Real Numbers and their Decimal Expansions Questions in English

(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,$0.2 \overline{35} = \frac{233}{990}$.

(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) 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.

(B) We know that the set of real numbers consists of all rational and irrational numbers.
Since every rational number belongs to the set of real numbers,it follows that every rational number is a real number.

(D) rational number is defined as a number that can be expressed in the form $p/q$,where $p$ and $q$ are integers and $q \neq 0$. The decimal expansion of a rational number is always either terminating or non-terminating repeating (recurring). $A$ non-terminating,non-repeating decimal expansion is a characteristic property of an irrational number. Therefore,the decimal representation of a rational number cannot be non-terminating,non-repeating.

(B) Let $x = 1.999... = 1.\overline{9} \quad \dots(1)$
Multiply both sides by $10$:
$10x = 19.999... = 19.\overline{9} \quad \dots(2)$
Subtract equation $(1)$ from equation $(2)$:
$10x - x = 19.999... - 1.999...$
$9x = 18$
Divide by $9$:
$x = \frac{18}{9} = 2$
Since $2$ can be written as $\frac{2}{1},$ where $p=2$ and $q=1$ are integers and $q \neq 0,$ the value is $2.$
Hence,option $(b)$ is the correct answer.

(N/A) $(i)$ $10.124124 \ldots$ is a decimal expansion which is non-terminating and recurring. Since a number with a non-terminating recurring decimal expansion can be expressed in the form $p/q$ where $p, q$ are integers and $q \neq 0$,it is a rational number.
$(ii)$ $1.010010001 \ldots$ is a decimal expansion which is non-terminating and non-recurring. Since it cannot be expressed in the form $p/q$ where $p, q$ are integers and $q \neq 0$,it is an irrational number.

(B) Let $x = 0.12333\ldots$ (Equation $1$)
Multiply by $10$ to shift the decimal point: $10x = 1.2333\ldots$ (Equation $2$)
Multiply Equation $1$ by $100$ to shift the decimal point past the non-repeating part: $100x = 12.333\ldots$ (Equation $3$)
Subtract Equation $2$ from Equation $3$:
$100x - 10x = 12.333\ldots - 1.2333\ldots$
$90x = 11.1$
$x = \frac{11.1}{90} = \frac{111}{900}$
Simplify the fraction by dividing both numerator and denominator by $3$:
$x = \frac{111 \div 3}{900 \div 3} = \frac{37}{300}$
Thus,$0.12 \overline{3} = \frac{37}{300}$.

(A) To find rational numbers between $0.1$ and $0.11$,we can express them as decimals with more digits.
$0.1 = 0.100$
$0.11 = 0.110$
Now,we can easily identify numbers between $0.100$ and $0.110$.
Three such rational numbers are $0.101, 0.102,$ and $0.103$.

(B) To express the decimal $0.2$ in the form $\frac{p}{q}$,we follow these steps:
$1$. Write the decimal as a fraction with a denominator of $10$ because there is one digit after the decimal point: $0.2 = \frac{2}{10}$.
$2$. Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor,which is $2$: $\frac{2 \div 2}{10 \div 2} = \frac{1}{5}$.
$3$. Thus,$0.2 = \frac{1}{5}$,where $p = 1$ and $q = 5$ are integers and $q \neq 0$.

(C) Let $x = 0.888 \ldots = 0.\overline{8} \quad \dots(1)$
Multiplying both sides by $10$,we get:
$10x = 8.888 \ldots = 8.\overline{8} \quad \dots(2)$
Subtracting equation $(1)$ from equation $(2)$:
$10x - x = 8.\overline{8} - 0.\overline{8}$
$9x = 8$
Therefore,$x = \frac{8}{9}$.

(D) Let $x = 5. \overline{2} = 5.2222 \ldots \quad \dots(1)$
Multiplying both sides by $10,$ we get
$10x = 52.2222 \ldots = 52. \overline{2} \quad \dots(2)$
Subtracting equation $(1)$ from equation $(2)$:
$10x - x = 52.2222 \ldots - 5.2222 \ldots$
$9x = 47$
$x = \frac{47}{9}$
Hence,$5. \overline{2} = \frac{47}{9}$.

(A) Let $x = 0.\overline{001} = 0.001001...$ $(1)$
Since there are three repeating digits,multiply both sides by $1000$:
$1000x = 1.001001...$ $(2)$
Subtracting equation $(1)$ from equation $(2)$:
$1000x - x = 1.001001... - 0.001001...$
$999x = 1$
$x = \frac{1}{999}$
Thus,the number $0.\overline{001}$ in the form $\frac{p}{q}$ is $\frac{1}{999}$.

(B) Let $x = 0.2555 \ldots = 0.2\overline{5} \quad ....(1)$
Multiply both sides by $10$:
$10x = 2.555 \ldots = 2.\overline{5} \quad ....(2)$
Multiply both sides by $100$:
$100x = 25.555 \ldots = 25.\overline{5} \quad ....(3)$
Subtracting equation $(2)$ from equation $(3)$:
$100x - 10x = 25.\overline{5} - 2.\overline{5}$
$90x = 23$
Therefore,$x = \frac{23}{90}$.

(C) Let $x = 0.1\overline{34} = 0.1343434... \quad ....(1)$
Multiply both sides by $10$ to shift the non-repeating part:
$10x = 1.343434... \quad ....(2)$
Multiply equation $(2)$ by $100$ to shift the repeating part:
$1000x = 134.343434... \quad ....(3)$
Subtract equation $(2)$ from equation $(3)$:
$1000x - 10x = 134.343434... - 1.343434...$
$990x = 133$
Therefore,$x = \frac{133}{990}$.

(D) Let $x = 0.00323232 \ldots = 0.00\overline{32} \quad \dots(1)$
Multiply equation $(1)$ by $100$ to shift the decimal point:
$100x = 0.323232 \ldots = 0.\overline{32} \quad \dots(2)$
Multiply equation $(2)$ by $100$ again to shift the decimal point past the repeating part:
$10000x = 32.323232 \ldots = 32.\overline{32} \quad \dots(3)$
Subtracting equation $(2)$ from equation $(3)$:
$10000x - 100x = 32.\overline{32} - 0.\overline{32}$
$9900x = 32$
$x = \frac{32}{9900}$
Simplifying the fraction by dividing both numerator and denominator by $4$:
$x = \frac{32 \div 4}{9900 \div 4} = \frac{8}{2475}$

(A) Let $x = 0.404040 \ldots = 0.\overline{40} \quad \dots(1)$
Since there are two repeating digits,multiply both sides by $100$:
$100x = 40.404040 \ldots = 40.\overline{40} \quad \dots(2)$
Subtracting equation $(1)$ from equation $(2)$:
$100x - x = 40.\overline{40} - 0.\overline{40}$
$99x = 40$
$x = \frac{40}{99}$
Thus,the number $0.404040 \ldots$ expressed in the form $\frac{p}{q}$ is $\frac{40}{99}$.

(N/A) Let $x = 0.142857142857 \ldots$ $(1)$
Since the repeating block has $6$ digits,multiply both sides by $1,000,000$:
$1,000,000 x = 142857.142857142857 \ldots$ $(2)$
Subtracting equation $(1)$ from equation $(2)$:
$1,000,000 x - x = 142857.142857 \ldots - 0.142857 \ldots$
$999,999 x = 142857$
Solving for $x$:
$x = \frac{142857}{999999}$
Dividing both numerator and denominator by $142857$:
$x = \frac{1}{7}$
Thus,$0.142857142857 \ldots = \frac{1}{7}$.

(N/A) To write $\frac{2}{11}$ in decimal form,we perform long division by dividing $2$ by $11$.
Dividing $2$ by $11$:
$2 \div 11 = 0.1818...$
As the remainder $2$ repeats,the quotient $0.18$ repeats indefinitely.
Thus,$\frac{2}{11} = 0.1818... = 0.\overline{18}$.
Since the decimal expansion does not terminate and the digits repeat,the decimal expansion of $\frac{2}{11}$ is a non-terminating recurring decimal.

(N/A) To write $\frac{121}{400}$ in decimal form,we perform long division:
$121 \div 400 = 0.3025$
Since the remainder becomes $0$ after a finite number of steps,the decimal expansion of $\frac{121}{400}$ is a terminating decimal.

(N/A) To find the decimal expansion of $\frac{5}{13}$,we perform long division of $5$ by $13$:
$5 \div 13 = 0.384615384615...$
Since the remainder repeats after a certain number of steps,the quotient also repeats.
Thus,$\frac{5}{13} = 0.\overline{384615}$.
The decimal form of $\frac{5}{13}$ is a non-terminating recurring decimal.

(A) Let $x = 0 . \overline{35}$.
$\therefore x = 0.353535 \ldots$ $(1)$
Since there are two repeating digits,multiply both sides by $100$:
$100x = 35.353535 \ldots$ $(2)$
Subtract equation $(1)$ from equation $(2)$:
$100x - x = 35.353535 \ldots - 0.353535 \ldots$
$99x = 35$
$\therefore x = \frac{35}{99}$
Thus,$0 . \overline{35} = \frac{35}{99}$.

(N/A) Let $x = 0.5 \overline{7}.$
$\therefore x = 0.5777...$ $(1)$
Multiply both sides by $10$ to shift the decimal point before the repeating part:
$10x = 5.777...$ $(2)$
Now,multiply equation $(2)$ by $10$ to shift the decimal point after one repeating digit:
$100x = 57.777...$ $(3)$
Subtract equation $(2)$ from equation $(3)$:
$100x - 10x = 57.777... - 5.777...$
$90x = 52$
$x = \frac{52}{90}$
Simplify the fraction by dividing both numerator and denominator by $2$:
$x = \frac{26}{45}$
Thus,$0.5 \overline{7} = \frac{26}{45}.$

(N/A) Let $x = 0.\overline{125}$.
$\therefore x = 0.125125\ldots$ $(1)$
Since there are three repeating digits,multiply both sides by $1000$.
$1000x = 125.125125\ldots$ $(2)$
Subtract equation $(1)$ from equation $(2)$:
$1000x - x = 125.125125\ldots - 0.125125\ldots$
$999x = 125$
$\therefore x = \frac{125}{999}$
Thus,$0.\overline{125} = \frac{125}{999}$.

(N/A) To convert $\frac{71}{125}$ into decimal form,we can divide the numerator by the denominator or multiply both by $8$ to get a power of $10$ in the denominator.
$\frac{71 \times 8}{125 \times 8} = \frac{568}{1000} = 0.568$
Since the remainder becomes $0$ after a finite number of steps,the decimal expansion is a Terminating decimal expansion.

(N/A) To convert $\frac{4}{13}$ into decimal form,we perform long division of $4$ by $13$.
$4 \div 13 = 0.307692307692...$
Since the sequence of digits $307692$ repeats indefinitely,the decimal expansion is $0.\overline{307692}$.
Therefore,the decimal expansion is a non-terminating recurring decimal.

(N/A) To convert $\frac{25}{8}$ into decimal form,we perform long division:
$25 \div 8 = 3.125$
Since the remainder becomes $0$ after a finite number of steps,the decimal expansion is a Terminating decimal expansion.

(N/A) To convert $\frac{37}{60}$ into decimal form,we perform long division:
$37 \div 60 = 0.61666...$
This can be written as $0.61\overline{6}$.
Since the digits after the decimal point repeat indefinitely,the decimal expansion is a non-terminating recurring decimal.

(A) To convert $\frac{29}{12}$ into decimal form,we perform long division:
$29 \div 12 = 2.41666...$
Since the digit $6$ repeats indefinitely,the decimal expansion is $2.41\overline{6}$.
Because the remainder never becomes zero and a digit repeats,it is a Non-terminating recurring decimal.

(A) Let $x = 0.\overline{4} = 0.4444...$ (Equation $1$).
Since there is one repeating digit after the decimal point, multiply both sides by $10$:
$10x = 4.4444...$ (Equation $2$).
Subtract Equation $1$ from Equation $2$:
$10x - x = 4.4444... - 0.4444...$
$9x = 4$
$x = \frac{4}{9}$.
Thus, $0.\overline{4}$ expressed in the form $\frac{p}{q}$ is $\frac{4}{9}$.

(B) Let $x = 0.\overline{83}$.
This can be written as $x = 0.838383...$ (Equation $1$).
Since there are two repeating digits, multiply both sides by $100$:
$100x = 83.838383...$ (Equation $2$).
Subtract Equation $1$ from Equation $2$:
$100x - x = 83.838383... - 0.838383...$
$99x = 83$.
Therefore, $x = \frac{83}{99}$.
Thus, $0.\overline{83} = \frac{83}{99}$.

(N/A) Let $x = 2.137137137...$ (Equation $1$).
Since there are $3$ repeating digits after the decimal point, multiply both sides by $1000$:
$1000x = 2137.137137137...$ (Equation $2$).
Subtract Equation $1$ from Equation $2$:
$1000x - x = 2137.137137... - 2.137137...$
$999x = 2135$.
Therefore, $x = \frac{2135}{999}$.

(A) Let $x = 0.5\overline{7} = 0.5777...$ (Equation $1$).
Multiply both sides by $10$ to shift the decimal point:
$10x = 5.777...$ (Equation $2$).
Multiply Equation $2$ by $10$ again to shift the decimal point past the repeating digit:
$100x = 57.777...$ (Equation $3$).
Subtract Equation $2$ from Equation $3$:
$100x - 10x = 57.777... - 5.777...$
$90x = 52$.
$x = \frac{52}{90}$.
Simplify the fraction by dividing the numerator and denominator by their greatest common divisor,$2$:
$x = \frac{26}{45}$.

(A) Let $x = 1.23444...$ (Equation $1$).
Multiply by $100$ to shift the decimal point: $100x = 123.444...$ (Equation $2$).
Multiply Equation $2$ by $10$ to shift the decimal point again: $1000x = 1234.444...$ (Equation $3$).
Subtract Equation $2$ from Equation $3$:
$1000x - 100x = 1234.444... - 123.444...$
$900x = 1111$.
Therefore,$x = \frac{1111}{900}$.

(B) Let $x = 0.7\overline{39} = 0.7393939...$ (Equation $1$).
Multiply both sides by $10$ to shift the decimal point: $10x = 7.393939...$ (Equation $2$).
Multiply Equation $2$ by $100$ to shift the decimal point past the repeating part: $1000x = 739.393939...$ (Equation $3$).
Subtract Equation $2$ from Equation $3$:
$1000x - 10x = 739.393939... - 7.393939...$
$990x = 732$.
$x = \frac{732}{990}$.
Simplify the fraction by dividing both numerator and denominator by their greatest common divisor,which is $6$:
$x = \frac{732 \div 6}{990 \div 6} = \frac{122}{165}$.

(A) Let $x = 0.\overline{35} = 0.353535...$ and $y = 0.\overline{28} = 0.282828...$
To convert $0.\overline{35}$ to a fraction: $100x = 35.3535...$,so $99x = 35$,which gives $x = \frac{35}{99}$.
To convert $0.\overline{28}$ to a fraction: $100y = 28.2828...$,so $99y = 28$,which gives $y = \frac{28}{99}$.
Adding the two fractions: $x + y = \frac{35}{99} + \frac{28}{99} = \frac{63}{99}$.
Dividing $63$ by $99$ gives $0.636363...$,which is $0.\overline{63}$.

(N/A) To add $0. \overline{3}$ and $0.4 \overline{7}$,first convert them into fractions.
$0. \overline{3} = \frac{3}{9} = \frac{1}{3}$
$0.4 \overline{7} = \frac{47 - 4}{90} = \frac{43}{90}$
Now,add the two fractions:
$\frac{1}{3} + \frac{43}{90} = \frac{30}{90} + \frac{43}{90} = \frac{73}{90}$
Converting $\frac{73}{90}$ back to decimal form:
$73 \div 90 = 0.8111... = 0.8 \overline{1}$

(N/A) Step $1$: Convert $0.\overline{52}$ to a fraction.
Let $x = 0.525252...$ $(i)$
$100x = 52.525252...$ (ii)
Subtracting $(i)$ from (ii): $99x = 52$,so $x = \frac{52}{99}$.
Step $2$: Convert $0.4\overline{6}$ to a fraction.
Let $y = 0.4666...$ (iii)
$10y = 4.666...$ (iv)
$100y = 46.666...$ $(v)$
Subtracting (iv) from $(v)$: $90y = 42$,so $y = \frac{42}{90} = \frac{7}{15}$.
Step $3$: Subtract the fractions.
$\frac{52}{99} - \frac{7}{15} = \frac{52 \times 5 - 7 \times 33}{495} = \frac{260 - 231}{495} = \frac{29}{495}$.
Step $4$: Convert to decimal.
$\frac{29}{495} = 0.0585858... = 0.0\overline{58}$.

(N/A) Let $x = 0.1\overline{6}$. This can be written as $x = 0.1666...$ (Equation $1$).
Multiply both sides by $10$ to shift the decimal point:
$10x = 1.6666...$ (Equation $2$).
Subtract Equation $1$ from Equation $2$:
$10x - x = (1.6666...) - (0.1666...)$
$9x = 1.5$
$x = \frac{1.5}{9}$
To remove the decimal,multiply the numerator and denominator by $10$:
$x = \frac{15}{90}$
Simplify the fraction by dividing both by $15$:
$x = \frac{1}{6}$.
Thus,$0.1\overline{6} = \frac{1}{6}$.