Real Numbers and their Decimal Expansions Questions in English

(N/A) Let $x = 0.142857142857...$ (Equation $1$).
Since there are $6$ repeating digits,multiply both sides of Equation $1$ by $10^6 = 1,000,000$:
$1,000,000x = 142857.142857142857...$ (Equation $2$).
Subtract Equation $1$ from Equation $2$:
$1,000,000x - x = 142857.142857... - 0.142857...$
$999,999x = 142857$.
$x = \frac{142857}{999999}$.
Dividing both numerator and denominator by $142857$,we get:
$x = \frac{1}{7}$.
Thus,$0.\overline{142857} = \frac{1}{7}$.

(N/A) Let $x = 0.\overline{076923}$.
This can be written as $x = 0.076923076923...$ (Equation $1$).
Since there are $6$ repeating digits, multiply both sides by $10^6 = 1,000,000$:
$1,000,000x = 76923.076923076923...$ (Equation $2$).
Subtract Equation $1$ from Equation $2$:
$1,000,000x - x = 76923.076923... - 0.076923...$
$999,999x = 76923$.
$x = \frac{76923}{999999}$.
Dividing both numerator and denominator by $76923$:
$x = \frac{76923 \div 76923}{999999 \div 76923} = \frac{1}{13}$.
Thus, $0.\overline{076923} = \frac{1}{13}$.

(N/A) To visualize $2.64444$ on the number line,we use the process of successive magnification:
$1$. Since $2.64444$ lies between $2$ and $3$,we divide the distance between $2$ and $3$ into $10$ equal parts and locate $2.6$ and $2.7$.
$2$. Now,$2.64444$ lies between $2.6$ and $2.7$. We magnify this region and divide it into $10$ equal parts to locate $2.64$ and $2.65$.
$3$. Next,$2.64444$ lies between $2.64$ and $2.65$. We magnify this region and divide it into $10$ equal parts to locate $2.644$ and $2.645$.
$4$. Continuing this process,$2.64444$ lies between $2.644$ and $2.645$. We magnify this region to locate $2.6444$ and $2.6445$.
$5$. Finally,we magnify the region between $2.6444$ and $2.6445$ to locate the point $2.64444$ on the number line.

(N/A) Step $1$: Locate $2.3$ and $2.4$ on the number line. Divide the distance between $2.3$ and $2.4$ into $10$ equal parts. The first mark represents $2.31$,the second $2.32$,and so on.
Step $2$: Locate $2.36$ and $2.37$ on the number line. Divide the distance between $2.36$ and $2.37$ into $10$ equal parts.
Step $3$: The $5$th mark between $2.36$ and $2.37$ represents $2.365$.

(N/A) To visualise $-4.126$ on the number line using successive magnification,we follow these steps:
$1$. Since $-4.126$ lies between $-4$ and $-5$,we divide the segment between $-4$ and $-5$ into $10$ equal parts and locate $-4.1$ and $-4.2$.
$2$. Now,$-4.126$ lies between $-4.1$ and $-4.2$. We magnify this segment and divide it into $10$ equal parts to locate $-4.12$ and $-4.13$.
$3$. Finally,$-4.126$ lies between $-4.12$ and $-4.13$. We magnify this segment and divide it into $10$ equal parts to locate the point representing $-4.126$ exactly.

To represent $3.\overline{42}$ up to $4$ decimal places, we need to locate $3.4242$ on the number line.
Step $1$: $3.4242$ lies between $3$ and $4$. Divide the interval $[3, 4]$ into $10$ equal parts and magnify the interval $[3.4, 3.5]$.
Step $2$: $3.4242$ lies between $3.4$ and $3.5$. Divide the interval $[3.4, 3.5]$ into $10$ equal parts and magnify the interval $[3.42, 3.43]$.
Step $3$: $3.4242$ lies between $3.42$ and $3.43$. Divide the interval $[3.42, 3.43]$ into $10$ equal parts and magnify the interval $[3.424, 3.425]$.
Step $4$: $3.4242$ lies between $3.424$ and $3.425$. Divide the interval $[3.424, 3.425]$ into $10$ equal parts. The point $3.4242$ is the $2$nd mark after $3.424$.

(D) To represent $5.4949$ on the number line using successive magnification:
$1$. Locate $5.4949$ between $5$ and $6$. Divide the interval into $10$ equal parts to find $5.4$ and $5.5$.
$2$. Magnify the interval between $5.4$ and $5.5$. Divide it into $10$ equal parts to locate $5.49$ and $5.50$.
$3$. Magnify the interval between $5.49$ and $5.50$. Divide it into $10$ equal parts to locate $5.494$ and $5.495$.
$4$. Magnify the interval between $5.494$ and $5.495$. Divide it into $10$ equal parts. The point $5.4949$ lies between $5.494$ and $5.495$,specifically at the $9^{th}$ subdivision.

(A) Let $x = 0.\overline{3} = 0.333\dots$ (Equation $1$).
Multiply both sides by $10$:
$10x = 3.333\dots$ (Equation $2$).
Subtract Equation $1$ from Equation $2$:
$10x - x = 3.333\dots - 0.333\dots$
$9x = 3$
$x = \frac{3}{9} = \frac{1}{3}$.
Therefore,$0.\overline{3} = \frac{1}{3}$.

(B) Let $x = 4.185185185...$ (Equation $1$)
Since there are $3$ repeating digits,multiply by $1000$:
$1000x = 4185.185185...$ (Equation $2$)
Subtract Equation $1$ from Equation $2$:
$1000x - x = 4185.185185... - 4.185185...$
$999x = 4181$
$x = \frac{4181}{999}$
Now,convert to a mixed fraction:
$x = 4 + \frac{181}{999}$
Wait,checking the repeating decimal $4.185185...$ as a fraction:
$x = 4 + \frac{185}{999}$
Simplify $\frac{185}{999}$ by dividing by $37$:
$185 \div 37 = 5$
$999 \div 37 = 27$
So,$x = 4 \frac{5}{27}$.

(C) Let $x = 0.3\overline{6} = 0.3666...$ (Equation $1$)
Multiply both sides by $10$: $10x = 3.666...$ (Equation $2$)
Multiply Equation $1$ by $100$: $100x = 36.666...$ (Equation $3$)
Subtract Equation $2$ from Equation $3$:
$100x - 10x = 36.666... - 3.666...$
$90x = 33$
$x = \frac{33}{90}$
Simplify the fraction by dividing both numerator and denominator by $3$:
$x = \frac{11}{30}$

(D) To find a rational number between $7$ and $8$,we evaluate each option:
$A) \frac{33}{5} = 6.6$ (This is less than $7$)
$B) \frac{51}{6} = 8.5$ (This is greater than $8$)
$C) \frac{60}{7} \approx 8.57$ (This is greater than $8$)
$D) \frac{61}{8} = 7.625$ (Since $7 < 7.625 < 8$,this is the correct rational number between $7$ and $8$).

(B) To determine the decimal expansion of the rational number $-\frac{11}{4}$,we perform the division:
$11 \div 4 = 2.75$.
Since the division results in a finite number of decimal places,the decimal expansion is terminating.
Therefore,$-\frac{11}{4} = -2.75$,which is a terminating decimal.

(N/A) The square root of a number $x$ is a value that,when multiplied by itself,gives $x$.
Since $11 \times 11 = 121$ and $(-11) \times (-11) = 121$,the square root of $121$ is $11$ or $-11$.

(B) To find the decimal expansion of $\frac{2}{3}$, we perform long division by dividing $2$ by $3$.
When we divide $2$ by $3$, we get $0.666...$ which can be written as $0.\overline{6}$.
Since the remainder never becomes $0$ and the digit $6$ repeats infinitely, the decimal expansion is non-terminating and recurring.

(TERMINATING) To determine the type of decimal expansion of $\frac{47}{50}$,we first simplify the fraction or examine the denominator.
Given fraction: $\frac{47}{50}$.
The denominator is $50 = 2^1 \times 5^2$.
Since the denominator is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers,the decimal expansion is terminating.
Calculation: $\frac{47}{50} = \frac{47 \times 2}{50 \times 2} = \frac{94}{100} = 0.94$.
Therefore,the decimal expansion is terminating.

(A) Let $x = 3.\overline{5} = 3.555...$ (Equation $1$).
Multiply both sides by $10$:
$10x = 35.555...$ (Equation $2$).
Subtract Equation $1$ from Equation $2$:
$10x - x = 35.555... - 3.555...$
$9x = 32$.
Therefore, $x = \frac{32}{9}$.

(B) Let $x = 0.272727...$ (Equation $1$)
Since there are two repeating digits, multiply both sides by $100$:
$100x = 27.272727...$ (Equation $2$)
Subtract Equation $1$ from Equation $2$:
$100x - x = 27.272727... - 0.272727...$
$99x = 27$
$x = \frac{27}{99}$
Simplify the fraction by dividing both numerator and denominator by $9$:
$x = \frac{3}{11}$

(B) To determine the decimal expansion of $\frac{1}{7}$,we perform long division of $1$ by $7$.
Dividing $1$ by $7$ gives $0.142857142857...$
Since the remainder never becomes $0$ and the sequence of digits $142857$ repeats indefinitely,the decimal expansion is non-terminating and recurring.

(A) number is rational if it can be expressed in the form $p/q$,where $p$ and $q$ are integers and $q \neq 0$.
Any repeating decimal is a rational number.
Let $x = 0.3\overline{7} = 0.3777...$
Multiply by $10$: $10x = 3.777...$
Multiply by $100$: $100x = 37.777...$
Subtract the two equations: $100x - 10x = 37.777... - 3.777...$
$90x = 34$
$x = 34/90 = 17/45$.
Since $0.3\overline{7}$ can be expressed as $17/45$,it is a rational number.

(N/A) The given number is $3.8232323 \ldots$.
Here,the digit $23$ is repeating after the decimal point.
Therefore,we can represent this repeating decimal by placing a bar over the repeating digits.
Thus,the short form is $3.8 \overline{23}$.