Use element argument to prove the statement. Assume that all sets are subsets of a universal set U, Statement: For all sets A, B, and C, A ∩ (B − C) = (A ∩ B) − (A ∩ C), I have atteched the example to use to solve this problem. Please help me becuase I could not figure it out. 10. For all sets A, B, and C, (AU B) NC S AU(BNC). Prnf: Let 4.B¢ C beLet xEAue) C. Then xe AUB ard rE c 2KE A or xeB) ad xeC, If xe A , Hen xe AulBnc)If xEB, ten xG BnC > xe AulBnc).sefs,14. For all sets A and B, AU (AN B) = A. nof Let 4a 6 be anysets.AUCANB) E A : Led xe AUCANB). by dehndan ofencón,XEA or xEA and xE B. Sggoe A. Then xe A and xeBSo xE AAE AULANB): Let xE A. Hen xE A OCANB) by detnchonunin.19. For all sets A, B, and C, if A C B and A C C then A C BnC. Prut: Let 4, baC beset oanyAEB & AEC. Let xeA. Then xEB CÁC B) and xeCCAEB). So, x c BnC.

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Use element argument to prove the statement. Assume that all sets are subsets of a universal set U, Statement: For all sets A, B, and C, A ∩ (B − C) = (A ∩ B) − (A ∩ C), I have atteched the example to use to solve this problem. Please help me becuase I could not figure it out.

Use element argument to prove the statement. Assume that all sets are subsets of a universal set U, Statement: For all sets A, B, and C, A ∩ (B − C) = (A ∩ B) − (A ∩ C), I have atteched the example to use to solve this problem. Please help me becuase I could not figure it out.

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter5: Rings, Integral Domains, And Fields

Section5.4: Ordered Integral Domains

Problem 11E: 11. (See Exercise 10.) According to Definition 5.29, is defined in by if and only if . Show that ...

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Question

10. For all sets A, B, and C, (AU B) NC S AU(BNC). Prnf: Let 4.B¢ C be
Let xEAue) C. Then xe AUB ard rE c 2
KE A or xeB) ad xeC, If xe A , Hen xe AulBnc)
If xEB, ten xG BnC > xe AulBnc).
sefs,
14. For all sets A and B, AU (AN B) = A. nof Let 4a 6 be any
sets.
AUCANB) E A : Led xe AUCANB). by dehndan ofencón,
XEA or xEA and xE B. Sggoe A. Then xe A and xeB
So xE A
AE AULANB): Let xE A. Hen xE A OCANB) by detnchon
unin.
19. For all sets A, B, and C, if A C B and A C C then A C BnC. Prut: Let 4, baC be
set o
any
AEB & AEC. Let xeA. Then xEB CÁC B) and xeC
CAEB). So, x c BnC.

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For the sets given in Exercise 9, is there a distributive relationship for intersection with respect to union? That is, does ABC=ABAC? A=1,2,3,4, B=2,4,6,8, and C=1,3,5,7,9.
11. (See Exercise 10.) According to Definition 5.29, is defined in by if and only if . Show that if and only if . 10. An ordered field is an ordered integral domain that is also a field. In the quotient field of an ordered integral domain define by . Prove that is a set of positive elements for and hence, that is an ordered field. Definition 5.29 Greater than Let be an ordered integral domain with as the set of positive elements. The relation greater than, denoted by is defined on elements and of by if and only if . The symbol is read “greater than.” Similarly, is read “less than.” We define if and only if. As direct consequences of the definition, we have if and only if and if and only if . The three properties of in definition 5.28 translate at once into the following properties of in . If and then . If and then . For each one and only one of the following statements is true: . The other basic properties of are stated in the next theorem. We prove the first two and leave the proofs of the others as exercises.
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