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Hw5sol - Solutions to homework assignment 5

Solutions to homework assignment 5
Course

Formal Mathematical Reasoning And Writing (MATH 323)

18 Documents
Students shared 18 documents in this course
Academic year: 2012/2013
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University of Oregon

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Math 323: Homework 5 Solutions David Glickenstein February 15, 2013 5) Let A 8g B 6g and C 7g : Then f ) (B C) n A 7g 5) Which of the following enable me to conclude that x 2 A n B? Note that it is equivalent that x 2 or x 2 B: a) x 2 A B: Yes, for then x 2 A: b) x 2 B n A: Yes, since x 2 B: c) x 2 A B: Yes, since then x 2 B: d) x 2 A B and x 2 A: Yes, since the second statement in the conjunction is e) x 2 A B and x 2 A B: No, since it may be that x 2 A and x 2 B: For instance, let A 2g, B 3g and x 1: 5) Proposition 1 If U A B and A B then A U n B: Proof. Suppose we have A and B as stated. Then we must show that A U n B and U n B A: First, suppose that x 2 A: Since A B x 2 so x 2 U n B: Now suppose that y 2 U n B: Then y 2 B: Since U A B and y 2 we must have that y 2 A: 5) Proposition 2 1 1 n : n2N Proof. We show that 1 1 n : If x 2 n2N 1 1 n then there exists an integer n 1 n2N we see that x 2 : such that x 2 1 n1 : Since 1 n1 Conversely, if x 2 then x 2 1 n1 for n thus x 2 1 1 n : n2N Proposition 3 1 1 n f1g : n2N Proof. Notice that 1 2 1 1 n for any n so f1g 1 1 n : Now let x 1 be a real number. n2N If x 1 or x 2 then clearly x 2 : If 1 x then there exists a number y such that 1 y x: 1 n1 : Thus If we take n to be any integer such that n y 1 1 then 1 n1 y and so x 2 1 n1 : x2 n2N 5) Let B : x 2 R and x 2g : 1 Proposition 4 B 1): B2B Proof. If x 2 B, then there exists y 2 such that x 2 1): Conversely, suppose x 2: Then B2B x 2 so x 2 B: B2B Proposition 5 B f2g : B2B Proof. Certainly 2 2 for everyh y 2: Conversely, suppose x 2. If x then x 2 : If x i then x x (x 2 2) and so x 2 x (x 2 2) : Thus x 2 B: B2B Bonus solutions: 5) Let A 8g B 6g and C 7g : Then a) A B 6g b) A B 8g c) A n B 8g d) B C 6g e) B n C 4g f ) (B C) n A 7g g) (A C) n B ? h) C n (A B) f7g 5) Let B 1 n1 : n 2 N : B 2) : Proposition 6 B2B Proof. If x 2 2) then since 2) 2 B, x is in the union. Thus, 2) B: B2B Conversely, suppose x 2 1 1 n for some n: Then since 1 1 n x 2 2) : Thus B 2) : B2B Proposition 7 B ?: B2B Proof. We just need to show that there are no real numbers that are in 1 n1 for every n 2 N. This is clear if x 1 or x 2: Otherwise, x 2 2) : In this case, there exists n such that 1 n1 x: (This can be shown taking n an integer larger than (x 1) :) It follows that x is not in 1 n1 and hence not in the intersection. 5) Proposition 8 B T Aj j2J Proof. Suppose x 2 B T T (B Aj ) j2J Aj : Then x 2 B or for all j 2 x 2 Aj : If x 2 then x 2 B Aj for all j: T If x 2 Aj for all j 2 then x 2 B Aj for all j 2 J: Hence x 2 (B Aj ) : j2J T Conversely, suppose x 2 (B Aj ) : Then for all j 2 J, x 2 B or x 2 Aj : Thus either x 2 B or x 2 Aj j2J T for all j: Hence x 2 B Aj j2J j2J 2

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Hw5sol - Solutions to homework assignment 5

Course: Formal Mathematical Reasoning And Writing (MATH 323)

18 Documents
Students shared 18 documents in this course

University: The University of Arizona

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Math 323: Homework 5 Solutions
David Glickenstein
February 15, 2013
5.4) Let A=f2;4;6;8g; B =f3;4;5;6g;and C=f5;6;7g:Then
f) (B[C)nA=f3;5;7g
5.17) Which of the following enable me to conclude that x =2AnB? Note that it is equivalent that x =2A
or x2B:
a) x =2A[B: Yes, for then x =2A:
b) x2BnA: Yes, since x2B:
c) x2A\B: Yes, since then x2B:
d) x2A[Band x =2A: Yes, since the second statement in the conjunction is su¢cient.
e) x2A[Band x =2A\B: No, since it may be that x2Aand x =2B: For instance, let A=f1;2g,
B=f2;3g;and x= 1:
5.19)
Proposition 1 If U=A[Band A\B=?;then A=UnB:
Proof. Suppose we have Aand Bas stated. Then we must show that AUnBand UnBA: First,
suppose that x2A: Since A\B=?; x =2B; so x2UnB: Now suppose that y2UnB: Then y =2B: Since
U=A[Band y2U; we must have that y2A:
5.25a)
Proposition 2 [
n2N1;1 + 1
n= [1;2] :
Proof. We …rst show that [
n2N1;1 + 1
n[1;2] :If x2[
n2N1;1 + 1
n;then there exists an integer n1
such that x21;1 + 1
n:Since 1 + 1
n2;we see that x2[1;2] :
Conversely, if x2[1;2] ;then x21;1 + 1
nfor n= 1;thus x2[
n2N1;1 + 1
n:
Proposition 3 \
n2N1;1 + 1
n=f1g:
Proof. Notice that 121;1 + 1
nfor any n1;so f1g \
n2N1;1 + 1
n:Now let x6= 1 be a real number.
If x < 1or x > 2then clearly x =2[1;2] :If 1< x 2;then there exists a number ysuch that 1< y < x:
If we take nto be any integer such that n > 1
y1;then 1 + 1
n< y < x; and so x =21;1 + 1
n:Thus
x =2\
n2N1;1 + 1
n:
5.25c) Let B=f[2; x] : x2Rand x > 2g:
1

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