Chapter 2.3, Problem 1E

Find the order of the element 3 in each group.

1. $\left({\mathbb{Z}}_4,+\right)$
2. $\left({\mathbb{Z}}_5,+\right)$
3. $\left({\mathbb{Z}}_6,+\right)$
4. $\left({\mathbb{Z}}_8,+\right)$
5. $\left({\mathbb{Z}}_9,+\right)$
6. $\left({\text{U}}_5,\cdot \right)$
7. $\left({\text{U}}_7,\cdot \right)$
8. $\left({\text{U}}_{11},\cdot \right)$

To determine

Find the union and intersection of the sets.

Answer to Problem 1E

The union: ${\displaystyle \underset{A\subseteq A}{\cup }A=\{1,2,3,4,5,6,7,8\}}$

The intersection: ${\displaystyle \underset{A\subseteq A}{\cap }A=\{4,5\}}$

Explanation of Solution

Given :

It is given in the question that the set is $\text{A }=\left\{\left\{\text{1},\text{2},\text{3},\text{4},\text{5}\right\},\left\{\text{2},\text{3},\text{4},\text{5},\text{6}\right\},\left\{\text{3},\text{4},\text{5},\text{6},\text{7}\right\},\left\{\text{4},\text{5},\text{6},\text{7},\text{8}\right\}\right\}$ .

Concept Used:

In this we have to use the concept of sets and induction.

Calculation: The union: ${\displaystyle \underset{A\subseteq A}{\cup }A=\{1,2,3,4,5,6,7,8\}}$

The intersection: ${\displaystyle \underset{A\subseteq A}{\cap }A=\{4,5\}}$

Conclusion:

The union: ${\displaystyle \underset{A\subseteq A}{\cup }A=\{1,2,3,4,5,6,7,8\}}$

The intersection: ${\displaystyle \underset{A\subseteq A}{\cap }A=\{4,5\}}$

To determine

Find the union and intersection of the sets.

Answer to Problem 1E

The union: ${\displaystyle \underset{A\subseteq A}{\cup }A=\{1,2,3,4,5,6,7,8,9,10,11,12,13\}}$

The intersection: ${\displaystyle \underset{A\subseteq A}{\cap }A=\left\{\varphi \right\}}$

Explanation of Solution

Given :

It is given in the question that the set is $\text{A}=\left\{\left\{\text{1}.\text{3}.\text{5}\right\},\left\{\text{2},\text{4},\text{6}\right\},\left\{\text{7},\text{9},\text{11},\text{13}\right\},\left\{\text{8},\text{1}0,\text{12}\right\}\right\}.$

Concept Used:

In this we have to use the concept of sets and induction.

Calculation:

The union: ${\displaystyle \underset{A\subseteq A}{\cup }A=\{1,2,3,4,5,6,7,8,9,10,11,12,13\}}$

The intersection: ${\displaystyle \underset{A\subseteq A}{\cap }A=\left\{\varphi \right\}}$

Conclusion:

The union: ${\displaystyle \underset{A\subseteq A}{\cup }A=\{1,2,3,4,5,6,7,8,9,10,11,12,13\}}$

The intersection: ${\displaystyle \underset{A\subseteq A}{\cap }A=\left\{\varphi \right\}}$

To determine

Find the union and intersection of the set. A = $\{A_n:n\in N\}$ ,where $A_n=\{5n,5n+1,5n+2,\mathrm{......},6n\}$ for each natural number n .

Answer to Problem 1E

The union: ${\displaystyle \underset{n\in N}{\cup }{A}_n=\{x=5n+m;n\in N,0\le m\le n\}}$

The intersection: ${\displaystyle \underset{A\subseteq A}{\cap }{A}_n=\left\{\varphi \right\}}$

Explanation of Solution

Given :

It is given in the question that the set is A = $\{A_n:n\in N\}$ ,where $A_n=\{5n,5n+1,5n+2,\mathrm{......},6n\}$ for each natural number n .

Concept Used:

In this we have to use the concept of sets and induction.

Calculation: The union: ${\displaystyle \underset{n\in N}{\cup }{A}_n=\{x=5n+m;n\in N,0\le m\le n\}}$

The intersection: ${\displaystyle \underset{A\subseteq A}{\cap }{A}_n=\left\{\varphi \right\}}$

Conclusion:

The union: ${\displaystyle \underset{n\in N}{\cup }{A}_n=\{x=5n+m;n\in N,0\le m\le n\}}$

The intersection: ${\displaystyle \underset{A\subseteq A}{\cap }{A}_n=\left\{\varphi \right\}}$

To determine

Find the union and intersection of the set.

Answer to Problem 1E

The union: ${\displaystyle \underset{n\in N}{\cup }{B}_n=N}$

The intersection: ${\displaystyle \underset{A\subseteq A}{\cap }{B}_n=\left\{\varphi \right\}}$

Explanation of Solution

Given :

It is given in the question that the set is B = $\{B_n:n\in N\}$ ,where $B_n=N-\{1,2,3,\mathrm{......},n\}$ for each natural number n .

Concept Used:

In this we have to use the concept of sets and induction.

Calculation:

The union: ${\displaystyle \underset{n\in N}{\cup }{B}_n=N}$

The intersection: ${\displaystyle \underset{A\subseteq A}{\cap }{B}_n=\left\{\varphi \right\}}$

Conclusion:

The union: ${\displaystyle \underset{n\in N}{\cup }{B}_n=N}$

The intersection: ${\displaystyle \underset{A\subseteq A}{\cap }{B}_n=\left\{\varphi \right\}}$

To determine

Find the union and intersection of the set. A is the family of all sets of integers that contain $10$ .

Answer to Problem 1E

The union: ${\displaystyle \underset{n\in N}{\cup }A=Z}$

The intersection: ${\displaystyle \underset{A\subseteq A}{\cap }A=\left\{10\right\}}$

Explanation of Solution

Given :

It is given in the question that the set is A is the family of all sets of integer that contain $10$ .

Concept Used:

In this we have to use the concept of sets and induction.

Calculation:

The union: ${\displaystyle \underset{n\in N}{\cup }A=Z}$

The intersection: ${\displaystyle \underset{A\subseteq A}{\cap }A=\left\{10\right\}}$

Conclusion:

The union: ${\displaystyle \underset{n\in N}{\cup }A=Z}$

The intersection: ${\displaystyle \underset{A\subseteq A}{\cap }A=\left\{10\right\}}$

To determine

Find the union and intersection of the set.

Answer to Problem 1E

The union: ${\displaystyle \underset{A\subset A}{\overset{10}{\cup }}{A}_n}=\{1,2,3,\mathrm{.....},19\}$

The intersection: ${\displaystyle \underset{n=1}{\overset{10}{\cap }}{A}_n}=\varphi$

Explanation of Solution

Given :

It is given in the question that the set is A = $\{A_n:n\in (1,2,3,\mathrm{....},10)\}$ ,where $A_1=\left\{1\right\},A_2=\{2,3\},A_3=\{3,4,5\},\mathrm{....}{A}_{10}=\{10,11,\mathrm{....},19\}$

Concept Used:

In this we have to use the concept of sets and induction.

Calculation:

The union: ${\displaystyle \underset{A\subset A}{\overset{10}{\cup }}{A}_n}=\{1,2,3,\mathrm{.....},19\}$

The intersection: ${\displaystyle \underset{n=1}{\overset{10}{\cap }}{A}_n}=\varphi$

Conclusion:

The union: ${\displaystyle \underset{A\subset A}{\overset{10}{\cup }}{A}_n}=\{1,2,3,\mathrm{.....},19\}$

The intersection: ${\displaystyle \underset{n=1}{\overset{10}{\cap }}{A}_n}=\varphi$

To determine

Find the union and intersection of the set.

Answer to Problem 1E

The union: ${\displaystyle \underset{n\in N}{\cup }0,\frac{1}{n}}=(0,1)$

The intersection: ${\displaystyle \underset{n\in N}{\cap }(0,\frac{1}{n}})=\varphi$

Explanation of Solution

Given :

It is given in the question that the set is A = $\{A_n:n\in N\}$ ,where $A_n=\{0,\frac{1}{n}\}$ , for each natural number n .

Concept Used:

In this we have to use the concept of sets and induction.

Calculation:

The union: ${\displaystyle \underset{n\in N}{\cup }0,\frac{1}{n}}=(0,1)$

The intersection: ${\displaystyle \underset{n\in N}{\cap }(0,\frac{1}{n}})=\varphi$

Conclusion:

The union: ${\displaystyle \underset{n\in N}{\cup }0,\frac{1}{n}}=(0,1)$

The intersection: ${\displaystyle \underset{n\in N}{\cap }(0,\frac{1}{n}})=\varphi$

To determine

Find the union and intersection of the set.

Answer to Problem 1E

The union: ${\displaystyle \underset{r\in (0,\infty )}{\cup }(-\pi ,r)}=(-\pi ,\infty )$

The intersection: ${\displaystyle \underset{r\in (0,\infty )}{\cap }(}-\pi ,r)=(-\pi ,0)$

Explanation of Solution

Given :

It is given in the question that the set is A = $\{A_r:r\in (0,\infty )\}$ ,where $A_r=\{-\pi ,r\}$ for each $r\in (0,\infty )$ .

Concept Used:

In this we have to use the concept of sets and induction.

Calculation:

The union: ${\displaystyle \underset{r\in (0,\infty )}{\cup }(-\pi ,r)}=(-\pi ,\infty )$

The intersection: ${\displaystyle \underset{r\in (0,\infty )}{\cap }(}-\pi ,r)=(-\pi ,0)$

Conclusion:

The union: ${\displaystyle \underset{r\in (0,\infty )}{\cup }(-\pi ,r)}=(-\pi ,\infty )$

The intersection: ${\displaystyle \underset{r\in (0,\infty )}{\cap }(}-\pi ,r)=(-\pi ,0)$

To determine

Find the union and intersection of the set.

Answer to Problem 1E

Explanation of Solution

Given :

It is given in the question that the set is A = $\{A_r:r\in R\}$ ,where $A_r=[\left|r\right|,2\left|r\right|+1]$ for each $r\in R$ .

Concept Used:

In this we have to use the concept of sets and induction.

Calculation:

The union: $\underset{r\in R}{\cup }\left[\left|r\right|,2\left|r\right|+1\right)=\left[0,\infty \right)$

The intersection: $\underset{r\in R}{\cap }\left[\left|r\right|,2\left|r\right|+1\right)=\left[0,1\right)$

Conclusion:

To determine

Find the union and intersection of the set.

Answer to Problem 1E

The union: $\underset{n\in N}{\cup }nZ=Z$

The intersection: $\underset{n\in N}{\cap }nZ=0$

Explanation of Solution

Given :

It is given in the question that the set is M = $\{nZ:n\in N\}$ ,where $nZ=\{\mathrm{....},-3n,-2n,-n,0,n,2n,3n,\mathrm{.....}\}$ for each $n\in N$ .

Concept Used:

In this we have to use the concept of sets and induction.

Calculation:

The union: $\underset{n\in N}{\cup }nZ=Z$

The intersection: $\underset{n\in N}{\cap }nZ=0$

Conclusion:

The union: $\underset{n\in N}{\cup }nZ=Z$

The intersection: $\underset{n\in N}{\cap }nZ=0$

To determine

Find the union and intersection of the set.

Answer to Problem 1E

The union: $\underset{n=3}{\overset{\infty }{\cup }}\left[\frac{1}{n},2+\frac{1}{n}\right]=\left(0,\frac{7}{3}\right]$

The intersection: $\underset{n=3}{\overset{\infty }{\cap }}\left[\frac{1}{n},2+\frac{1}{n}\right]=\left[\frac{1}{3},2\right]$

Explanation of Solution

Given :

It is given in the question that the set is A = $\{A_n:n\ge 3\}$ ,where $A_n=[\frac{1}{n},2+\frac{1}{n}]$ for each $n\in N$

Concept Used:

In this we have to use the concept of sets and induction.

Calculation:

The union: $\underset{n=3}{\overset{\infty }{\cup }}\left[\frac{1}{n},2+\frac{1}{n}\right]=\left(0,\frac{7}{3}\right]$

The intersection: $\underset{n=3}{\overset{\infty }{\cap }}\left[\frac{1}{n},2+\frac{1}{n}\right]=\left[\frac{1}{3},2\right]$

Conclusion:

The union: $\underset{n=3}{\overset{\infty }{\cup }}\left[\frac{1}{n},2+\frac{1}{n}\right]=\left(0,\frac{7}{3}\right]$

The intersection: $\underset{n=3}{\overset{\infty }{\cap }}\left[\frac{1}{n},2+\frac{1}{n}\right]=\left[\frac{1}{3},2\right]$

To determine

Find the union and intersection of the set.

Answer to Problem 1E

The union: $\underset{n\in Z}{\cup }\left[n,n+1\right)=R$

The intersection: $\underset{n\in Z}{\cap }\left[n,n+1\right)=\cancel{0}$

Explanation of Solution

Given :

It is given in the question that the set is C = $\{C_n:n\in Z\}$ ,where $C_n=[n,n+1]$ for each $n\in Z$ .

Concept Used:

In this we have to use the concept of sets and induction.

Calculation:

The union: $\underset{n\in Z}{\cup }\left[n,n+1\right)=R$

The intersection: $\underset{n\in Z}{\cap }\left[n,n+1\right)=\cancel{0}$

Conclusion:

The union: $\underset{n\in Z}{\cup }\left[n,n+1\right)=R$

The intersection: $\underset{n\in Z}{\cap }\left[n,n+1\right)=\cancel{0}$

To determine

Find the union and intersection of the set.

Answer to Problem 1E

The union: $\underset{n\in Z}{\cup }\left(n,n+1\right)=R-Z$

The intersection: $\underset{n\in Z}{\cap }\left(n,n+1\right)=\cancel{0}$

Explanation of Solution

Given :

It is given in the question that the set is A = $\{A_n:n\in Z\}$ ,where $A_n=[n,n+1]$ for each $n\in Z$ .

Concept Used:

In this we have to use the concept of sets and induction.

Calculation:

The union: $\underset{n\in Z}{\cup }\left(n,n+1\right)=R-Z$

The intersection: $\underset{n\in Z}{\cap }\left(n,n+1\right)=\cancel{0}$

Conclusion:

The union: $\underset{n\in Z}{\cup }\left(n,n+1\right)=R-Z$

The intersection: $\underset{n\in Z}{\cap }\left(n,n+1\right)=\cancel{0}$

To determine

Find the union and intersection of the set.

Answer to Problem 1E

The union: $\underset{n\in N}{\cup }\left(-n,\frac{1}{n}\right)=\left(-\infty ,1\right]$

The intersection: $\underset{n\in N}{\cap }\left(-n,\frac{1}{n}\right)=\left(-1,0\right]$

Explanation of Solution

Given :

It is given in the question that the set is D = $\{D_n:n\in N\}$ ,where $D_n=[-n,\frac{1}{n}]$ for $n\in N$ .

Concept Used:

In this we have to use the concept of sets and induction.

Calculation:

The union: $\underset{n\in N}{\cup }\left(-n,\frac{1}{n}\right)=\left(-\infty ,1\right]$

The intersection: $\underset{n\in N}{\cap }\left(-n,\frac{1}{n}\right)=\left(-1,0\right]$

Conclusion:

The union: $\underset{n\in N}{\cup }\left(-n,\frac{1}{n}\right)=\left(-\infty ,1\right]$

The intersection: $\underset{n\in N}{\cap }\left(-n,\frac{1}{n}\right)=\left(-1,0\right]$

To determine

Find the union and intersection of the set.

Answer to Problem 1E

The union: $\underset{pisprime}{\cup }pN=N-\left\{1\right\}$

The intersection: $\underset{pisprime}{\cap }pN=\cancel{0}$

Explanation of Solution

Given :

It is given in the question that the set is A = $\{pN:pisaprime\}$ ,where $pN=[np:n\in N]$ for each prime p .

Concept Used:

In this we have to use the concept of sets and induction.

Calculation:

The union: $\underset{pisprime}{\cup }pN=N-\left\{1\right\}$

The intersection: $\underset{pisprime}{\cap }pN=\cancel{0}$

Conclusion:

The union: $\underset{pisprime}{\cup }pN=N-\left\{1\right\}$

The intersection: $\underset{pisprime}{\cap }pN=\cancel{0}$

To determine

Find the union and intersection of the set.

Answer to Problem 1E

The union: $\underset{n\in Z}{\cup }\left\{\left(x,y\right):0\le x\le 1,0\le y\le x^n\right\}=\left\{\left(x,y\right):0\le x\le 1,0\le y\le x\right\}$

The intersection: $\underset{n\in Z}{\cap }\left\{\left(x,y\right):0\le x\le 1,0\le y\le x^n\right\}=\left\{\left(x,y\right):(0\le x\le 1,y=0)^(x=1,0\le y<1)\right\}$

Explanation of Solution

Given :

It is given in the question that the set is T = $\{T_n:n\in Z\}$ , where $T_n=\{(x,y\in R\times R:0\le x\le 1,0\le y\le x^n]$ for each $n\in Z$ .

Concept Used:

In this we have to use the concept of sets and induction.

Calculation: The union: $\underset{n\in Z}{\cup }\left\{\left(x,y\right):0\le x\le 1,0\le y\le x^n\right\}=\left\{\left(x,y\right):0\le x\le 1,0\le y\le x\right\}$

The intersection: $\underset{n\in Z}{\cap }\left\{\left(x,y\right):0\le x\le 1,0\le y\le x^n\right\}=\left\{\left(x,y\right):(0\le x\le 1,y=0)^(x=1,0\le y<1)\right\}$

Conclusion:

The union: $\underset{n\in Z}{\cup }\left\{\left(x,y\right):0\le x\le 1,0\le y\le x^n\right\}=\left\{\left(x,y\right):0\le x\le 1,0\le y\le x\right\}$

The intersection: $\underset{n\in Z}{\cap }\left\{\left(x,y\right):0\le x\le 1,0\le y\le x^n\right\}=\left\{\left(x,y\right):(0\le x\le 1,y=0)^(x=1,0\le y<1)\right\}$

To determine

Find the union and intersection of the set.

Answer to Problem 1E

The union: $\underset{n\in Z}{\cup }\left\{\left(x,y\right):0\le x\le 1,x^n\le y\le x^{\frac{1}{n}}\right\}=\left\{\left(x,y\right):0\le x\le 1,0\le y\le 1\right\}$

The intersection: $\underset{n\in Z}{\cap }\left\{\left(x,y\right):0\le x\le 1,x^n\le y\le x^{\frac{1}{n}}\right\}=\left\{\left(x,y\right):0\le x\le 1,y=x\right\}$

Explanation of Solution

Given :

It is given in the question that the set is V = $\{V_n:n\in N\}$ ,where $V_n=\{(x,y\in R\times R:0\le x\le 1,x^n\le y\le \sqrt[n]{x}]$ for each $n\in N$ .

Concept Used:

In this we have to use the concept of sets and induction.

Calculation: The union: $\underset{n\in Z}{\cup }\left\{\left(x,y\right):0\le x\le 1,x^n\le y\le x^{\frac{1}{n}}\right\}=\left\{\left(x,y\right):0\le x\le 1,0\le y\le 1\right\}$

The intersection: $\underset{n\in Z}{\cap }\left\{\left(x,y\right):0\le x\le 1,x^n\le y\le x^{\frac{1}{n}}\right\}=\left\{\left(x,y\right):0\le x\le 1,y=x\right\}$

Conclusion:

The union: $\underset{n\in Z}{\cup }\left\{\left(x,y\right):0\le x\le 1,x^n\le y\le x^{\frac{1}{n}}\right\}=\left\{\left(x,y\right):0\le x\le 1,0\le y\le 1\right\}$

The intersection: $\underset{n\in Z}{\cap }\left\{\left(x,y\right):0\le x\le 1,x^n\le y\le x^{\frac{1}{n}}\right\}=\left\{\left(x,y\right):0\le x\le 1,y=x\right\}$

To determine

Find the union and intersection of the set.

Answer to Problem 1E

The union: $\underset{n\in N}{\cup }R-\left[0,n\right]=R-\left[0,1\right]$

The intersection: $\underset{n\in N}{\cap }R-\left[0,n\right]=R\_$ ( $R\_$ , all negative real number)

Explanation of Solution

Given :

It is given in the question that the set is E = $\{E_n:n\in N\}$ , where $E_n=R-\{0,n\}$ for each natural number n .

Concept Used:

In this we have to use the concept of sets and induction.

Calculation:

The union: $\underset{n\in N}{\cup }R-\left[0,n\right]=R-\left[0,1\right]$

The intersection: $\underset{n\in N}{\cap }R-\left[0,n\right]=R\_$ ( $R\_$ , all negative real number)

Conclusion:

The union: $\underset{n\in N}{\cup }R-\left[0,n\right]=R-\left[0,1\right]$

The intersection: $\underset{n\in N}{\cap }R-\left[0,n\right]=R\_$ ( $R\_$ , all negative real number)