Question #183099

III. PROBLEM SOLVING.

A. SET. Let A, B and C are sets and U be universal set.

U = {-1, 0, 1, 2, 3, 4, 5, 6, a, b, c, d, e}

A = {-1, 1, 2, 4}

B = {0, 2, 4, 6}

C = {b, c, d}

Find for the following. Show complete solutions.

1. 𝐡 βˆͺ 𝐢

2. 𝐴 βˆ’ 𝐡 π‘₯ 𝐢

3. π‘ƒπ‘œπ‘€π‘’π‘Ÿ 𝑠𝑒𝑑 π‘œπ‘“ 𝐢

4. |𝑃(𝐡)|

B. SEQUENCES. Consider the sequence {Sn} defined by Sn = 2n – 5, where 𝒏 β‰₯ βˆ’πŸ. Find for:

1. βˆ‘1𝑖=βˆ’1 𝑆𝑖

2. ∏ 𝑆𝑖 4𝑖=2

C. RELATION. Consider X = {-3, -2, -1, 0, 1} defined by (x,y) ∈ R if x β‰₯ y.

Find for:

1. Elements of R (3 pts)

2. Domain and Range of R (2 pts)

3. Draw the digraph (3 pts)

4. Identify the properties of R (2pts)


1
Expert's answer
2021-05-07T09:28:39-0400

Solution:

(A):

U = {-1, 0, 1, 2, 3, 4, 5, 6, a, b, c, d, e}

A = {-1, 1, 2, 4}

B = {0, 2, 4, 6}

C = {b, c, d}

1. 𝐡 βˆͺ 𝐢 = {0, 2, 4, 6, b, c, d}

2. 𝐡×𝐢={(0,b),(0,c),(0,d),(2,b),(2,c),(2,d),(4,b),(4,c),(4,d),(6,b),(6,c),(6,d)}𝐡 \times 𝐢=\{(0,b),(0,c),(0,d),(2,b),(2,c),(2,d),(4,b),(4,c),(4,d),(6,b),(6,c),(6,d)\}

Now, Aβˆ’BΓ—C={βˆ’1,1,2,4}A-B\times C=\{-1, 1, 2, 4\}

3. π‘ƒπ‘œπ‘€π‘’π‘Ÿ 𝑠𝑒𝑑 π‘œπ‘“ 𝐢={Ο•,{b},{c},{d},{b,c},{c,d},{b,d},{b,c,d}}=\{\phi,\{b\}, \{c\},\{d\},\{b,c\},\{c,d\},\{b,d\},\{b,c,d\}\}

4. |𝑃(𝐡)|=2n=2^n , where n is the number of elements in set B.

∣P(B)∣=24=16|P(B)|=2^4=16

(B):

Sn=2nβˆ’5,nβ‰₯βˆ’1S_n=2n-5,n\ge-1

Ξ£βˆ’11Si=Sβˆ’1+S0+S1=2(βˆ’1)βˆ’5+2(0)βˆ’5+2(1)βˆ’5=βˆ’2+0+2βˆ’15=βˆ’15\Sigma_{-1}^1 S_i=S_{-1}+S_0+S_1 \\=2(-1)-5+2(0)-5+2(1)-5 \\=-2+0+2-15 \\=-15

Ξ 24Si=S2Γ—S3Γ—S4=(4βˆ’5)(6βˆ’5)(8βˆ’5)=(βˆ’1)(1)(3)=βˆ’3\Pi_2^4S_i=S_2\times S_3 \times S_4=(4-5)(6-5)(8-5)=(-1)(1)(3)=-3

(C):

Consider X = {-3, -2, -1, 0, 1} defined by (x,y) βˆˆ R if x β‰₯ y.

1.

R={(βˆ’3,βˆ’3),(βˆ’2,βˆ’2),(βˆ’1,βˆ’1),(0,0),(1,1),(βˆ’2,βˆ’3),(βˆ’1,βˆ’3),(0,βˆ’3),(1,βˆ’3),(βˆ’1,βˆ’2),(0,βˆ’2),(1,βˆ’2),(0,βˆ’1),(1,βˆ’1)(1,0)}R=\{(-3,-3),(-2,-2),(-1,-1),(0,0),(1,1),(-2,-3),(-1,-3),\\(0,-3),(1,-3),(-1,-2),(0,-2),(1,-2),(0,-1),(1,-1)(1,0)\}

2. Domain of R={βˆ’3,βˆ’2,βˆ’1,0,1}=\{-3,-2,-1,0,1\}

And range of R={βˆ’3,βˆ’2,βˆ’1,0,1}=\{-3,-2,-1,0,1\}

3. Digraph of R:



4.

Reflexive:

It is clearly reflexive as (a,a)∈R,βˆ€a∈X(a,a)\in R, \forall a\in X

Symmetric:

It is clearly not symmetric as aβ‰₯ba\ge b but bβ‰₯ab\ge a is not true, βˆ€a,b∈X\forall a,b \in X

Moreover, (0,βˆ’2)∈R(0,-2)\in R but (βˆ’2,0)∉R(-2,0)\not\in R

Transitive:

aβ‰₯b,bβ‰₯cβ‡’aβ‰₯ca\ge b, b\ge c \Rightarrow a\ge c which is true βˆ€a,b,c∈X\forall a,b,c \in X

Hence, it is transitive.


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