Chapter 4.3, Problem 15E

a.

To determine

Assign a grade of $A\left(correct\right),C\left(partiallycorrect\right),orF\left(failure\right)$ .

Answer to Problem 15E

  $Failure\left(F\right)$

Explanation of Solution

Given information:

Claim: The function $f:R\times R\to R$ given by $f\left(x,y\right)=2x-3y$ is a surjection.

Proof: Suppose that $\left(x,y\right)\in R\times R$ . Then $x\in R,so2x\in R$ . Also, $y\in R,so3y\in R$ . Therefore $2x-3y\in R$ . Thus, $f\left(x,y\right)\in R$ , so $f$ is a surjection.

Calculation:

Here, we have the claim $f:R\times R\to R$ ; $f\left(x,y\right)=2x-3y$ is a surjection.

We have given the proof $\left(x,y\right)\in R\times R$ then $x\in R,2x\in R$

Now, $y\in R,3y\in R$ . Thus, $2x-3y\in R$ . Hence, the given function is surjection.

Now, the proof of the function shows only that $Rng\left(f\right)\subseteq R$ but it does not show that all real numbers will hold and can be written in the form of $2x-3y$ .

Hence, the grade of the assignment is $Failure\left(F\right)$ .

b.

To determine

Assign a grade of $A\left(correct\right),C\left(partiallycorrect\right),orF\left(failure\right)$ .

Answer to Problem 15E

  $Correct\left(A\right)$

Explanation of Solution

Given information:

Claim: The function $f:[1,\infty )\to \left(0,\infty \right)$ defined by $f\left(x\right)=\frac{1}{x}$ maps onto $\left(0,\infty \right)$ .

Proof: Suppose that $w\in \left(0,\infty \right)$ . Choose $x=\frac{1}{w}$ . Then $f\left(x\right)=\frac{1}{\frac{1}{w}}=w$ . Therefore, the function $f$ is onto $\left(0,\infty \right)$ .

Calculation:

Here, we have the claim $f:[1,\infty )\to \left(0,\infty \right)$ defined by $f\left(x\right)=\frac{1}{x}$ maps onto $\left(0,\infty \right)$ .

We have given the proof $w\in \left(0,\infty \right)$ and $x=\frac{1}{w}$ is chosen therefore $f\left(x\right)=\frac{1}{\frac{1}{w}}=w$ where the function $f$ is onto $(0,\infty ]$ .

Now, it is verified that the value of the function $f:[1,\infty )\to (0,\infty ]$ maps onto $(0,\infty ]$ .

Hence, the grade of the assignment is $Correct\left(A\right)$ .

c.

To determine

Assign a grade of $A\left(correct\right),C\left(partiallycorrect\right),orF\left(failure\right)$ .

Answer to Problem 15E

  $partiallycorrect\left(C\right)$

Explanation of Solution

Given information:

Claim: If $f:A\underrightarrow{onto}Bandg:B\underrightarrow{onto}C$ then $gof:A\to C$ maps onto $C$ .

Proof: Suppose that $a\in A$ . Then $f\left(a\right)\in B$ . Because $f\left(a\right)\in B,g\left(f\left(a\right)\right)\in C$ . Therefore, $\left(gof\right)\left(a\right)=g\left(f\left(a\right)\right)\in C$ so $gof$ is onto $C$ .

Calculation:

Here, we have the claim $f:A\underrightarrow{onto}Bandg:B\underrightarrow{onto}C$ then $gof:A\to C$ maps onto $C$ .

We have given the proof $a\in A$ then $f\left(a\right)\in B$ . As $f\left(a\right)\in B$ then $g\left(f\left(a\right)\right)\in C$ .

Now, we will suppose that $a\in A$ . Let $c\in C,thenb\in B$ such that $g\left(b\right)=canda\in A$ . Therefore, $f\left(a\right)=b$ .

Thus, $\left(gof\right)\left(a\right)=g\left(f\left(a\right)\right)\in Csogof$ maps onto $C$ .

Now, we will consider:

  $\begin{array}{l}\left(gof\right)\left(a\right)=g\left(f\left(a\right)\right)\\ g\left(f\left(a\right)\right)=g\left(b\right)\\ g\left(b\right)=c\end{array}$

Where, $a\in A,b\in B,c\in C$

Thus, $gofisontoC$

Since, as there can be few cases where the condition might not hold, for example when the set $C$ does not hold the elements corresponding to the given conditions for the elements in set $A$ .

Hence, the grade of the assignment is $partiallycorrect\left(C\right)$ .

d.

To determine

Assign a grade of $A\left(correct\right),C\left(partiallycorrect\right),orF\left(failure\right)$ .

Answer to Problem 15E

  $correct\left(A\right)$

Explanation of Solution

Given information:

Claim: The function $f:R\to R$ is given $f\left(x\right)=2x+7$ is onto $R$ .

Proof: Suppose that $f$ is not onto $R$ . Then there exists $b\in R$ with $b\notin Rng\left(f\right)$ . Thus, for all real numbers $x$ , $b\ne 2x+7$ . But $a=\frac{1}{2}\left(b-7\right)$ is a real number, and $f\left(a\right)=b$ . This is a contradiction. Thus $f$ is onto $R$ .

Calculation:

Here, we have the claim that the function $f:R\to R$ given by $2x+7$ is onto $R$ .

We have given the proof that $f$ is onto $R$ then there exists $b\in R$ with $b\notin Rng\left(f\right)$ .

Now, for all real numbers $x,b\ne 2x+7$ but $a=\frac{1}{2}\left(b-7\right)$ is a real number and $f\left(a\right)=b$ which is a contradiction thus, $f$ is onto $R$ .

Hence, the grade of the assignment is $correct\left(A\right)$ .

e.

To determine

Assign a grade of $A\left(correct\right),C\left(partiallycorrect\right),orF\left(failure\right)$ .

Answer to Problem 15E

  $correct\left(A\right)$

Explanation of Solution

Given information:

Claim: Let $I$ be the interval $\left(0,1\right)$ . The function $f:I\times I\to I$ given by $f\left(x,y\right)=x^y$ is a surjection.

Proof: Let $t\in I$ . Then $0<t<1$ , so $0<t^2<t<1$ , so $t^2\notin I$ . Choose $x=t^{2}andy=\frac{1}{2}\in I$ . Then $f\left(x,y\right)=x^y={\left(t^2\right)}^{\frac{1}{2}}=t$ . Therefore, $I\subseteq Rng\left(f\right)$ so the function $f$ is onto $I$

Calculation:

Here, we have the claim that with interval $I=\left(0,1\right)$ . The function $f:I\times I\to I,f\left(x,y\right)=x^y$ is a surjection.

We have given the proof that $t\in I$ then $0<t<1$ . So $0<t^2<t<1$ .

Now, $t^2\in I$ . Choose $x=t^2;y=\frac{1}{2}\in I$

Thus, $f\left(x,y\right)=x^y={\left(t^2\right)}^{\frac{1}{2}}=t$

Therefore, $I\subseteq Rng\left(f\right)$ so the function $f$ is onto $I$ .

Hence, the grade of the assignment is $correct\left(A\right)$ .

f.

To determine

Assign a grade of $A\left(correct\right),C\left(partiallycorrect\right),orF\left(failure\right)$ .

Answer to Problem 15E

  $correct\left(A\right)$

Explanation of Solution

Given information:

Claim: If $f:A\underrightarrow{1-1}Bandg:B\underrightarrow{1-1}C$ then $gof:A\underrightarrow{1-1}C$ .

Proof: We must show that if $\left(x,y\right)and\left(z,y\right)$ are elements of $gof,thenx=z$ . If $\left(x,y\right)\in gof$ then there is $u\in B$ such that $\left(x,u\right)\in fand\left(u,y\right)\in g$ . If $\left(z,y\right)\in gof$ then there is $v\in B$ imply $u=v$ because $g$ is one-to-one. Then $\left(x,u\right)\in f$ , $\left(z,v\right)\in f$ , and $u=v$ therefore, $x=z$ because $f$ is one-to-one. Hence, $\left(x,y\right)and\left(z,y\right)ingof$ imply $x=z$ . Therefore $gof$ is one-to-one.

Calculation:

Here, we have the claim that $f:A\underrightarrow{1-1}Bandg:B\underrightarrow{1-1}C$ then $gof:A\underrightarrow{1-1}C$ .

We have given the proof that suppose $\left(x,y\right)and\left(z,y\right)$ are elements of $gof$ then $x=z$ .

Now, if $\left(x,y\right)\in gof$ then there is $u\in B$ such that $\left(x,u\right)\in fand\left(u,y\right)\in g$ . If $\left(z,y\right)\in gof$ then $v\in B$ such that $\left(z,v\right)\in fand\left(v,y\right)\in g$ that is $u=v$ because $g$ is one-to-one. Then $\left(x,u\right)\in f,\left(z,v\right)\in fandu=v,x=z$ . Thus, $gof$ is one-to-one.

Hence, the grade of the assignment is $correct\left(A\right)$ .

g.

To determine

Assign a grade of $A\left(correct\right),C\left(partiallycorrect\right),orF\left(failure\right)$ .

Answer to Problem 15E

  $correct\left(A\right)$

Explanation of Solution

Given information:

Claim: The function $f:R\to R$ is given by $f\left(x\right)=2x+7$ is one-to-one.

Proof: Suppose that $x_{1}and{x}_2$ are real numbers with $f\left(x_1\right)\ne f\left(x_2\right)$ . Then $2x_1+7\ne 2x_2+7$ and thus $2x_1\ne 2x_2$ . Hence, $x_1\ne x_2$ which shows that $f$ is one-to-one.

Calculation:

Here, we have the claim that the function $f:R\to R,f\left(x\right)=2x+7$ is one-to-one.

Now, we have given the proof that suppose $x_{1}and{x}_2$ are real numbers with $f\left(x_1\right)\ne f\left(x_2\right)$ then $2x_1+7\ne 2x_2+7$ thus $2x_1\ne 2x_{2}and{x}_1\ne x_2$ which shows that $f$ is one to one.

Hence, the grade of the assignment is $correct\left(A\right)$ .

h.

To determine

Assign a grade of $A\left(correct\right),C\left(partiallycorrect\right),orF\left(failure\right)$ .

Answer to Problem 15E

  $partiallycorrect\left(C\right)$

Explanation of Solution

Given information:

Claim: The function $f$ in part $\left(e\right)$ is an injection.

Proof: Suppose that $\left(x,y\right)and\left(x,z\right)$ are in $I\times Iandf\left(x,y\right)=f\left(x,z\right)$ . Then $x^y=x^z$ . Dividing by $x^z$ , we have $x^{y-z}=x^0=1$ . Because $x\ne 1$ and $x^{y-z}=1,y-z$ must be $0$ . Therefore $y=z$ . This shows that $\left(x,y\right)=\left(x,z\right)$ so $f$ is an injection.

Calculation:

Here, we have the claim that with interval $I=\left(0,1\right)$ . The function $f:I\times I\to I,f\left(x,y\right)=x^y$ is an injection.

Now, we have given proof that suppose $\left(x,y\right),\left(y,z\right)\in I\times Iandf\left(x,y\right)=f\left(x,z\right)$ .

Now, $x^y=x^z$ divide both sides by $x^z$ and get $x^{y-z}=x^0=1$ . As $x\ne 1and{x}^{y-z}=x^0=1theny-z$ must be $0$ . Therefore $y=z$ this shows that $\left(x,y\right)=\left(x,z\right)$ thus, $f$ is an injection.

Now, let $f\left(x,y\right)=f\left(x,z\right)forsomex,y\in R$

  $\begin{array}{l}f\left(x,y\right)=f\left(x,z\right)\\ x^y=x^z\\ y=z\end{array}$

Now, simply we will take the power common which satisfied the one-to-one function.

Hence, the grade of the assignment is $partiallycorrect\left(C\right)$ .

i.

To determine

Assign a grade of $A\left(correct\right),C\left(partiallycorrect\right),orF\left(failure\right)$ .

Answer to Problem 15E

  $correct\left(A\right)$

Explanation of Solution

Given information:

Claim: The function $f$ in part $\left(e\right)$ is not an injection.

Proof: Both $\left(\frac{1}{2},\frac{1}{2}\right)and\left(\frac{1}{4},\frac{1}{4}\right)areinI\times I$ . But $f\left(\frac{1}{4},\frac{1}{4}\right)={\left(\frac{1}{4}\right)}^{\frac{1}{4}}={\left({\left(\frac{1}{2}\right)}^2\right)}^{\frac{1}{4}}={\left(\frac{1}{2}\right)}^{\frac{1}{2}}=f\left(\frac{1}{2},\frac{1}{2}\right)$ .

Calculation:

Here, we have the claim that with interval $I=\left(0,1\right)$ . The function $f:I\times I\to I,f\left(x,y\right)=x^y$ is not an injection.

Now, we have given proof that suppose $f\left(\frac{1}{2},\frac{1}{2}\right)=f\left(\frac{1}{4},\frac{1}{4}\right)areinI\times I$ .

Now,

  $\begin{array}{l}f\left(\frac{1}{4},\frac{1}{4}\right)={\left(\frac{1}{4}\right)}^{\frac{1}{4}}\\ f\left(\frac{1}{4},\frac{1}{4}\right)={\left({\left(\frac{1}{2}\right)}^2\right)}^{\frac{1}{4}}\\ f\left(\frac{1}{4},\frac{1}{4}\right)={\left(\frac{1}{2}\right)}^{\frac{1}{2}}\\ f\left(\frac{1}{4},\frac{1}{4}\right)=f\left(\frac{1}{2},\frac{1}{2}\right)\end{array}$

Now, as the value of the function for two different values is same, that implies the function is not one to one, but is many to one.

Therefore, the function is not an injection.

Hence, the grade of the assignment is $correct\left(A\right)$ .

j.

To determine

Assign a grade of $A\left(correct\right),C\left(partiallycorrect\right),orF\left(failure\right)$ .

Answer to Problem 15E

  $correct\left(A\right)$

Explanation of Solution

Given information:

Claim: The function $f:Z_{10}\to Z_{10}$ given by $f\left(\overline{x}\right)=\overline{3x+1}$ is onto $Z_{10}$ .

Proof: Let $\overline{w}\in Z_{10}$ . Then $w\in Z$ and $3$ divides exactly one of $w,w-1orw-2$ .

Case 1: If $3$ divides $w-1$ , choose $x=\frac{w-1}{3}$ . Then $f\left(\overline{x}\right)=f\overline{\left(\frac{w-1}{3}\right)}=\overline{\left(3\frac{w-1}{3}+1\right)}=\overline{w}$ .

Case 2: If $3$ divides $w$ , choose $x=\frac{w}{3}+3$ . Then $f\left(\overline{x}\right)=f\overline{\left(\frac{w}{3}+3\right)}=\overline{\left(3\left(\frac{w}{3}+3\right)\right)+1}=\overline{w+10}=\overline{w}$ .

Case 3: If $3$ divides $w-2$ , choose $x=\frac{w-2}{3}+7$ . Then,

  $f\left(\overline{x}\right)=f\overline{\left(\frac{w-2}{3}+7\right)}=\overline{\left(3\left(\frac{w-2}{3}+7\right)\right)+1}=\overline{w+20}=\overline{w}$ .

Calculation:

Here, we have the claim that the function $f:Z_{10}\to Z_{10}$ given by $f\left(\overline{x}\right)=\overline{3x+1}$ is onto $Z_{10}$ .

Now, we have given proof that suppose $\overline{w}\in Z_{10}$ then $w\in Z$ and $3$ divides exactly one of $w,w-1orw-2$ .

Now, considering different cases:

Case 1: If $3$ divides $w-1$ , choose $x=\frac{w-1}{3}$ . Then $f\left(\overline{x}\right)=f\overline{\left(\frac{w-1}{3}\right)}=\overline{\left(3\frac{w-1}{3}+1\right)}=\overline{w}$ .

Case 2: If $3$ divides $w$ , choose $x=\frac{w}{3}+3$ . Then $f\left(\overline{x}\right)=f\overline{\left(\frac{w}{3}+3\right)}=\overline{\left(3\left(\frac{w}{3}+3\right)\right)+1}=\overline{w+10}=\overline{w}$ .

Case 3: If $3$ divides $w-2$ , choose $x=\frac{w-2}{3}+7$ . Then,

  $f\left(\overline{x}\right)=f\overline{\left(\frac{w-2}{3}+7\right)}=\overline{\left(3\left(\frac{w-2}{3}+7\right)\right)+1}=\overline{w+20}=\overline{w}$ .

Now, in every case $w$ is in $Rng\left(f\right)$ so $f$ maps onto $Z_{10}$ .

Hence, the grade of the assignment is $correct\left(A\right)$ .