Mix Examples - Number Systems Questions in English

(A) number is called rational if it can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.
Since $\frac{22}{7}$ is expressed as a ratio of two integers $22$ and $7$,where $7 \neq 0$,it satisfies the definition of a rational number.
Therefore,$\frac{22}{7}$ is a rational number.

(D) Natural numbers are the set of counting numbers starting from $1$, i.e., $N = \{1, 2, 3, ...\}.$
Whole numbers are the set of natural numbers including zero, i.e., $W = \{0, 1, 2, 3, ...\}.$
Comparing the two sets, the number $0$ is present in the set of whole numbers but is not included in the set of natural numbers.
Therefore, the correct answer is $0$.

(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) To prove the expression,we simplify the left-hand side $(LHS)$ step by step using the laws of exponents:
$1$. Simplify the numerator and denominator of the first fraction: $\frac{x^{ab-ac}}{x^{ba-bc}}$.
$2$. Simplify the term inside the parenthesis: $\left(\frac{x^b}{x^a}\right)^c = (x^{b-a})^c = x^{bc-ac}$.
$3$. Now,the expression becomes: $\frac{x^{ab-ac}}{x^{ab-bc}} \div x^{bc-ac}$.
$4$. Using the division rule $x^m / x^n = x^{m-n}$,the first part is: $x^{(ab-ac) - (ab-bc)} = x^{ab-ac-ab+bc} = x^{bc-ac}$.
$5$. Finally,divide by the second term: $x^{bc-ac} \div x^{bc-ac} = x^{(bc-ac) - (bc-ac)} = x^0 = 1$.
Thus,$LHS = RHS = 1$.

(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.

(A) To determine whether $\sqrt{8^{2}+15^{2}}$ is a rational or an irrational number,we first simplify the expression inside the square root.
Step $1$: Calculate the squares of the numbers.
$8^{2} = 64$
$15^{2} = 225$
Step $2$: Add the results.
$64 + 225 = 289$
Step $3$: Find the square root of the sum.
$\sqrt{289} = 17$
Since $17$ can be expressed as $\frac{17}{1}$,which is in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$,it is a rational number.

(B) First,simplify the expression inside the square root: $8 + 15 = 23$.
Therefore,the expression becomes $\sqrt{23}$.
Since $23$ is not a perfect square,its square root cannot be expressed as a fraction $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.
Thus,$\sqrt{23}$ is an irrational 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 $3.8 \overline{23}$.