Chapter 3.2, Problem 6E

a.

To determine

Prove that the relation is an equivalence relation.

Answer to Problem 6E

Equivalence Relation

  $\begin{array}{l}\left[0\right]=\boxed{Q}\\ \left[1\right]=\boxed{\left[0\right]}\\ \left[\sqrt{2}\right]=\boxed{\left\{\sqrt{2}+x:x\in Q\right\}}\end{array}$

Explanation of Solution

Given information:

The relation $SonR$ given by $xSyifx-y\in Q$ . Give the equivalence class of $0;of1;of\sqrt{2}$ .

Calculation:

Consider, the $SonR$ given by $xSyifx-y\in Q$ and the equivalence class of $0;of1;of\sqrt{2}$ .

Now check that the relation is reflexive, symmetric and transitive.

  $\begin{array}{l}\mathrm{Re}flexive:\\ \left(\forall x\in R\right)\left(x-x=0\in Q\iff xSx\right)\\ Symmetric:\\ \left(\forall x,y\in R\right)\left(xSy\iff x-y=\in Q\iff y-x\in Q\iff ySx\right)\\ Transitive:\\ \left(\forall x,y,z\in R\right)\left(xSy\wedge ySz\right)\iff \left(x-y=\in Q\wedge y-z=\in Q\right)\\ \iff \left(x-z=\left(x-y\right)-\left(z-y\right)\in Q\right)\iff \left(xSz\right)\end{array}$

The relation is reflexive, symmetric and transitive.

Hence, this is an equivalence relation.

Equivalence class of:

  $\begin{array}{l}\left[0\right]=\boxed{Q}\\ \left[1\right]=\boxed{\left[0\right]}\\ \left[\sqrt{2}\right]=\boxed{\left\{\sqrt{2}+x:x\in Q\right\}}\end{array}$

b.

To determine

Find an element of $106$ that is less than $50$ ; between $150and300$ ; greater than $1000$ . Also find three elements in the equivalence class $635$ .

Answer to Problem 6E

Equivalence Relation

  $\boxed{9}\in \left[106\right]\wedge 9\wedge 50$

  $\boxed{208}\in \left[106\right]\wedge 150<208<300$

  $\boxed{1007}\in \left[106\right]\wedge 1007>1000$

  $\begin{array}{l}\boxed{32}\in \left[635\right]\wedge 32<50\\ \boxed{236}\in \left[635\right]\wedge 150<236<300\\ \boxed{1037}\in \left[635\right]\wedge 1037>1000\end{array}$

Explanation of Solution

Given information:

The relation $RonNgivenbymRn$ iff $mandn$ have the same digit in the tens places.

Calculation:

Now check that the relation is reflexive, symmetric and transitive.

  $\begin{array}{l}\mathrm{Re}flexive:\\ \left(\forall x\right)\left(xhasthesamedigitintenplaceasx\right)\\ Symmetric:\\ \left(\forall x,y\right)\left(\begin{array}{l}xhasthesamedigitintenplaceasy\\ thenyhasthesamedigitintenplaceasx\end{array}\right)\\ Transitive:\\ \left(\forall x,y,z\right)\left(\begin{array}{l}xhasthesamedigitintenplaceasy\\ thenyhasthesamedigitintenplaceasz\\ thenxhasthesamedigitintenplaceasz\end{array}\right)\end{array}$

The relation is reflexive, symmetric and transitive.

Hence, this is an equivalence relation.

The element of $106$ that is less than $50$ :

  $\boxed{9}\in \left[106\right]\wedge 9\wedge 50$

The element of $106$ that lies between $150and300$ :

  $\boxed{208}\in \left[106\right]\wedge 150<208<300$

The element of $106$ that is greater than $1000$ :

  $\boxed{1007}\in \left[106\right]\wedge 1007>1000$

Three elements in the equivalence class $635$ :

  $\begin{array}{l}\boxed{32}\in \left[635\right]\wedge 32<50\\ \boxed{236}\in \left[635\right]\wedge 150<236<300\\ \boxed{1037}\in \left[635\right]\wedge 1037>1000\end{array}$

c.

To determine

Give the equivalence class of $3;of-\frac{2}{3};of0$ .

Answer to Problem 6E

  $\begin{array}{l}\left[3\right]=\boxed{\left\{3,\frac{1}{3}\right\}}\\ \left[-\frac{2}{3}\right]=\boxed{\left\{-\frac{2}{3},\frac{-3}{2}\right\}}\\ \left[0\right]=\boxed{\left\{0\right\}}\end{array}$

Explanation of Solution

Given information:

The relation $VonRgivenbyxVy$ iff $x=yorxy=1$ .

Calculation:

Now check that the relation is reflexive, symmetric and transitive.

  $\begin{array}{l}\mathrm{Re}flexive:\\ \left(\forall x\in R\right)\left(xVx\right)\\ Symmetric:\\ \left(\forall x,y\in R\right)\left(xVy\iff x=y\vee xy=1\iff yVx\right)\\ Transitive:\\ \left(\forall x,y,z\in R\right)\left(\begin{array}{l}\left(xVy\wedge yVz\right)\iff \left(x=y\vee xy=1\right)\wedge \left(y=zVzy=1\right)\iff \\ \left(x=y=z\right)\vee \left(xy=1\right)\vee \left(x=z\right)\iff xVz\end{array}\right)\end{array}$

The relation is reflexive, symmetric and transitive.

Hence, this is an equivalence relation.

The equivalence class of :

  $\begin{array}{l}\left[3\right]=\boxed{\left\{3,\frac{1}{3}\right\}}\\ \left[-\frac{2}{3}\right]=\boxed{\left\{-\frac{2}{3},\frac{-3}{2}\right\}}\\ \left[0\right]=\boxed{\left\{0\right\}}\end{array}$

d.

To determine

Name three elements in each of these classes: $7,10,72$ .

Answer to Problem 6E

  $\begin{array}{l}\left[\overline{7}\right]:\boxed{3,99,51}\\ \left[\overline{10}\right]:\boxed{6,198,102}\\ \left[\overline{72}\right]:\boxed{8,24,40}\end{array}$

Explanation of Solution

Given information:

On $N-\left\{1\right\}$ , the relation $R$ given by $aRb$ iff the prime factorizations of $aandb$ have the same number of $2\text{'}s$ . For example, $48R80$ because $48=2^4\cdot 3and80=2^4\cdot 5$ .

Calculation:

Now check that the relation is reflexive, symmetric and transitive.

  $\begin{array}{l}\mathrm{Re}flexive:\\ \left(\forall x\in N-\left\{1\right\}\right)\left(xRx\right)\\ Symmetric:\\ \left(\forall x,y\in N-\left\{1\right\}\right)\left(xandyhavethesamenumberof2\text{'}s\right)\\ xRy\iff yRx\\ Transitive:\\ \left(\forall x,y,z\in N-\left\{1\right\}\right)\left(\begin{array}{l}xandyhavethesamenumberof2\text{'}s\\ yandzhavethesamenumberof2\text{'}s\end{array}\right)\\ \left(xRy\wedge yRz\right)\iff \left(xRz\right)\end{array}$

The relation is reflexive, symmetric and transitive.

Hence, this is an equivalence relation.

The equivalence class of :

  $\begin{array}{l}\left[\overline{7}\right]:\boxed{3,99,51}\\ \left[\overline{10}\right]:\boxed{6,198,102}\\ \left[\overline{72}\right]:\boxed{8,24,40}\end{array}$

e.

To determine

Describe the equivalence class of $\left(1,2\right);of\left(4,0\right)$ .

Answer to Problem 6E

  $\begin{array}{l}\left[1,2\right]:\boxed{\left\{\left(x,y\right)\in R\times R;x^2+y^2=5\right\}}\\ \left(circlearound\left(0,0\right)witharadiusof\sqrt{5}\right)\\ \left[\left(4,0\right)\right]:\boxed{\left\{\left(x,y\right)\in R\times R;x^2+y^2=16\right\}}\\ \left(circlearound\left(0,0\right)witharadiusof4\right)\end{array}$

Explanation of Solution

Given information:

The relation $TonR\times Rgivenby\left(x,y\right)T\left(a,b\right)$ iff $x^2+y^2=a^2+b^2$ .

Calculation:

Now check that the relation is reflexive, symmetric and transitive.

  $\begin{array}{l}\mathrm{Re}flexive:\\ \left(\forall \left(x,y\right)\in R\times R\right)\left(\left(x,y\right)T\left(x,y\right)\right)\\ x^2+y^2=x^2+y^2\\ Symmetric:\\ \left(\forall \left(x,y\right),\left(z,v\right)\in R\times R\right)\left(\left(x,y\right)T\left(z,v\right)\right)\\ x^2+y^2=z^2+v^2=x^2+y^2\\ Transitive:\\ \left(\forall \left(x,y\right),\left(z,v\right),\left(w,u\right)\in R\times R\right)\left(x,y\right)\wedge \left(z,v\right)\left(w,u\right)\iff \left(x,y\right)T\left(w,u\right)\\ x^2+y^2=z^2+v^2=w^2+u^2\end{array}$

The relation is reflexive, symmetric and transitive.

Hence, this is an equivalence relation.

The equivalence class of :

  $\begin{array}{l}\left[1,2\right]:\boxed{\left\{\left(x,y\right)\in R\times R;x^2+y^2=5\right\}}\\ \left(circlearound\left(0,0\right)witharadiusof\sqrt{5}\right)\\ \left[\left(4,0\right)\right]:\boxed{\left\{\left(x,y\right)\in R\times R;x^2+y^2=16\right\}}\\ \left(circlearound\left(0,0\right)witharadiusof4\right)\end{array}$

f.

To determine

Find the number of elements in $Xand3\left(X\right)/R$ .

Answer to Problem 6E

  $\begin{array}{l}\left[\left\{m\right\}\right]=\left\{\left\{m\right\},\left\{p\right\},\left\{q\right\},\left\{r\right\},\left\{s\right\}\right\}\\ \left[\left\{m,n,p,q,r\right\}\right]=\left\{\begin{array}{l}\left\{m,n,p,q,r\right\},\left\{n,p,q,r,s\right\},\left\{m,p,q,r,s\right\},\\ \left\{m,n,q,r,s\right\},\left\{m,n,p,r,s\right\},\left\{m,n,p,q,s\right\}\end{array}\right\}\\ \left[X\right]has\left(\begin{array}{l}6\\ 5\end{array}\right)=6elements.\\ P\left(X\right)/Rhas7elements.\end{array}$

Explanation of Solution

Given information:

For the set $X=\left\{m,n,p,q,r,s\right\}$ ,let $R$ be the relation on $P\left(X\right)givenbyARB$ iff A and B have the same number of elements. List all the elements in $\left\{\overline{m}\right\};in\left\{\overline{m,n,p,q,r}\right\}$

Calculation:

Now check that the relation is reflexive, symmetric and transitive.

  $\begin{array}{l}\mathrm{Re}flexive:\\ \left(\forall A\in P\left(x\right)\right)\left(ARA\right)\\ \overline{\overline{A}}=\overline{\overline{A}}\\ Symmetric:\\ \left(\forall A,B\in P\left(x\right)\right)\left(ARB\iff BRA\right)\\ \overline{\overline{A}}=\overline{\overline{B}}\\ Transitive:\\ \left(\forall A,B,C\in P\left(x\right)\right)\left(ARB\wedge BRC\iff ARC\right)\\ \overline{\overline{A}}=\overline{\overline{B}}=\overline{\overline{C}}\end{array}$

The relation is reflexive, symmetric and transitive.

Hence, this is an equivalence relation.

The equivalence class of :

  $\begin{array}{l}\left[\left\{m\right\}\right]=\left\{\left\{m\right\},\left\{p\right\},\left\{q\right\},\left\{r\right\},\left\{s\right\}\right\}\\ \left[\left\{m,n,p,q,r\right\}\right]=\left\{\begin{array}{l}\left\{m,n,p,q,r\right\},\left\{n,p,q,r,s\right\},\left\{m,p,q,r,s\right\},\\ \left\{m,n,q,r,s\right\},\left\{m,n,p,r,s\right\},\left\{m,n,p,q,s\right\}\end{array}\right\}\\ \left[X\right]has\left(\begin{array}{l}6\\ 5\end{array}\right)=6elements.\\ P\left(X\right)/Rhas7elements.\end{array}$

g.

To determine

Describe all ordered pairs in the equivalence class of $\left(0,0\right)$ ; in the class of $\left(1,0\right)$ .

Answer to Problem 6E

  $\begin{array}{l}\boxed{\left\{\left(4,1\right),\left(-4,-1\right),\left(2,1\right),\left(-2,-1\right)\subset \left[\left(3,0\right)\right]\right\}}\\ \boxed{\left[\left(0,0\right)\right]=\left\{\left(x,y\right)\in R\times R:x=y\right\}}\\ \boxed{\left[\left(0,1\right)\right]=\left\{\left(x,y\right)\in R\times R:y=x\pm 1\right\}}\end{array}$

Explanation of Solution

Given information:

The relation $PonR\times Rgivenby\left(x,y\right)P\left(z,w\right)$ iff $\left|x-y\right|=\left|z-w\right|$ Name at least one ordered pair in each quadrant that is related to $\left(3,0\right)$ .

Calculation:

Now check that the relation is reflexive, symmetric and transitive.

  $\begin{array}{l}\mathrm{Re}flexive:\\ \left(\forall \left(x,y\right)\in R\times R\right)\left(x,y\right)P\left(x,y\right)\\ \left|x-y\right|=\left|x-y\right|\\ Symmetric:\\ \left(\forall \left(x,y\right),\left(z,v\right)\in R\times R\right)\left(\left(x,y\right)P\left(z,v\right)\right)\\ \left|x-y\right|=\left|z-v\right|=\left|x-y\right|\\ Transitive:\\ \left(\forall \left(x,y\right),\left(z,v\right),\left(w,u\right)\in R\times R\right)\left(\left(x,y\right)P\left(z,v\right)\wedge \left(z,u\right)P\left(w,u\right)\right)\iff \left(\left(x,y\right)P\left(w,u\right)\right)\\ \left|x-y\right|=\left|z-v\right|=\left|w-u\right|\end{array}$

The relation is reflexive, symmetric and transitive.

Hence, this is an equivalence relation.

The equivalence class of :

  $\begin{array}{l}\boxed{\left\{\left(4,1\right),\left(-4,-1\right),\left(2,1\right),\left(-2,-1\right)\subset \left[\left(3,0\right)\right]\right\}}\\ \boxed{\left[\left(0,0\right)\right]=\left\{\left(x,y\right)\in R\times R:x=y\right\}}\\ \boxed{\left[\left(0,1\right)\right]=\left\{\left(x,y\right)\in R\times R:y=x\pm 1\right\}}\end{array}$

h.

To determine

Name three elements in each of these classes: $\overline{x^2},\overline{4x^3+10x}$ . Describe $\overline{x^3}and\overline{7}$ .

Answer to Problem 6E

  $\begin{array}{l}\left\{x^2+1,x^2+\sqrt{2},x^2\right\}\subset \left[x^2\right]\\ \left\{4x^3+10x+1,4x^3+10x+\sqrt{2},4x^2+10x\right\}\subset \left[4x^3+10x\right]\\ \left\{x^3+1,x^3+\sqrt{2},x^3\right\}\subset \left[x^3\right]\\ \left\{1,\sqrt{2},0\right\}\subset \left[7\right]\end{array}$

Explanation of Solution

Given information:

Let $R$ be the relation on the set of all differentiable functions defined by $fRg$ iff $fandg$ have the same first derivative; that is, $f\text{'}=g\text{'}$ .

Calculation:

Now check that the relation is reflexive, symmetric and transitive.

  $\begin{array}{l}\mathrm{Re}flexive:\\ \left(\forall f\right)\left(fRf\right)\\ f\text{'}=f\text{'}\\ Symmetric:\\ \left(\forall f,g\right)\left(fRg\iff gRf\right)\\ f\text{'}=g\text{'}=f\text{'}\\ Transitive:\\ \left(\forall f,g,h\right)\left(\left(fRg\wedge gRh\right)\iff fRh\right)\\ f\text{'}=g\text{'}=h\text{'}\end{array}$

The relation is reflexive, symmetric and transitive.

Hence, this is an equivalence relation.

The equivalence class of :

  $\begin{array}{l}\left\{x^2+1,x^2+\sqrt{2},x^2\right\}\subset \left[x^2\right]\\ \left\{4x^3+10x+1,4x^3+10x+\sqrt{2},4x^2+10x\right\}\subset \left[4x^3+10x\right]\\ \left\{x^3+1,x^3+\sqrt{2},x^3\right\}\subset \left[x^3\right]\\ \left\{1,\sqrt{2},0\right\}\subset \left[7\right]\end{array}$

i.

To determine

Describe the equivalence class of $0;of\pi /2;of\pi /4$ .

Answer to Problem 6E

  $\begin{array}{l}\left[0\right]=\left\{2k\pi :k\in Z\right\}\\ \left[\pi /2\right]=\left\{\pi /2+2k\pi :k\in Z\right\}\\ \left[\pi /4\right]=\left\{\pi /4+2k\pi :k\in Z\right\}\cup \left\{3\pi /4+2k\pi :k\in Z\right\}\end{array}$

Explanation of Solution

Given information:

The relation $TonRgivenbyxTy$ iff $\sin x=\sin y$ .

Calculation:

Now check that the relation is reflexive, symmetric and transitive.

  $\begin{array}{l}\mathrm{Re}flexive:\\ \left(\forall xR\right)\left(xTx\right)\\ \sin x=\sin x\\ Symmetric:\\ \left(\forall x,y\in R\right)\left(xTy\iff \sin x=\sin y\iff yTx\right)\\ Transitive:\\ \left(\forall x,y,z\in R\right)\left(\left(xTy\wedge yTz\right)\iff \left(\sin x=\sin y=\sin z\right)\iff xTz\right)\end{array}$

The relation is reflexive, symmetric and transitive.

Hence, this is an equivalence relation.

The equivalence class of :

  $\begin{array}{l}\left[0\right]=\left\{2k\pi :k\in Z\right\}\\ \left[\pi /2\right]=\left\{\pi /2+2k\pi :k\in Z\right\}\\ \left[\pi /4\right]=\left\{\pi /4+2k\pi :k\in Z\right\}\cup \left\{3\pi /4+2k\pi :k\in Z\right\}\end{array}$