Mix Examples - 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 $\frac{7}{6}, \frac{4}{3}, \frac{3}{2}, \frac{5}{3}, \text{ and } \frac{11}{6}$.

(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) $(i)$ 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.

(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.

(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.

(C) Between any two distinct rational numbers,there exist infinitely many rational numbers. This property is known as the density property of rational numbers.

(A) The product of two irrational numbers can be either rational or irrational.
For example:
$1$. If we take $\sqrt{2} \times \sqrt{2} = 2$,which is a rational number.
$2$. If we take $\sqrt{2} \times \sqrt{3} = \sqrt{6}$,which is an irrational number.
Therefore,the product is sometimes rational and sometimes irrational.

(FALSE, FALSE) $(i)$ The given statement is false. $A$ rational number is defined as a number that can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$. Here,$\frac{\sqrt{2}}{3}$ has $p = \sqrt{2}$,which is an irrational number,not an integer. Therefore,$\frac{\sqrt{2}}{3}$ is an irrational number.
$(ii)$ The given statement is false. By definition,integers are whole numbers (...,$-2, -1, 0, 1, 2, ...$). Between any two consecutive integers,such as $3$ and $4$,there are no other integers.

(N/A) $(i)$ The statement is false. There are infinitely many rational numbers between any two distinct rational numbers. Therefore,the number of rational numbers between $15$ and $18$ is infinite.
$(ii)$ The statement is true. Numbers that cannot be expressed in the form $\frac{p}{q}$ (where $p, q$ are integers and $q \neq 0$) are called irrational numbers. For example,$\sqrt{2}$,$\sqrt{3}$,and $\pi$ are such numbers.

(N/A) $(i)$ The statement is false. Consider the irrational number $\sqrt[4]{2}$. Its square is $(\sqrt[4]{2})^{2} = \sqrt{2}$,which is an irrational number.
$(ii)$ The statement is false. We can simplify the expression as $\frac{\sqrt{12}}{\sqrt{3}} = \sqrt{\frac{12}{3}} = \sqrt{4} = 2$. Since $2$ can be written as $\frac{2}{1}$,it is a rational number.

(N/A) $(i)$ $\sqrt{196} = 14$. Since $14$ can be expressed in the form $\frac{p}{q}$ where $p, q$ are integers and $q \neq 0$ (i.e.,$\frac{14}{1}$),it is a rational number.
$(ii)$ $3\sqrt{18} = 3\sqrt{9 \times 2} = 3 \times 3\sqrt{2} = 9\sqrt{2}$. Since $\sqrt{2}$ is an irrational number and the product of a non-zero rational number and an irrational number is always irrational,$9\sqrt{2}$ is an irrational number.

(N/A) $(i)$ $0.5918$ is a terminating decimal expansion. Any number that can be expressed as a terminating decimal is a rational number because it can be written in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$. Here,$0.5918 = \frac{5918}{10000}$,which is rational.
$(ii)$ Simplify the expression: $(1+\sqrt{5})-(4+\sqrt{5}) = 1 + \sqrt{5} - 4 - \sqrt{5} = 1 - 4 = -3$. Since $-3$ can be written as $\frac{-3}{1}$,which is in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$,it is a rational number.

(N/A) To find rational numbers between $-1$ and $-2$,we can express them as fractions with a common denominator.
Let the numbers be $-1 = -10/10$ and $-2 = -20/10$.
Any three numbers between $-20/10$ and $-10/10$ are rational numbers.
For example,$-11/10, -12/10, -13/10$ are three rational numbers between $-1$ and $-2$.
Alternatively,these can be written as $-1.1, -1.2, -1.3$.

(B) To find rational numbers between $\frac{1}{4}$ and $\frac{1}{5}$,we first make their denominators equal.
Find the Least Common Multiple $(LCM)$ of $4$ and $5$,which is $20$.
$\frac{1}{4} = \frac{1 \times 5}{4 \times 5} = \frac{5}{20}$
$\frac{1}{5} = \frac{1 \times 4}{5 \times 4} = \frac{4}{20}$
To find three rational numbers,we multiply the numerator and denominator by $(3+1) = 4$.
$\frac{5}{20} = \frac{5 \times 4}{20 \times 4} = \frac{20}{80}$
$\frac{4}{20} = \frac{4 \times 4}{20 \times 4} = \frac{16}{80}$
The rational numbers between $\frac{16}{80}$ and $\frac{20}{80}$ are $\frac{17}{80}, \frac{18}{80}, \frac{19}{80}$.
Thus,three rational numbers between $\frac{1}{4}$ and $\frac{1}{5}$ are $\frac{17}{80}, \frac{18}{80}, \frac{19}{80}$ (or $\frac{9}{40}$).

(A) rational number is a number that can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$. The average of $2$ and $3$ is $\frac{2+3}{2} = 2.5$,which is a rational number.
An irrational number is a number that cannot be expressed in the form $\frac{p}{q}$ and has a non-terminating,non-recurring decimal expansion.
An example of an irrational number between $2$ and $3$ is $2.1010010001...$ (a non-terminating and non-recurring decimal).

(N/A) To find a rational number between $\frac{1}{3}$ and $\frac{1}{2}$,we can express them with a common denominator:
$\frac{1}{3} = \frac{1 \times 4}{3 \times 4} = \frac{4}{12}$ and $\frac{1}{2} = \frac{1 \times 6}{2 \times 6} = \frac{6}{12}$.
Since $\frac{4}{12} < \frac{5}{12} < \frac{6}{12}$,the rational number $\frac{5}{12}$ lies between $\frac{1}{3}$ and $\frac{1}{2}$.
To find an irrational number,we convert the fractions to decimal form:
$\frac{1}{3} = 0.3333\ldots$ and $\frac{1}{2} = 0.5$.
An irrational number is a non-terminating and non-recurring decimal. We can choose a number like $0.414114111\ldots$,which is greater than $0.3333\ldots$ and less than $0.5$.

(N/A) First,convert the fractions to decimal form:
$\frac{-2}{5} = -0.4$ and $\frac{1}{2} = 0.5$
$1$. Rational number:
$A$ rational number is any number that can be expressed as $\frac{p}{q}$ where $p, q$ are integers and $q \neq 0$. Any terminating decimal is a rational number. We can choose $0$ (or $0.1$,$0.2$,etc.) as it lies between $-0.4$ and $0.5$.
Thus,$0$ is a rational number between $\frac{-2}{5}$ and $\frac{1}{2}$.
$2$. Irrational number:
An irrational number is a non-terminating and non-recurring decimal. We can construct one by following a pattern that does not repeat. For example,$0.1010010001...$ lies between $-0.4$ and $0.5$.
Thus,$0.1010010001...$ is an irrational number between $\frac{-2}{5}$ and $\frac{1}{2}$.

(N/A) rational number is a number that can be expressed in the form $p/q$ where $p$ and $q$ are integers and $q \neq 0$. Any terminating decimal is a rational number.
$1.$ To find a rational number between $0.15$ and $0.16$,we can choose any terminating decimal such as $0.151$. Since $0.15 < 0.151 < 0.16$,$0.151$ is a rational number.
$2.$ An irrational number is a number whose decimal expansion is non-terminating and non-recurring. To find an irrational number between $0.15$ and $0.16$,we can construct a pattern that does not repeat.
$0.15101101110...$ is an irrational number because it is non-terminating and non-recurring,and it lies between $0.15$ and $0.16$.

The decimal expansions are $\sqrt{2} \approx 1.4142135 \ldots$ and $\sqrt{3} \approx 1.7320508 \ldots$
$1$. Rational Number: $A$ terminating decimal like $1.5$ lies between $1.4142135 \ldots$ and $1.7320508 \ldots$. Since $1.5 = \frac{15}{10} = \frac{3}{2}$,it is a rational number.
$2$. Irrational Number: $A$ non-terminating and non-recurring decimal like $1.575575557 \ldots$ lies between $1.4142135 \ldots$ and $1.7320508 \ldots$. Thus,$1.575575557 \ldots$ is an irrational number.

(N/A) rational number is any number that can be expressed in the form $p/q$ where $p$ and $q$ are integers and $q \neq 0$. Since $2.357 < 3 < 3.121$,the integer $3$ is a rational number between $2.357$ and $3.121$.
An irrational number is a number whose decimal expansion is non-terminating and non-recurring. $A$ number like $3.101101110\dots$ satisfies $2.357 < 3.101101110\dots < 3.121$,therefore,it is an irrational number between the given values.

(N/A) rational number is a number that can be expressed in the form $p/q$,where $p$ and $q$ are integers and $q \neq 0$. $A$ terminating decimal is a rational number.
$1$. Rational number: We can choose $0.0005$. Since $0.0005 = 5/10000 = 1/2000$,it is a rational number lying between $0.0001$ and $0.001$.
$2$. Irrational number: An irrational number is a non-terminating and non-recurring decimal. We can construct one by following a pattern: $0.0001010010001...$ This number is clearly greater than $0.0001$ and less than $0.001$,and it does not terminate or repeat,making it an irrational number.

(N/A) To find a rational number between $0.484848$ and $3.623623$,we can choose any terminating decimal or integer within this range. For example,$1$ is a rational number because it can be expressed as $\frac{1}{1}$.
To find an irrational number between $0.484848$ and $3.623623$,we need a non-terminating and non-recurring decimal. For example,$1.909009000\dots$ is an irrational number because its decimal expansion neither terminates nor repeats.

(N/A) To represent the given numbers on the number line:
$1$. $7$: This is a positive integer. Locate $7$ on the number line to the right of $0$.
$2$. $7.2$: This is $7 + 0.2$. Divide the segment between $7$ and $8$ into $10$ equal parts. The second mark after $7$ represents $7.2$.
$3$. $\frac{-3}{2} = -1.5$: This is a negative rational number. It lies exactly halfway between $-1$ and $-2$ on the left side of $0$.
$4$. $\frac{-12}{5} = -2.4$: This is a negative rational number. Divide the segment between $-2$ and $-3$ into $5$ equal parts. The fourth mark to the left of $-2$ represents $-2.4$.

(N/A) Presentation of $\sqrt{5}$ on the number line:
We write $5$ as the sum of the squares of two natural numbers:
$5 = 1 + 4 = 1^{2} + 2^{2}$
On the number line,take $OA = 2$ units.
Draw $BA = 1$ unit,perpendicular to $OA$. Join $OB$.
By Pythagoras theorem,$OB = \sqrt{2^{2} + 1^{2}} = \sqrt{5}$.
Using a compass with center $O$ and radius $OB$,draw an arc which intersects the number line at point $C$. Then,$C$ corresponds to $\sqrt{5}$.
Presentation of $\sqrt{10}$ on the number line:
We write $10$ as the sum of the squares of two natural numbers:
$10 = 1 + 9 = 1^{2} + 3^{2}$
On the number line,take $OA = 3$ units.
Draw $BA = 1$ unit,perpendicular to $OA$. Join $OB$.
By Pythagoras theorem,$OB = \sqrt{3^{2} + 1^{2}} = \sqrt{10}$.
Using a compass with center $O$ and radius $OB$,draw an arc which intersects the number line at point $C$. Then,$C$ corresponds to $\sqrt{10}$.
Presentation of $\sqrt{17}$ on the number line:
We write $17$ as the sum of the squares of two natural numbers:
$17 = 1 + 16 = 1^{2} + 4^{2}$
On the number line,take $OA = 4$ units.
Draw $BA = 1$ unit,perpendicular to $OA$. Join $OB$.
By Pythagoras theorem,$OB = \sqrt{4^{2} + 1^{2}} = \sqrt{17}$.
Using a compass with center $O$ and radius $OB$,draw an arc which intersects the number line at point $C$. Then,$C$ corresponds to $\sqrt{17}$.

(N/A) $1$. Draw a line segment $AB = 8.1$ units on a number line.
$2$. From point $B$,mark a distance of $1$ unit and label the new point as $C$. Now,$AC = 8.1 + 1 = 9.1$ units.
$3$. Find the midpoint of $AC$ and label it as $O$.
$4$. Draw a semicircle with center $O$ and radius $OC$ (or $OA$).
$5$. Draw a line perpendicular to $AC$ at point $B$,which intersects the semicircle at point $D$. The length of $BD$ is $\sqrt{8.1}$.
$6$. With $B$ as the center and $BD$ as the radius,draw an arc that intersects the number line at point $E$. The distance $BE$ represents $\sqrt{8.1}$ on the number line.

(C) First, convert each term into a fraction:
$0.6 = \frac{6}{10} = \frac{3}{5}$
For $x = 0.\overline{7} = 0.777\ldots$
$10x = 7.777\ldots$
Subtracting $x$ from $10x$ gives $9x = 7$, so $x = \frac{7}{9}$.
For $y = 0.4\overline{7} = 0.4777\ldots$
$10y = 4.777\ldots$
$100y = 47.777\ldots$
Subtracting $10y$ from $100y$ gives $90y = 43$, so $y = \frac{43}{90}$.
Now, add the fractions:
$\frac{6}{10} + \frac{7}{9} + \frac{43}{90} = \frac{54}{90} + \frac{70}{90} + \frac{43}{90} = \frac{54 + 70 + 43}{90} = \frac{167}{90}$.
Thus, the expression is $\frac{167}{90}$.

(A) To find rational numbers between $-\frac{3}{4}$ and $-\frac{1}{3}$,first make the denominators equal by finding the Least Common Multiple $(LCM)$ of $4$ and $3$,which is $12$.
$-\frac{3}{4} = -\frac{3 \times 3}{4 \times 3} = -\frac{9}{12}$
$-\frac{1}{3} = -\frac{1 \times 4}{3 \times 4} = -\frac{4}{12}$
Since there are not enough integers between $-9$ and $-4$,multiply the numerator and denominator by a larger factor,such as $2$,to get more space.
$-\frac{9}{12} = -\frac{18}{24}$
$-\frac{4}{12} = -\frac{8}{24}$
Now,we can choose five rational numbers between $-\frac{18}{24}$ and $-\frac{8}{24}$,which are $-\frac{17}{24}, -\frac{16}{24}, -\frac{15}{24}, -\frac{14}{24}, -\frac{13}{24}$.

(B) The statement is False.
Natural numbers are the set of counting numbers starting from $1, 2, 3, ...$.
Whole numbers are the set of natural numbers including $0$,i.e.,$0, 1, 2, 3, ...$.
Since $0$ is a whole number but is not a natural number,the statement that every whole number is a natural number is false.

(A) The set of whole numbers is defined as $W = \{0, 1, 2, 3, \dots\}$.
The set of integers is defined as $Z = \{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\}$.
Since every element of the set of whole numbers is present in the set of integers,every whole number is indeed an integer.
Therefore,the statement is True.

(FALSE) The statement is False.
Whole numbers are the set of numbers ${0, 1, 2, 3, ...}$.
Integers are the set of numbers ${..., -3, -2, -1, 0, 1, 2, 3, ...}$.
Since negative numbers (e.g.,$-1, -2, -3$) are integers but are not whole numbers,the statement that 'each integer is a whole number' is incorrect.

(D) To determine the nature of the number $\frac{\sqrt{50}}{\sqrt{98}}$,we simplify the expression:
$\frac{\sqrt{50}}{\sqrt{98}} = \sqrt{\frac{50}{98}}$
Divide both the numerator and the denominator by $2$:
$\sqrt{\frac{50 \div 2}{98 \div 2}} = \sqrt{\frac{25}{49}}$
Since $\sqrt{25} = 5$ and $\sqrt{49} = 7$,the expression simplifies to $\frac{5}{7}$.
$A$ number that can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$,is a rational number.
Since $\frac{5}{7}$ fits this definition,the number is rational.

(D) To determine the nature of the number $\sqrt{5^{2}+12^{2}}$,we first calculate the value inside the square root.
Step $1$: Calculate the squares: $5^{2} = 25$ and $12^{2} = 144$.
Step $2$: Add the values: $25 + 144 = 169$.
Step $3$: Find the square root: $\sqrt{169} = 13$.
Since $13$ is a positive integer,it belongs to the set of natural numbers. Therefore,the correct option is $D$.

(C) The given expression is $\sqrt{5} \times \sqrt{5}$.
By the property of square roots,$\sqrt{a} \times \sqrt{a} = a$ for any $a \ge 0$.
Therefore,$\sqrt{5} \times \sqrt{5} = 5$.
The number $5$ is a positive integer,which belongs to the set of natural numbers $\{1, 2, 3, \dots\}$.
Thus,$\sqrt{5} \times \sqrt{5}$ is a natural 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 $0$.