epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

Solutions for the Exercises at the End of the Sections and the Review Exercises 143
Second part: yl U (B U C) <Z{A U B) U (^ U C). Let x G A U (B U C).
Then, either x e Aox x e{B^ C). This imphes x e Aox x eB ox
X € C, Therefore, either (x e A ox x e B) ox(x € A ox x £ C). Thus, we
can conclude that xe(AUB)U{AU C).
2. First part: A C.B. Let x e A. Then x is a multiple of 2 and of 3.
Therefore, x = 2n with n integer number. Because x is divisible by 3 as
well, while 2 is not, we can conclude that n is divisible by 3. So
X = 2n = 2(3m) = 6m with m integer number. Therefore, x is divisible
by 6. Then, x eB.
Second part: B C.A. Let x e B. Then x is a multiple of 6. Thus, we can
write X = 6t for some integer number t. Then x is divisible by 2 and 3,
because 6 is divisible by 2 and 3. So, x e A.
3. First part: {A\J B)'c,A'f\B'. Let x€(^UBy. This implies that
x^{AVJB). Therefore, x^A and x^B (because if x belonged to
either A or B, then it would belong to their union). Thus, x e A' and
X € &. This implies that xe{A'C\ B').
Second part: A' fl B'^{A U B). Let x e (.4' fi B'). Then x € .4' and
X G F. Therefore, x^A and x^J5. This implies that x^{AyjB). So, we
can conclude that x e (AU By.
4. First part: (AnB)n C^A n(Bn C). Let xe(AnB)nC. Thus,
X e(AnB) and x G C. This implies that x e A and x G B and x e C.
Then, x G ^ and x G (B fl C). Therefore, xeAn{BnC).
Second part: An(Bn C)£(A nB)nC. Let x£An(Bn C). Then
X e A and x G (B fi C). This implies that x G .4 and x e B and x e C.
Thus, X G (^ n B) and X G C. Therefore, xe(AnB)nC.
5. The two sets are not equal. The number 144 is in A, as 144= 16 x 9,
and 144 = 36 x 4, but 144 is not in B. So, A ^ B.
6. a. One can use a Venn diagram to get a better grasp of the sets
involved. The following is a representation of the set AU(Bn C).