Chapter 2.2, Problem 13E

a.

To determine

To list:ordered pairs in $A\times B$ and $B\times A$

Answer to Problem 13E

  $A\times B=\left\{\begin{array}{l}\left(1,a\right),\left(1,e\right),\left(1,k\right),\left(1,n\right),\left(1,r\right),\\ \left(3,a\right),\left(3,e\right),\left(3,k\right),\left(3,n\right),\left(3,r\right),\\ \left(5,a\right),\left(5,e\right),\left(5,k\right),\left(5,n\right),\left(5,r\right)\end{array}\right\}$

  $B\times A=\left\{\begin{array}{l}\left(a,1\right),\left(e,1\right),\left(k,1\right),\left(n,1\right),\left(r,1\right),\\ \left(a,3\right),\left(e,3\right),\left(k,3\right),\left(n,3\right),\left(r,3\right),\\ \left(a,5\right),\left(e,5\right),\left(k,5\right),\left(n,5\right),\left(r,5\right)\end{array}\right\}$

Explanation of Solution

Given:

  $A=\left\{1,3,5\right\}$

  $B=\left\{a,e,k,n,r\right\}$

Definition Used:

Cross product of two sets A and B:

  $A\times B=\left\{\left(a,b\right)\text{: }a\in A\text{ and }b\in B\right\}$

Given $A=\left\{1,3,5\right\}$ and $B=\left\{a,e,k,n,r\right\}$

Then by definition of cross product,

  $A\times B=\left\{\begin{array}{l}\left(1,a\right),\left(1,e\right),\left(1,k\right),\left(1,n\right),\left(1,r\right),\\ \left(3,a\right),\left(3,e\right),\left(3,k\right),\left(3,n\right),\left(3,r\right),\\ \left(5,a\right),\left(5,e\right),\left(5,k\right),\left(5,n\right),\left(5,r\right)\end{array}\right\}$

And

  $B\times A=\left\{\begin{array}{l}\left(a,1\right),\left(e,1\right),\left(k,1\right),\left(n,1\right),\left(r,1\right),\\ \left(a,3\right),\left(e,3\right),\left(k,3\right),\left(n,3\right),\left(r,3\right),\\ \left(a,5\right),\left(e,5\right),\left(k,5\right),\left(n,5\right),\left(r,5\right)\end{array}\right\}$

b.

To determine

To list: ordered pairs in $A\times B$ and $B\times A$

Answer to Problem 13E

  $A\times B=\left\{\begin{array}{l}\left(1,q\right),\left(2,q\right),\left(\left\{1,2\right\},q\right),\\ \left(1,\left\{t\right\}\right),\left(2,\left\{t\right\}\right),\left(\left\{1,2\right\},\left\{t\right\}\right),\\ \left(1,\pi \right),\left(2,\pi \right),\left(\left\{1,2\right\},\pi \right)\end{array}\right\}$

  $B\times A=\left\{\begin{array}{l}\left(q,1\right),\left(q,2\right),\left(q,\left\{1,2\right\}\right),\\ \left(\left\{t\right\},1\right),\left(\left\{t\right\},2\right),\left(\left\{t\right\},\left\{1,2\right\}\right),\\ \left(\pi ,1\right),\left(\pi ,2\right),\left(\pi ,\left\{1,2\right\}\right)\end{array}\right\}$

Explanation of Solution

Given:

  $A=\left\{1,2,\left\{1,2\right\}\right\}$

  $B=\left\{q,\left\{t\right\},\pi \right\}$

Definition Used:

Cross product of two sets A and B:

  $A\times B=\left\{\left(a,b\right)\text{: }a\in A\text{ and }b\in B\right\}$

Given $A=\left\{1,2,\left\{1,2\right\}\right\}$ and $B=\left\{q,\left\{t\right\},\pi \right\}$

Then by definition of cross product,

  $A\times B=\left\{\begin{array}{l}\left(1,q\right),\left(2,q\right),\left(\left\{1,2\right\},q\right),\\ \left(1,\left\{t\right\}\right),\left(2,\left\{t\right\}\right),\left(\left\{1,2\right\},\left\{t\right\}\right),\\ \left(1,\pi \right),\left(2,\pi \right),\left(\left\{1,2\right\},\pi \right)\end{array}\right\}$

And

  $B\times A=\left\{\begin{array}{l}\left(q,1\right),\left(q,2\right),\left(q,\left\{1,2\right\}\right),\\ \left(\left\{t\right\},1\right),\left(\left\{t\right\},2\right),\left(\left\{t\right\},\left\{1,2\right\}\right),\\ \left(\pi ,1\right),\left(\pi ,2\right),\left(\pi ,\left\{1,2\right\}\right)\end{array}\right\}$

c.

To determine

To list: ordered pairs in $A\times B$ and $B\times A$

Answer to Problem 13E

  $A\times B=\left\{\begin{array}{l}\left(\varnothing ,\left(\varnothing ,\{\varnothing \}\right)\right),\left(\varnothing ,\{\varnothing \}\right),\left(\varnothing ,\left(\{\varnothing \},\varnothing \right)\right),\\ \left(\{\varnothing \},\left(\varnothing ,\{\varnothing \}\right)\right),\left(\{\varnothing \},\{\varnothing \}\right),\left(\{\varnothing \},\left(\{\varnothing \},\varnothing \right)\right),\\ \left(\left\{\varnothing ,\{\varnothing \}\right\},\left(\varnothing ,\{\varnothing \}\right)\right),\left(\left\{\varnothing ,\{\varnothing \}\right\},\{\varnothing \}\right),\left(\left\{\varnothing ,\{\varnothing \}\right\},\left(\{\varnothing \},\varnothing \right)\right)\end{array}\right\}$

  $B\times A=\left\{\begin{array}{l}\left(\left(\varnothing ,\{\varnothing \}\right),\varnothing \right),\left(\{\varnothing \},\{\varnothing \}\right),\left(\left(\{\varnothing \},\varnothing \right),\left\{\varnothing ,\{\varnothing \}\right\}\right),\\ \left(\left(\varnothing ,\{\varnothing \}\right),\varnothing \right),\left(\{\varnothing \},\{\varnothing \}\right),\left(\left(\{\varnothing \},\varnothing \right),\left\{\varnothing ,\{\varnothing \}\right\}\right),\\ \left(\left(\varnothing ,\{\varnothing \}\right),\varnothing \right),\left(\{\varnothing \},\{\varnothing \}\right),\left(\left(\{\varnothing \},\varnothing \right),\left\{\varnothing ,\{\varnothing \}\right\}\right)\end{array}\right\}$

Explanation of Solution

Given:

  $A=\left\{\varnothing ,\{\varnothing \},\left\{\varnothing ,\{\varnothing \}\right\}\right\}$

  $B=\left\{\left(\varnothing ,\{\varnothing \}\right),\{\varnothing \},\left(\{\varnothing \},\varnothing \right)\right\}$

Definition Used:

Cross product of two sets A and B:

  $A\times B=\left\{\left(a,b\right)\text{: }a\in A\text{ and }b\in B\right\}$

Given $A=\left\{\varnothing ,\{\varnothing \},\left\{\varnothing ,\{\varnothing \}\right\}\right\}$ and $B=\left\{\left(\varnothing ,\{\varnothing \}\right),\{\varnothing \},\left(\{\varnothing \},\varnothing \right)\right\}$

Then by definition of cross product,

  $A\times B=\left\{\begin{array}{l}\left(\varnothing ,\left(\varnothing ,\{\varnothing \}\right)\right),\left(\varnothing ,\{\varnothing \}\right),\left(\varnothing ,\left(\{\varnothing \},\varnothing \right)\right),\\ \left(\{\varnothing \},\left(\varnothing ,\{\varnothing \}\right)\right),\left(\{\varnothing \},\{\varnothing \}\right),\left(\{\varnothing \},\left(\{\varnothing \},\varnothing \right)\right),\\ \left(\left\{\varnothing ,\{\varnothing \}\right\},\left(\varnothing ,\{\varnothing \}\right)\right),\left(\left\{\varnothing ,\{\varnothing \}\right\},\{\varnothing \}\right),\left(\left\{\varnothing ,\{\varnothing \}\right\},\left(\{\varnothing \},\varnothing \right)\right)\end{array}\right\}$

And

  $B\times A=\left\{\begin{array}{l}\left(\left(\varnothing ,\{\varnothing \}\right),\varnothing \right),\left(\{\varnothing \},\{\varnothing \}\right),\left(\left(\{\varnothing \},\varnothing \right),\left\{\varnothing ,\{\varnothing \}\right\}\right),\\ \left(\left(\varnothing ,\{\varnothing \}\right),\varnothing \right),\left(\{\varnothing \},\{\varnothing \}\right),\left(\left(\{\varnothing \},\varnothing \right),\left\{\varnothing ,\{\varnothing \}\right\}\right),\\ \left(\left(\varnothing ,\{\varnothing \}\right),\varnothing \right),\left(\{\varnothing \},\{\varnothing \}\right),\left(\left(\{\varnothing \},\varnothing \right),\left\{\varnothing ,\{\varnothing \}\right\}\right)\end{array}\right\}$

d.

To determine

To list: ordered pairs in $A\times B$ and $B\times A$

Answer to Problem 13E

  $A\times B=\left\{\left(\left(2,4\right),\left(4,1\right)\right),\left(\left(3,1\right),\left(4,1\right)\right),\left(\left(2,4\right),\left(2,3\right)\right),\left(\left(3,1\right),\left(2,3\right)\right)\right\}$

  $B\times A=\left\{\left(\left(4,1\right),\left(2,4\right)\right),\left(\left(2,3\right),\left(2,4\right)\right),\left(\left(4,1\right),\left(3,1\right)\right),\left(\left(2,3\right),\left(3,1\right)\right)\right\}$

Explanation of Solution

Given:

  $A=\left\{\left(2,4\right),\left(3,1\right)\right\}$

  $B=\left\{\left(4,1\right),\left(2,3\right)\right\}$

Definition Used:

Cross product of two sets A and B:

  $A\times B=\left\{\left(a,b\right)\text{: }a\in A\text{ and }b\in B\right\}$

Given $A=\left\{\left(2,4\right),\left(3,1\right)\right\}$ and $B=\left\{\left(4,1\right),\left(2,3\right)\right\}$

Then by definition of cross product,

  $A\times B=\left\{\left(\left(2,4\right),\left(4,1\right)\right),\left(\left(3,1\right),\left(4,1\right)\right),\left(\left(2,4\right),\left(2,3\right)\right),\left(\left(3,1\right),\left(2,3\right)\right)\right\}$

And

  $B\times A=\left\{\left(\left(4,1\right),\left(2,4\right)\right),\left(\left(2,3\right),\left(2,4\right)\right),\left(\left(4,1\right),\left(3,1\right)\right),\left(\left(2,3\right),\left(3,1\right)\right)\right\}$