(3) A common way to prove that two sets A and B are equal is to show that AC B and B C A. Write down a careful proof that A = B, where A = {x Є R : x² − x − 2} and B = {−1,2}, following the same structure as in the example below. Example: Let C = - {x ЄR: x2 4x+3=0} and D = {1,3}. We prove that C that CCD and that D C C: = D by showing - Proof that C C D: Let x E C. Then x Є R and x² - 4x + 3 = 0. Hence (x − 1)(x − 3) = 0, so x 10 or x 3 = 0. Therefore x Proof that DC C: Let x E D, so x we have 1 Є C. Since 3 Є R and 32 x = C. = = 1 or x = 1 or x = 3, so x E D. = 3. Since 1 R and 1² - 4 × 1+3=1−4+3 = 0, - 4 × 3+3=9-12+3 = 0, we have 3 Є C. In either case, Since CCD and DC C, we have C = D. ■□. 5 marks: 2 for AC B, 2 for BC A, 1 for conclusion

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.1: Real Numbers
Problem 36E
Question
(3) A common way to prove that two sets A and B are equal is to show that AC B and B C A.
Write down a careful proof that A
=
B, where
A = {x Є R : x² − x − 2} and B = {−1,2},
following the same structure as in the example below.
Example: Let C
=
-
{x ЄR: x2 4x+3=0} and D = {1,3}. We prove that C
that CCD and that D C C:
= D by showing
-
Proof that C C D: Let x E C. Then x Є R and x² - 4x + 3 = 0. Hence (x − 1)(x − 3) = 0, so
x 10 or x 3 = 0. Therefore x
Proof that DC C: Let x E D, so x
we have 1 Є C. Since 3 Є R and 32
x = C.
=
=
1 or x
= 1 or x
= 3, so x E D.
=
3. Since 1 R and 1² - 4 × 1+3=1−4+3 = 0,
- 4 × 3+3=9-12+3 = 0, we have 3 Є C. In either case,
Since CCD and DC C, we have C = D. ■□.
5 marks: 2 for AC B, 2 for BC A, 1 for conclusion
Transcribed Image Text:(3) A common way to prove that two sets A and B are equal is to show that AC B and B C A. Write down a careful proof that A = B, where A = {x Є R : x² − x − 2} and B = {−1,2}, following the same structure as in the example below. Example: Let C = - {x ЄR: x2 4x+3=0} and D = {1,3}. We prove that C that CCD and that D C C: = D by showing - Proof that C C D: Let x E C. Then x Є R and x² - 4x + 3 = 0. Hence (x − 1)(x − 3) = 0, so x 10 or x 3 = 0. Therefore x Proof that DC C: Let x E D, so x we have 1 Є C. Since 3 Є R and 32 x = C. = = 1 or x = 1 or x = 3, so x E D. = 3. Since 1 R and 1² - 4 × 1+3=1−4+3 = 0, - 4 × 3+3=9-12+3 = 0, we have 3 Є C. In either case, Since CCD and DC C, we have C = D. ■□. 5 marks: 2 for AC B, 2 for BC A, 1 for conclusion
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