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## Here's the Solution to this Question

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. 𝐵 ∪ 𝐶 = {0, 2, 4, 6, b, c, d}

2. $𝐵 \times 𝐶=\{(0,b),(0,c),(0,d),(2,b),(2,c),(2,d),(4,b),(4,c),(4,d),(6,b),(6,c),(6,d)\}$

Now, $A-B\times C=\{-1, 1, 2, 4\}$

3. 𝑃𝑜𝑤𝑒𝑟 𝑠𝑒𝑡 𝑜𝑓 𝐶$=\{\phi,\{b\}, \{c\},\{d\},\{b,c\},\{c,d\},\{b,d\},\{b,c,d\}\}$

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

$|P(B)|=2^4=16$

(B):

$S_n=2n-5,n\ge-1$

$\Sigma_{-1}^1 S_i=S_{-1}+S_0+S_1 \\=2(-1)-5+2(0)-5+2(1)-5 \\=-2+0+2-15 \\=-15$

(C):

Consider X = {-3, -2, -1, 0, 1} defined by (x,y) ∈ R if x ≥ y.

1.

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

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

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

3. Digraph of R: 4.

Reflexive:

It is clearly reflexive as $(a,a)\in R, \forall a\in X$

Symmetric:

It is clearly not symmetric as $a\ge b$ but $b\ge a$ is not true, $\forall a,b \in X$

Moreover, $(0,-2)\in R$ but $(-2,0)\not\in R$

Transitive:

$a\ge b, b\ge c \Rightarrow a\ge c$ which is true $\forall a,b,c \in X$

Hence, it is transitive.