Textbook - Number Systems Questions in English

(N/A) $(i)$ False,because $0$ is a whole number but not a natural number.
$(ii)$ True,because every integer $m$ can be expressed in the form $\frac{m}{1}$,where $m$ is an integer and $1$ is a non-zero integer,so it is a rational number.
$(iii)$ False,because $\frac{3}{5}$ is a rational number but it is not an integer.

(A) We can approach this problem in at least two ways.
Method $1$: To find a rational number between $r$ and $s$,we can use the formula $\frac{r+s}{2}$.
First number: $\frac{1+2}{2} = \frac{3}{2}$.
Second number: $\frac{1 + 3/2}{2} = \frac{5/2}{2} = \frac{5}{4}$.
Third number: $\frac{1 + 5/4}{2} = \frac{9/4}{2} = \frac{9}{8}$.
Fourth number: $\frac{3/2 + 2}{2} = \frac{7/2}{2} = \frac{7}{4}$.
Fifth number: $\frac{7/4 + 2}{2} = \frac{15/4}{2} = \frac{15}{8}$.
Method $2$: To find $n$ rational numbers between two numbers,we can express them with denominator $n+1$.
Here $n=5$,so we use denominator $5+1=6$.
$1 = \frac{6}{6}$ and $2 = \frac{12}{6}$.
The five rational numbers are $\frac{7}{6}, \frac{8}{6}, \frac{9}{6}, \frac{10}{6}, \text{ and } \frac{11}{6}$.
Simplifying these,we get $\frac{7}{6}, \frac{4}{3}, \frac{3}{2}, \frac{5}{3}, \text{ and } \frac{11}{6}$.

(A) Yes,zero is a rational number. $A$ rational number is defined as any number that can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \ne 0$. Since zero can be written as $\frac{0}{1}$,$\frac{0}{2}$,$\frac{0}{3}$,etc.,where the numerator $p = 0$ (an integer) and the denominator $q$ is any non-zero integer,it satisfies the definition of a rational number.

(N/A) There are infinitely many rational numbers between $3$ and $4$.
To find $n = 6$ rational numbers,we can express $3$ and $4$ as fractions with a denominator of $n + 1 = 7$.
$3 = \frac{3 \times 7}{7} = \frac{21}{7}$
$4 = \frac{4 \times 7}{7} = \frac{28}{7}$
Thus,six rational numbers between $\frac{21}{7}$ and $\frac{28}{7}$ are $\frac{22}{7}, \frac{23}{7}, \frac{24}{7}, \frac{25}{7}, \frac{26}{7}, \text{ and } \frac{27}{7}$.

(N/A) There are infinitely many rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$.
To find $5$ rational numbers,we can multiply the numerator and denominator of both fractions by $(5 + 1) = 6$.
$\frac{3}{5} = \frac{3 \times 6}{5 \times 6} = \frac{18}{30}$
$\frac{4}{5} = \frac{4 \times 6}{5 \times 6} = \frac{24}{30}$
Therefore,$5$ rational numbers between $\frac{18}{30}$ and $\frac{24}{30}$ are:
$\frac{19}{30}, \frac{20}{30}, \frac{21}{30}, \frac{22}{30}, \frac{23}{30}$.

(N/A) $(i)$ True; the collection of whole numbers is ${0, 1, 2, 3, ...}$ and the collection of natural numbers is ${1, 2, 3, ...}$. Since all natural numbers are present in the set of whole numbers,the statement is true.
$(ii)$ False; integers include negative numbers,zero,and positive numbers,whereas whole numbers only include zero and positive numbers. For example,$-3$ is an integer but not a whole number.
$(iii)$ False; rational numbers include fractions and decimals,whereas whole numbers are non-negative integers. For example,$\frac{1}{5}$ is a rational number but not a whole number.

It is easy to see how the Greeks might have discovered $\sqrt{2}$. Consider a square $OABC$,with each side $1$ unit in length. Then you can see by the Pythagoras theorem that $OB = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$.
To represent $\sqrt{2}$ on the number line,place the square $OABC$ on the number line such that the vertex $O$ coincides with zero and the side $OA$ lies along the positive direction of the number line.
We have just seen that $OB = \sqrt{2}$. Using a compass with centre $O$ and radius $OB$,draw an arc intersecting the number line at the point $P$. Then the point $P$ corresponds to $\sqrt{2}$ on the number line.

(N/A) To locate $\sqrt{3}$ on the number line,follow these steps:
$1$. First,locate $\sqrt{2}$ on the number line by constructing a right-angled triangle with base $1$ unit and height $1$ unit. The hypotenuse will be $\sqrt{1^2 + 1^2} = \sqrt{2}$.
$2$. Now,construct a line segment $BD$ of unit length ($1$ unit) perpendicular to the hypotenuse $OB$ (where $OB = \sqrt{2}$).
$3$. Join $OD$. In the right-angled triangle $\triangle OBD$,by the Pythagoras theorem:
$OD^2 = OB^2 + BD^2$
$OD^2 = (\sqrt{2})^2 + 1^2 = 2 + 1 = 3$
$OD = \sqrt{3}$.
$4$. Using a compass,with centre $O$ and radius $OD$,draw an arc that intersects the number line at point $Q$. The point $Q$ represents $\sqrt{3}$ on the number line.
In the same way,you can locate $\sqrt{n}$ for any positive integer $n$,after $\sqrt{n-1}$ has been located.

(N/A) $(i)$ True; the set of real numbers consists of both rational and irrational numbers. Therefore,every irrational number is a real number.
$(ii)$ False; negative numbers on the number line cannot be expressed as the square root of any natural number $m$,as the square root of any natural number is always non-negative.
$(iii)$ False; real numbers include both rational and irrational numbers. For example,$2$ is a real number but it is a rational number,not an irrational number.

(B) No,the square roots of all positive integers are not irrational.
For example,consider the square roots of perfect squares such as $\sqrt{4}$ and $\sqrt{9}$.
We know that $\sqrt{4} = 2$ and $\sqrt{9} = 3$.
Since $2$ and $3$ can be expressed in the form $\frac{p}{q}$ (where $p$ and $q$ are integers and $q \neq 0$),they are rational numbers.
Therefore,the square roots of all positive integers are not irrational.

(N/A) To represent $\sqrt{5}$ on the number line,we use the Pythagorean theorem: $\sqrt{5} = \sqrt{2^2 + 1^2}$.
$1$. Draw a number line and mark a point $O$ representing $0$ and a point $A$ representing $2$ units from $O$.
$2$. At point $A$,construct a perpendicular line segment $AB$ of length $1$ unit.
$3$. Join $O$ and $B$. In the right-angled triangle $\triangle OAB$,by the Pythagorean theorem:
$OB^2 = OA^2 + AB^2$
$OB^2 = 2^2 + 1^2 = 4 + 1 = 5$
$OB = \sqrt{5}$
$4$. Now,with $O$ as the center and $OB$ as the radius,draw an arc that intersects the number line at point $C$.
$5$. The distance $OC$ is equal to $OB$,which is $\sqrt{5}$. Thus,point $C$ represents $\sqrt{5}$ on the number line.

(N/A) The construction of the square root spiral is based on the Pythagorean theorem,which states that in a right-angled triangle,the square of the hypotenuse is equal to the sum of the squares of the other two sides $(h^2 = a^2 + b^2)$.
$1$. Start with point $O$ and draw $OP_1 = 1$ unit.
$2$. Draw $P_1P_2$ perpendicular to $OP_1$ such that $P_1P_2 = 1$ unit. In $\triangle OP_1P_2$,by Pythagoras theorem,$OP_2 = \sqrt{OP_1^2 + P_1P_2^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.
$3$. Draw $P_2P_3$ perpendicular to $OP_2$ such that $P_2P_3 = 1$ unit. In $\triangle OP_2P_3$,$OP_3 = \sqrt{OP_2^2 + P_2P_3^2} = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{2 + 1} = \sqrt{3}$.
$4$. Similarly,draw $P_3P_4$ perpendicular to $OP_3$ such that $P_3P_4 = 1$ unit. In $\triangle OP_3P_4$,$OP_4 = \sqrt{OP_3^2 + P_3P_4^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.
$5$. By continuing this process,the length of the hypotenuse $OP_n$ will be $\sqrt{n}$.

(N/A) For $\frac{10}{3} = 3.333... = 3.\overline{3}$. The remainder is always $1$,which repeats. The divisor is $3$.
For $\frac{7}{8} = 0.875$. The remainder becomes $0$ after a finite number of steps. The divisor is $8$.
For $\frac{1}{7} = 0.\overline{142857}$. The remainders are $3, 2, 6, 4, 5, 1$ which repeat. The divisor is $7$.
Observations:
$(i)$ The remainders either become $0$ after a certain stage,or start repeating themselves.
$(ii)$ The number of entries in the repeating string of remainders is less than the divisor.
$(iii)$ If the remainders repeat,then we get a repeating block of digits in the quotient.

(N/A) number is rational if it can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \ne 0$.
Given the number $3.142678$,we can remove the decimal point by dividing by the appropriate power of $10$.
Since there are $6$ digits after the decimal point,we multiply and divide by $1,000,000$.
$3.142678 = \frac{3142678}{1000000}$.
This fraction can be simplified by dividing both the numerator and the denominator by their greatest common divisor,which is $2$.
$\frac{3142678 \div 2}{1000000 \div 2} = \frac{1571339}{500000}$.
Since $1571339$ and $500000$ are integers and $500000 \ne 0$,the number $3.142678$ is a rational number.

(N/A) Let $x = 0.3333...$ (Equation $1$).
Multiply both sides by $10$:
$10x = 3.3333...$ (Equation $2$).
Since $x = 0.3333...$,we can write $10x = 3 + 0.3333... = 3 + x$.
Subtracting $x$ from both sides:
$10x - x = 3$
$9x = 3$
$x = \frac{3}{9} = \frac{1}{3}$.
Thus,$0.\overline{3}$ can be expressed as $\frac{1}{3}$,where $p = 1$ and $q = 3$ are integers and $q \ne 0$.

(C) Let $x = 1.272727 \ldots$
Since two digits are repeating,we multiply $x$ by $100$ to get:
$100x = 127.2727 \ldots$
We can write this as:
$100x = 126 + 1.272727 \ldots$
Since $x = 1.272727 \ldots$,we substitute $x$ into the equation:
$100x = 126 + x$
Subtracting $x$ from both sides:
$100x - x = 126$
$99x = 126$
$x = \frac{126}{99}$
Dividing both numerator and denominator by their greatest common divisor,$9$:
$x = \frac{14}{11}$
Thus,$1.\overline{27} = \frac{14}{11}$,where $p = 14$ and $q = 11$ are integers and $q \neq 0$.

(N/A) Let $x = 0.2 \overline{35}$.
Here,the digit $2$ does not repeat,but the block $35$ repeats.
Since two digits are repeating,we multiply $x$ by $100$ to get:
$100x = 23.53535 \ldots$
We can write this as:
$100x = 23.3 + 0.23535 \ldots$
Since $x = 0.23535 \ldots$,we substitute $x$ into the equation:
$100x = 23.3 + x$
Subtract $x$ from both sides:
$99x = 23.3$
Convert the decimal to a fraction:
$99x = \frac{233}{10}$
Divide by $99$:
$x = \frac{233}{990}$
Thus,$0.2 \overline{35} = \frac{233}{990}$.

(A) We know that $\frac{1}{7} = 0.\overline{142857}$ and $\frac{2}{7} = 0.\overline{285714}$.
To find an irrational number between these two values,we need a number that is non-terminating and non-recurring.
Any number of the form $0.15015001500015...$ satisfies this condition because it lies between $0.142857...$ and $0.285714...$ and does not repeat in a fixed pattern.
Thus,one such irrational number is $0.15015001500015...$.

(N/A) $(i)$ $\frac{36}{100} = 0.36$. This is a terminating decimal expansion.
$(ii)$ $\frac{1}{11} = 0.090909 \ldots = 0.\overline{09}$. This is a non-terminating repeating decimal expansion.
$(iii)$ $4 \frac{1}{8} = \frac{33}{8} = 4.125$. This is a terminating decimal expansion.
$(iv)$ $\frac{3}{13} = 0.230769230769 \ldots = 0.\overline{230769}$. This is a non-terminating repeating decimal expansion.
$(v)$ $\frac{2}{11} = 0.18181818 \ldots = 0.\overline{18}$. This is a non-terminating repeating decimal expansion.
$(vi)$ $\frac{329}{400} = 0.8225$. This is a terminating decimal expansion.

(N/A) We are given that $\frac{1}{7} = 0.\overline{142857}$.
To find the decimal expansions of the given fractions without long division,we multiply the value of $\frac{1}{7}$ by the respective numerators:
$\frac{2}{7} = 2 \times \frac{1}{7} = 2 \times 0.\overline{142857} = 0.\overline{285714}$
$\frac{3}{7} = 3 \times \frac{1}{7} = 3 \times 0.\overline{142857} = 0.\overline{428571}$
$\frac{4}{7} = 4 \times \frac{1}{7} = 4 \times 0.\overline{142857} = 0.\overline{571428}$
$\frac{5}{7} = 5 \times \frac{1}{7} = 5 \times 0.\overline{142857} = 0.\overline{714285}$
$\frac{6}{7} = 6 \times \frac{1}{7} = 6 \times 0.\overline{142857} = 0.\overline{857142}$
Thus,by multiplying the decimal expansion of $\frac{1}{7}$ by the respective numerators,we can determine the decimal expansions of these rational numbers without performing long division.

(N/A) $(i)$ Let $x = 0.\overline{6} = 0.6666\ldots$
Since there is one repeating digit,we multiply both sides by $10$:
$10x = 6.6666\ldots$
Subtracting $x$ from $10x$:
$10x - x = 6.6666\ldots - 0.6666\ldots$
$9x = 6$
$x = \frac{6}{9} = \frac{2}{3}$
Thus,$0.\overline{6} = \frac{2}{3}$.
$(ii)$ Let $x = 0.4\overline{7} = 0.4777\ldots$
Multiply by $10$:
$10x = 4.777\ldots$ $(1)$
Multiply by $100$:
$100x = 47.777\ldots$ $(2)$
Subtracting $(1)$ from $(2)$:
$100x - 10x = 47.777\ldots - 4.777\ldots$
$90x = 43$
$x = \frac{43}{90}$
Thus,$0.4\overline{7} = \frac{43}{90}$.
$(iii)$ Let $x = 0.\overline{001} = 0.001001\ldots$ $(1)$
Since there are three repeating digits,multiply by $1000$:
$1000x = 1.001001\ldots$ $(2)$
Subtracting $(1)$ from $(2)$:
$1000x - x = 1.001001\ldots - 0.001001\ldots$
$999x = 1$
$x = \frac{1}{999}$
Thus,$0.\overline{001} = \frac{1}{999}$.

(A) Let $x = 0.9999 \ldots$ $(1)$
Multiply both sides by $10$,we have:
$10x = 9.9999 \ldots$ $(2)$
Subtracting equation $(1)$ from equation $(2)$,we get:
$10x - x = (9.9999 \ldots) - (0.9999 \ldots)$
$9x = 9$
$x = \frac{9}{9} = 1$
Thus,$0.9999 \ldots = 1$.
As $0.9999 \ldots$ continues infinitely,there is no gap between $1$ and $0.9999 \ldots$. Hence,both are equal.

(A) In the decimal expansion of a rational number $\frac{p}{q}$,the number of entries in the repeating block of digits is always less than the divisor $q$.
In the fraction $\frac{1}{17}$,the divisor is $17$.
Therefore,the maximum number of digits in the repeating block is $17 - 1 = 16$.
By performing long division,we find:
$\frac{1}{17} = 0.\overline{0588235294117647}$
Thus,there are $16$ digits in the repeating block of the decimal expansion of $\frac{1}{17}$.

(N/A) Let us examine the decimal expansion of the following terminating rational numbers:
$\frac{3}{2} = \frac{3 \times 5}{2 \times 5} = \frac{15}{10} = 1.5$ [Denominator $= 2 = 2^1$]
$\frac{1}{5} = \frac{1 \times 2}{5 \times 2} = \frac{2}{10} = 0.2$ [Denominator $= 5 = 5^1$]
$\frac{7}{8} = \frac{7 \times 125}{8 \times 125} = \frac{875}{1000} = 0.875$ [Denominator $= 8 = 2^3$]
$\frac{8}{125} = \frac{8 \times 8}{125 \times 8} = \frac{64}{1000} = 0.064$ [Denominator $= 125 = 5^3$]
$\frac{13}{20} = \frac{13 \times 5}{20 \times 5} = \frac{65}{100} = 0.65$ [Denominator $= 20 = 2^2 \times 5^1$]
$\frac{17}{16} = \frac{17 \times 625}{16 \times 625} = \frac{10625}{10000} = 1.0625$ [Denominator $= 16 = 2^4$]
We observe that the prime factorization of $q$ (i.e.,the denominator) has only powers of $2$,powers of $5$,or powers of both. Thus,$q$ must be of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.

(N/A) number whose decimal expansion is non-terminating and non-recurring is an irrational number.
Three examples of such numbers are:
$1$. $\sqrt{2} = 1.414213562 \ldots$
$2$. $\sqrt{3} = 1.732050808 \ldots$
$3$. $\sqrt{5} = 2.236067978 \ldots$

(N/A) First,convert the rational numbers into decimal form:
$\frac{5}{7} = 0.\overline{714285}$
$\frac{9}{11} = 0.\overline{81}$
An irrational number is a non-terminating and non-repeating decimal.
We need to find three numbers between $0.714285...$ and $0.818181...$ that follow a non-terminating,non-repeating pattern.
Three such examples are:
$1) 0.73073007300073...$
$2) 0.75075007500075...$
$3) 0.79079007900079...$

(N/A) $(i)$ $\sqrt{23} \approx 4.79583152331 \ldots$
Since the decimal expansion is non-terminating and non-recurring,it is an irrational number.
$(ii)$ $\sqrt{225} = 15 = \frac{15}{1}$
Since it can be expressed in the form $\frac{p}{q}$ where $p, q$ are integers and $q \neq 0$,it is a rational number.
$(iii)$ $0.3796 = \frac{3796}{10000} = \frac{949}{2500}$
Since the decimal expansion is terminating,it is a rational number.
$(iv)$ $7.478478 \ldots = 7.\overline{478}$
Since the decimal expansion is non-terminating but recurring,it is a rational number.
$(v)$ $1.101001000100001 \ldots$
Since the decimal expansion is non-terminating and non-repeating,it is an irrational number.

(N/A) To visualize $5.3\overline{7}$ on the number line up to $5$ decimal places,we use the process of successive magnification:
$1$. We know that $5.3\overline{7}$ lies between $5$ and $6$. We divide the distance between $5$ and $6$ into $10$ equal parts and locate $5.3$ and $5.4$. $5.3\overline{7}$ lies between $5.3$ and $5.4$ [Fig $(i)$].
$2$. Next,we divide the distance between $5.3$ and $5.4$ into $10$ equal parts to locate $5.37$ and $5.38$. $5.3\overline{7}$ lies between $5.37$ and $5.38$ [Fig $(ii)$].
$3$. We then divide the distance between $5.37$ and $5.38$ into $10$ equal parts to locate $5.377$ and $5.378$. $5.3\overline{7}$ lies between $5.377$ and $5.378$ [Fig $(iii)$].
$4$. Finally,we divide the distance between $5.377$ and $5.378$ into $10$ equal parts to locate $5.3777$ and $5.3778$. By magnifying this portion,we can represent $5.37777$ on the number line [Fig $(iv)$].

(N/A) $3.765$ lies between $3$ and $4$.
Let us divide the interval $(3, 4)$ into $10$ equal parts.
Since $3.765$ lies between $3.7$ and $3.8$,we again magnify the interval $[3.7, 3.8]$ by dividing it further into $10$ parts to focus on the distance between $3.76$ and $3.77$.
The number $3.765$ lies between $3.76$ and $3.77$. Therefore,we further magnify the interval $[3.76, 3.77]$ into $10$ equal parts.
Now,the point corresponding to $3.765$ is clearly located,as shown in the final step of the magnification process.

(N/A) We can visualize the number $4.\overline{26}$ or $4.2626\ldots$ on the number line using the process of successive magnification.
$I$. The number $4.2626\ldots$ lies between $4$ and $5$. We divide the interval $[4, 5]$ into $10$ equal parts and locate $4.2$ and $4.3$.
$II$. The number $4.2626\ldots$ lies between $4.2$ and $4.3$. We magnify the interval $[4.2, 4.3]$ and divide it into $10$ equal parts to locate $4.26$ and $4.27$.
$III$. The number $4.2626\ldots$ lies between $4.26$ and $4.27$. We magnify the interval $[4.26, 4.27]$ and divide it into $10$ equal parts to locate $4.262$ and $4.263$.
$IV$. The number $4.2626\ldots$ lies between $4.262$ and $4.263$. We magnify the interval $[4.262, 4.263]$ and divide it into $10$ equal parts. We can now locate $4.2626$ on the number line.

(N/A) To determine if a number is irrational,we check if it can be expressed as a non-terminating,non-recurring decimal.
$1$. $7 \sqrt{5}$: Since $\sqrt{5} \approx 2.236$ is an irrational number,the product of a non-zero rational number $(7)$ and an irrational number $(\sqrt{5})$ is always irrational. Thus,$7 \sqrt{5} \approx 15.652...$ is irrational.
$2$. $\frac{7}{\sqrt{5}}$: Rationalizing the denominator,we get $\frac{7 \sqrt{5}}{5} \approx 3.1304...$. Since the quotient of a non-zero rational number and an irrational number is irrational,$\frac{7}{\sqrt{5}}$ is irrational.
$3$. $\sqrt{2}+21$: Since $\sqrt{2} \approx 1.414$ is irrational,the sum of an irrational number and a rational number $(21)$ is always irrational. Thus,$\sqrt{2}+21 \approx 22.414...$ is irrational.
$4$. $\pi-2$: Since $\pi \approx 3.1415...$ is an irrational number,the difference between an irrational number and a rational number $(2)$ is always irrational. Thus,$\pi-2 \approx 1.1415...$ is irrational.
Conclusion: All the given numbers are irrational.

(A) To add the given expressions,group the like terms (terms with the same square root factor):
$(2 \sqrt{2} + 5 \sqrt{3}) + (\sqrt{2} - 3 \sqrt{3})$
$= (2 \sqrt{2} + \sqrt{2}) + (5 \sqrt{3} - 3 \sqrt{3})$
$= (2 + 1) \sqrt{2} + (5 - 3) \sqrt{3}$
$= 3 \sqrt{2} + 2 \sqrt{3}$

(B) To multiply $6 \sqrt{5}$ by $2 \sqrt{5}$,we follow these steps:
$1$. Multiply the rational coefficients: $6 \times 2 = 12$.
$2$. Multiply the square root terms: $\sqrt{5} \times \sqrt{5} = 5$.
$3$. Multiply the results together: $12 \times 5 = 60$.
Thus,$6 \sqrt{5} \times 2 \sqrt{5} = 60$.

(C) To divide $8 \sqrt{15}$ by $2 \sqrt{3}$,we write it as a fraction:
$\frac{8 \sqrt{15}}{2 \sqrt{3}}$
We can simplify the expression by splitting $\sqrt{15}$ into $\sqrt{3} \times \sqrt{5}$:
$= \frac{8 \times \sqrt{3} \times \sqrt{5}}{2 \sqrt{3}}$
Cancel out the common term $\sqrt{3}$ from the numerator and the denominator:
$= \frac{8}{2} \times \sqrt{5}$
$= 4 \sqrt{5}$

$(i)$ Using the distributive property: $(5+\sqrt{7})(2+\sqrt{5}) = 5(2) + 5(\sqrt{5}) + \sqrt{7}(2) + \sqrt{7}(\sqrt{5}) = 10 + 5\sqrt{5} + 2\sqrt{7} + \sqrt{35}$.
$(ii)$ Using the identity $(a+b)(a-b) = a^2 - b^2$: $(5+\sqrt{5})(5-\sqrt{5}) = 5^2 - (\sqrt{5})^2 = 25 - 5 = 20$.
$(iii)$ Using the identity $(a+b)^2 = a^2 + 2ab + b^2$: $(\sqrt{3}+\sqrt{7})^2 = (\sqrt{3})^2 + 2(\sqrt{3})(\sqrt{7}) + (\sqrt{7})^2 = 3 + 2\sqrt{21} + 7 = 10 + 2\sqrt{21}$.
$(iv)$ Using the identity $(a-b)(a+b) = a^2 - b^2$: $(\sqrt{11}-\sqrt{7})(\sqrt{11}+\sqrt{7}) = (\sqrt{11})^2 - (\sqrt{7})^2 = 11 - 7 = 4$.

(A) To rationalise the denominator of $\frac{1}{\sqrt{2}}$,we multiply both the numerator and the denominator by $\sqrt{2}$.
$\frac{1}{\sqrt{2}} = \frac{1 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}$
Since $\sqrt{2} \times \sqrt{2} = 2$,we get:
$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$
Thus,the rationalised form is $\frac{\sqrt{2}}{2}$.

(B) To rationalise the denominator of $\frac{1}{2+\sqrt{3}}$,we multiply the numerator and the denominator by the conjugate of the denominator,which is $(2-\sqrt{3})$.
$\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} = \frac{2-\sqrt{3}}{(2)^2 - (\sqrt{3})^2}$
Using the identity $(a+b)(a-b) = a^2 - b^2$,we get:
$= \frac{2-\sqrt{3}}{4-3} = \frac{2-\sqrt{3}}{1} = 2-\sqrt{3}$.

(B) To rationalise the denominator,we multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{3}+\sqrt{5})$.
$\frac{5}{\sqrt{3}-\sqrt{5}} = \frac{5}{\sqrt{3}-\sqrt{5}} \times \frac{\sqrt{3}+\sqrt{5}}{\sqrt{3}+\sqrt{5}}$
Using the algebraic identity $(a-b)(a+b) = a^2 - b^2$ in the denominator:
$= \frac{5(\sqrt{3}+\sqrt{5})}{(\sqrt{3})^2 - (\sqrt{5})^2}$
$= \frac{5(\sqrt{3}+\sqrt{5})}{3 - 5}$
$= \frac{5(\sqrt{3}+\sqrt{5})}{-2}$
$= -\frac{5}{2}(\sqrt{3}+\sqrt{5})$

(A) To rationalise the denominator of $\frac{1}{7+3 \sqrt{2}}$,we multiply the numerator and the denominator by the conjugate of the denominator,which is $(7-3 \sqrt{2})$.
$\frac{1}{7+3 \sqrt{2}} = \frac{1}{7+3 \sqrt{2}} \times \frac{7-3 \sqrt{2}}{7-3 \sqrt{2}}$
Using the identity $(a+b)(a-b) = a^2 - b^2$ in the denominator:
$= \frac{7-3 \sqrt{2}}{(7)^2 - (3 \sqrt{2})^2}$
$= \frac{7-3 \sqrt{2}}{49 - (9 \times 2)}$
$= \frac{7-3 \sqrt{2}}{49 - 18}$
$= \frac{7-3 \sqrt{2}}{31}$

(N/A) $(i)$ $2-\sqrt{5}$: Since it is the difference between a rational number and an irrational number,it is an irrational number.
$(ii)$ $(3+\sqrt{23})-\sqrt{23} = 3+\sqrt{23}-\sqrt{23} = 3$. Since $3$ can be expressed as $\frac{3}{1}$,it is a rational number.
$(iii)$ $\frac{2 \sqrt{7}}{7 \sqrt{7}} = \frac{2}{7}$. Since this is in the form $\frac{p}{q}$ where $p, q$ are integers and $q \neq 0$,it is a rational number.
$(iv)$ $\frac{1}{\sqrt{2}}$: The quotient of a rational number and an irrational number is always an irrational number. Therefore,it is an irrational number.
$(v)$ $2 \pi$: The product of a non-zero rational number and an irrational number is always an irrational number. Therefore,$2 \pi$ is an irrational number.

(N/A) $(i)$ $(3+\sqrt{3})(2+\sqrt{2}) = 3(2+\sqrt{2}) + \sqrt{3}(2+\sqrt{2})$
$= 6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6}$
$(ii)$ Using the identity $(a+b)(a-b) = a^2 - b^2$:
$(3+\sqrt{3})(3-\sqrt{3}) = (3)^2 - (\sqrt{3})^2 = 9 - 3 = 6$
$(iii)$ Using the identity $(a+b)^2 = a^2 + b^2 + 2ab$:
$(\sqrt{5}+\sqrt{2})^2 = (\sqrt{5})^2 + (\sqrt{2})^2 + 2(\sqrt{5})(\sqrt{2}) = 5 + 2 + 2\sqrt{10} = 7 + 2\sqrt{10}$
$(iv)$ Using the identity $(a-b)(a+b) = a^2 - b^2$:
$(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2}) = (\sqrt{5})^2 - (\sqrt{2})^2 = 5 - 2 = 3$

(N/A) When we measure the length of a line with a scale or any other device,we only obtain an approximate rational value. This implies that at least one of $c$ or $d$ is irrational. Therefore,the ratio $\frac{c}{d}$ is irrational,which makes $\pi$ an irrational number. Thus,there is no contradiction in stating that $\pi$ is irrational.

(N/A) To represent $\sqrt{9.3}$ on the number line,follow these steps:
$1$. Draw a line segment $AB = 9.3 \text{ units}$ on a line $l$.
$2$. From point $B$,mark a point $C$ such that $BC = 1 \text{ unit}$. Now,$AC = 9.3 + 1 = 10.3 \text{ units}$.
$3$. Find the midpoint of $AC$,let it be $O$. Draw a semicircle with center $O$ and radius $OA = OC$.
$4$. Draw a perpendicular line at point $B$ to the line $l$,which intersects the semicircle at point $D$. The length of $BD$ is $\sqrt{9.3}$.
$5$. With $B$ as the center and $BD$ as the radius,draw an arc on the number line $l$ to intersect it at point $E$. The distance $BE$ represents $\sqrt{9.3}$ on the number line.

(N/A) $(i)$ $\frac{1}{\sqrt{7}} = \frac{1 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}} = \frac{\sqrt{7}}{7}$
$(ii)$ $\frac{1}{\sqrt{7}-\sqrt{6}} = \frac{1 \times (\sqrt{7}+\sqrt{6})}{(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6})} = \frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^2 - (\sqrt{6})^2} = \frac{\sqrt{7}+\sqrt{6}}{7-6} = \sqrt{7}+\sqrt{6}$
$(iii)$ $\frac{1}{\sqrt{5}+\sqrt{2}} = \frac{1 \times (\sqrt{5}-\sqrt{2})}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})} = \frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^2 - (\sqrt{2})^2} = \frac{\sqrt{5}-\sqrt{2}}{5-2} = \frac{\sqrt{5}-\sqrt{2}}{3}$
$(iv)$ $\frac{1}{\sqrt{7}-2} = \frac{1 \times (\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)} = \frac{\sqrt{7}+2}{(\sqrt{7})^2 - (2)^2} = \frac{\sqrt{7}+2}{7-4} = \frac{\sqrt{7}+2}{3}$

To simplify these expressions,we use the laws of exponents:
$(i)$ Using the product rule $a^m \cdot a^n = a^{m+n}$:
$2^{\frac{2}{3}} \cdot 2^{\frac{1}{3}} = 2^{\left(\frac{2}{3} + \frac{1}{3}\right)} = 2^{\frac{3}{3}} = 2^1 = 2$.
$(ii)$ Using the power of a power rule $(a^m)^n = a^{m \cdot n}$:
$\left(3^{\frac{1}{5}}\right)^4 = 3^{\left(\frac{1}{5} \cdot 4\right)} = 3^{\frac{4}{5}}$.
$(iii)$ Using the quotient rule $\frac{a^m}{a^n} = a^{m-n}$:
$\frac{7^{\frac{1}{5}}}{7^{\frac{1}{3}}} = 7^{\left(\frac{1}{5} - \frac{1}{3}\right)} = 7^{\left(\frac{3-5}{15}\right)} = 7^{-\frac{2}{15}}$.
$(iv)$ Using the product rule for different bases with the same exponent $a^n \cdot b^n = (a \cdot b)^n$:
$13^{\frac{1}{5}} \cdot 17^{\frac{1}{5}} = (13 \times 17)^{\frac{1}{5}} = 221^{\frac{1}{5}}$.

$(i)$ $64^{\frac{1}{2}} = (8^2)^{\frac{1}{2}} = 8^{2 \times \frac{1}{2}} = 8^1 = 8$
$(ii)$ $32^{\frac{1}{5}} = (2^5)^{\frac{1}{5}} = 2^{5 \times \frac{1}{5}} = 2^1 = 2$
$(iii)$ $125^{\frac{1}{3}} = (5^3)^{\frac{1}{3}} = 5^{3 \times \frac{1}{3}} = 5^1 = 5$

$(i)$ $9^{\frac{3}{2}} = (3^2)^{\frac{3}{2}} = 3^{2 \times \frac{3}{2}} = 3^3 = 27$
$(ii)$ $32^{\frac{2}{5}} = (2^5)^{\frac{2}{5}} = 2^{5 \times \frac{2}{5}} = 2^2 = 4$
$(iii)$ $16^{\frac{3}{4}} = (2^4)^{\frac{3}{4}} = 2^{4 \times \frac{3}{4}} = 2^3 = 8$
$(iv)$ $125^{-\frac{1}{3}} = (5^3)^{-\frac{1}{3}} = 5^{3 \times (-\frac{1}{3})} = 5^{-1} = \frac{1}{5}$

(N/A) $(i)$ Using the law of exponents $a^m \cdot a^n = a^{m+n}$:
$2^{2/3} \cdot 2^{1/5} = 2^{(2/3 + 1/5)} = 2^{(10+3)/15} = 2^{13/15}$
$(ii)$ Using the law of exponents $(a^m)^n = a^{m \cdot n}$:
$(1/3^3)^7 = (3^{-3})^7 = 3^{-3 \cdot 7} = 3^{-21}$
$(iii)$ Using the law of exponents $a^m / a^n = a^{m-n}$:
$11^{1/2} / 11^{1/4} = 11^{(1/2 - 1/4)} = 11^{(2-1)/4} = 11^{1/4}$
$(iv)$ Using the law of exponents $a^m \cdot b^m = (a \cdot b)^m$:
$7^{1/2} \cdot 8^{1/2} = (7 \cdot 8)^{1/2} = 56^{1/2}$