Math Problem Statement

Solve it

Solve it

Solution

Let's solve the questions step by step:


1. Verify the Set Theory Properties

We need to verify:

  1. (AB)C=ACBC(A \cap B)^C = A^C \cup B^C
  2. (AB)C=ACBC(A \cup B)^C = A^C \cap B^C

Given Sets:

  • A={1,3,5}A = \{1, 3, 5\},
  • B={2,4,6,8}B = \{2, 4, 6, 8\},
  • C={2,5,10}C = \{2, 5, 10\},
  • U={1,2,3,4,5,6,7,8,9,10}U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} (Universal Set).

Step 1: Verification of (AB)C=ACBC(A \cap B)^C = A^C \cup B^C:

  1. Compute ABA \cap B:
    AB={}A \cap B = \{\} (no common elements).

  2. Compute (AB)C(A \cap B)^C:
    (AB)C=U(AB)=U={1,2,3,4,5,6,7,8,9,10}(A \cap B)^C = U - (A \cap B) = U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}.

  3. Compute ACA^C:
    AC=UA={2,4,6,7,8,9,10}A^C = U - A = \{2, 4, 6, 7, 8, 9, 10\}.

  4. Compute BCB^C:
    BC=UB={1,3,5,7,9,10}B^C = U - B = \{1, 3, 5, 7, 9, 10\}.

  5. Compute ACBCA^C \cup B^C:
    ACBC={2,4,6,7,8,9,10}{1,3,5,7,9,10}=U.A^C \cup B^C = \{2, 4, 6, 7, 8, 9, 10\} \cup \{1, 3, 5, 7, 9, 10\} = U.

Since (AB)C=U(A \cap B)^C = U and ACBC=UA^C \cup B^C = U, the property holds.

Step 2: Verification of (AB)C=ACBC(A \cup B)^C = A^C \cap B^C:

  1. Compute ABA \cup B:
    AB={1,2,3,4,5,6,8}A \cup B = \{1, 2, 3, 4, 5, 6, 8\}.

  2. Compute (AB)C(A \cup B)^C:
    (AB)C=U(AB)={7,9,10}(A \cup B)^C = U - (A \cup B) = \{7, 9, 10\}.

  3. Compute ACBCA^C \cap B^C:
    ACBC={2,4,6,7,8,9,10}{1,3,5,7,9,10}={7,9,10}A^C \cap B^C = \{2, 4, 6, 7, 8, 9, 10\} \cap \{1, 3, 5, 7, 9, 10\} = \{7, 9, 10\}.

Since (AB)C={7,9,10}(A \cup B)^C = \{7, 9, 10\} and ACBC={7,9,10}A^C \cap B^C = \{7, 9, 10\}, the property holds.


2. Functions f(x)=x+1f(x) = x + 1, g(y)=2yg(y) = 2y

We are tasked to find fg(x)f \circ g(x) and gf(x)g \circ f(x).

Step 1: Compute fg(x)f \circ g(x):

(fg)(x)=f(g(x))=f(2x)=2x+1.(f \circ g)(x) = f(g(x)) = f(2x) = 2x + 1.

Step 2: Compute gf(x)g \circ f(x):

(gf)(x)=g(f(x))=g(x+1)=2(x+1)=2x+2.(g \circ f)(x) = g(f(x)) = g(x + 1) = 2(x + 1) = 2x + 2.

So:

  • fg(x)=2x+1f \circ g(x) = 2x + 1,
  • gf(x)=2x+2g \circ f(x) = 2x + 2.

3. Adjacency Matrix

The problem involves finding the adjacency matrix for two graphs:

First Graph:

Vertices: V1,V2,V3,V4,V5V_1, V_2, V_3, V_4, V_5

Edges (connections):

  • V1V2V_1 \leftrightarrow V_2,
  • V1V3V_1 \leftrightarrow V_3,
  • V1V4V_1 \leftrightarrow V_4,
  • V3V4V_3 \leftrightarrow V_4,
  • V5V_5 is isolated.

Adjacency matrix:

0 & 1 & 1 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$ #### **Second Graph:** Vertices: $$V_1, V_2, V_3, V_4$$ Edges (connections): - $$V_1 \leftrightarrow V_2$$, - $$V_1 \leftrightarrow V_4$$, - $$V_2 \leftrightarrow V_3$$, - $$V_3 \leftrightarrow V_4$$. Adjacency matrix: $$\begin{bmatrix} 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \end{bmatrix}$$ --- Would you like me to explain any specific part in more detail? --- ### **Relative Questions:** 1. How does De Morgan’s Laws apply to complement operations in sets? 2. What are the differences between $$f \circ g(x)$$ and $$g \circ f(x)$$ in function composition? 3. How do you interpret an adjacency matrix for an undirected graph? 4. What is the significance of isolated vertices in a graph? 5. How would the adjacency matrix change if the graph was directed? **Tip**: Always double-check the elements of a set when performing union, intersection, or complement operations to avoid missing elements.

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Math Problem Analysis

Mathematical Concepts

Set Theory
Functions
Graph Theory
Adjacency Matrix
Complementation

Formulas

(A ∩ B)^C = A^C ∪ B^C
(A ∪ B)^C = A^C ∩ B^C
f(x) = x + 1
g(y) = 2y
f ∘ g(x) = 2x + 1
g ∘ f(x) = 2x + 2

Theorems

De Morgan’s Laws
Function Composition
Adjacency Matrix Representation

Suitable Grade Level

Grades 10-12