I have the following proof, i understand the injective proof but I have no clue how and why they do the surjective proof, is there a certain plan or steps I have to follow? Please explain in more detail and show why each step is taken Result 10.9 The function f : R − {2} → R − {3} defined by3xx-2is bijective.3aProof Here it is necessary to show that f is both one-to-one and onto. We begin with the firstof these. Assume that ƒ(a) = f(b), where a, b = R − {2}. ThenMulti-plying both sides by (a − 2)(b − 2), we obtain 3a(b − 2) = 3b(a − 2). Simplifying, wehave 3ab 6a = 3ab - 6b. Adding −3ab to both sides and dividing by −6, we obtaina = b. Thus, f is one-to-one.3bb-2a-2To show that f is onto, let r = R - {3}. We show that there exists x ЄR - {2} such2rthat f(x) =r. Choose x =Thenr-33 (2)2r-3implying that f is onto. Therefore f is bijective.f(x) = f ƒ (₂2rf(x) =r - 3=6r2r – 2(r-3)=6r== r,

Answered: Image /qna-images/answer/cf4c7c60-e211-470b-acb8-f91acf5511cc.jpg

I have the following proof, i understand the injective proof but I have no clue how and why they do the surjective proof, is there a certain plan or steps I have to follow? Please explain in more detail and show why each step is taken

I have the following proof, i understand the injective proof but I have no clue how and why they do the surjective proof, is there a certain plan or steps I have to follow? Please explain in more detail and show why each step is taken

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Result 10.9 The function f : R − {2} → R − {3} defined by
3x
x-2
is bijective.
3a
Proof Here it is necessary to show that f is both one-to-one and onto. We begin with the first
of these. Assume that ƒ(a) = f(b), where a, b = R − {2}. Then
Multi-
plying both sides by (a − 2)(b − 2), we obtain 3a(b − 2) = 3b(a − 2). Simplifying, we
have 3ab 6a = 3ab - 6b. Adding −3ab to both sides and dividing by −6, we obtain
a = b. Thus, f is one-to-one.
3b
b-2
a-2
To show that f is onto, let r = R - {3}. We show that there exists x ЄR - {2} such
2r
that f(x) =r. Choose x =
Then
r-3
3 (2)
2
r-3
implying that f is onto. Therefore f is bijective.
f(x) = f ƒ (₂
2r
f(x) =
r - 3
=
6r
2r – 2(r-3)
=
6r
=
= r,

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