Tag: representation of irrational numbers on number line

Questions Related to representation of irrational numbers on number line

Which of the following numbers lie between $1$ and $3$?

maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

  1. $\dfrac13$

  2. $\sqrt{2}$

  3. $\sqrt{10}$

  4. $\dfrac{8}3$

Show answer
Correct Option: B,D
Explanation:

Converting all option in decimal format (approximately)

A. $\frac{1}{3}=0.33$
B.$\sqrt {2  } =1.41$
C.$ \sqrt { 10 }=3.16 $
D. $\frac{8}{3}=2.67$
It is clear that option B and D lies between 1 and 3
So correct answer will be option B and D

Between any $2$ real numbers, __________ can always be represented on a number line.

maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

  1. an integer

  2. an irrational number

  3. a natural number

  4. a rational number

Show answer
Correct Option: B,D
Explanation:

Between any two real number, there is always many rational number.

And between any two rationals there is always an irrational numbers.
Thus, we can represent rationals and irrationals between any two reals.
Hence, option B and D is correct.

Following are the steps to represent $\sqrt5$  on number line.
Arrange them in order.
1) Draw OC on line with $l(OC)=l(OB)$,
2) Draw $AB \perp OA\ and\ l(AB) =1$
3) Take $l(OA)=2$
4) $l(OC)=\sqrt5$, C is required point on real line.

maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

  1. $1,2,3,4$

  2. $2,4,1,3$

  3. $3,2,4,1$

  4. $3,2,1,4$

Show answer
Correct Option: D
Explanation:

The correct order of representing $\sqrt { 5 } $ on number line is 

Step 1. Take $l(OA) = 2$.
Step 2. Draw $AB$$\perp $$OA\ and\  l(AB)$ $= 1$
Step 3. Draw $OC$ on line with $l(OC) = l(OB)$
Step 4. $l(OC) =$ $\sqrt { 5 } $, $C$ is the required point on real line
Therefore, option(D) is correct.

Which of the following irrational numbers lie between $6$ and $8$?

maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

  1. $\sqrt{49}$

  2. $\sqrt{19}$

  3. $\sqrt{47}$

  4. $\sqrt{62}$

Show answer
Correct Option: C,D
Explanation:

$6^{2} = 36$

$7^{2} = 49$
$8^{2} = 64$

$\Rightarrow \sqrt47$ and $\sqrt62$ are only irrational numbers from the stated that lie in between $6$ and $8$.

$\sqrt49 = 7$ which isn't irrational.

The number $\sqrt{10}$ lies between $2$ integers $a$ and $b$ such that $b-a = 1$. Then $b+a = \, ?$

maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

  1. $4$

  2. $5$

  3. $7$

  4. None of these

Show answer
Correct Option: C
Explanation:

$1^{2} = 1$

$2^{2} = 4$
$3^{2} = 9$
$4^{2} = 16$

$\Rightarrow \sqrt10$ lies between $3$ and $4$

$4 = a , 3 = b$ 

$\therefore a+b = 7$

Which one of the following is not true?

maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

  1. $\sqrt{2}$ is an irrational number

  2. $\sqrt{17}$ is a irrational number

  3. $0.10110011100011110...$ is an irrational number

  4. $\sqrt[4]{16}$ is an irrational number

Show answer
Correct Option: D
Explanation:

$2^{4} =16$


$\because \sqrt[4]{16} = 2$, which is a rational number.

The greater number between $\sqrt{17}-\sqrt{12}$ and $\sqrt{11}-\sqrt{6}$ is ____.

maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

  1. $\sqrt{17}-\sqrt{12}$

  2. $\sqrt{11}-\sqrt{6}$

  3. Both are equal

  4. Cannot comare

Show answer
Correct Option: B
Explanation:

$\sqrt{17}=4.12\ \sqrt {12}=3.46\ \therefore\sqrt{17}-\sqrt{12}=0.66$

$\sqrt{11}=3.32\ \sqrt6=2.45\ \therefore\sqrt{11}-\sqrt6=0.87$

$\sqrt{11}-\sqrt6>\sqrt{17}-\sqrt{12}$

The value of $0.\overline{2}$ in the form $\frac{p}{q}$ , where p and q are integers and $q\ne 0$ is :

maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

  1. $\dfrac{1}{5}$

  2. $\dfrac{2}{9}$

  3. $\dfrac{2}{5}$

  4. $\dfrac{1}{8}$

Show answer
Correct Option: B
Explanation:

$Let\quad x=.222....\ On\quad multiplying\quad by\quad 10\quad on\quad both\quad sides\quad \ 10x=2.222....\ On\quad subtracting\quad both\quad equations\quad \ 9x=2\ x=\dfrac { 2 }{ 9 } $

Hence,correct answer is option B.

Which of the following is/are correct?

maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

  1. Every integer is a rational number.

  2. The sum of a rational number and an irrational number is an irrational number.

  3. Every real number is rational.

  4. Every point on the number line is associated with a real number

Show answer
Correct Option: A,B,D
Explanation:

Yes every integer can be represented in the form of $p/q$ where q is 1 for integers, so every integer is a rational number.
The sum of rational and irrational numbers is always irrational.
No, every real number is not rational. Real numbers are classified as rational numbers and irrational numbers.
Any number on the number line is a real number because that number can be either rational or irrational. Rational and irrational numbers together form real numbers.

To represent a rational number $\sqrt{2}$ on number line, take sides of right triangle as:

maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

  1. $1$ and $1$

  2. $1$ and $2$

  3. $2$ and $0$

  4. $-1$ and $-1$

Show answer
Correct Option: A
Explanation:
Notice that $\sqrt{2}= \sqrt{(1^2 + 1^2)}$. So, we can form a length of $\sqrt{2}$ units using two mutually perpendicular sides of length $1$ unit each. (Since, $1,1,\sqrt{2}$ form sides of a right angled triangle by Pythagoras theorem).