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Goshia [24]
3 years ago
14

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 way

s.
(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.)
Mathematics
1 answer:
AysviL [449]3 years ago
7 0

Answer:

Check.

Step-by-step explanation:

a. Let f: \mathbb{Z} \rightarrow O such that f(n) = 2n.

Injective:

Let n, m \in \mathbb{Z} 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  \in O, so n = 2k for some integer k. Then, f(k) = n and therefore f is surjective.

b. Let  g: O \rightarrow \mathbb{Z} such that g(n) = n/2. Then,

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

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

Therefore, g is the inverse of f.

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Also, the finished pattern was 1092 square centimeters in area. Hence:

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From both equations:

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This is further explained below.

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