is injective.

Surjective:

Let n  , so n = 2k for some integer k. Then, f(k) = n and therefore f is surjective.

b. Let   such that g(n) = n/2. Then,

g(f(n)) = g(2n) = 2n/2 = n, and for m

f(g(m)) = f(m/2) = 2*m/2 = m.

Therefore, g is the inverse of f.

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.