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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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Answer:

305 degree

Step-by-step explanation:

As AD and CE are the diameters of circle P, so that they intersect each other at point P.

=> So that ∡CPA and ∡DPE are two vertical angles.

=> ∡CPA = ∡DPE = 93 degree

As we can see, ∡CPA = ∡CPB + ∡BPA =38 + ∡BPA

=> ∡BPA = ∡CPA - 38 = 93 - 38 = 55 degree

As P is the centre of the circle, so that ∡BPA is equal to the measure of the arc AB, equal to 55 degree.

In the circle P, the total measure of arc AEB and arc AB are 360 degree

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