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Representation of irrational numbers on number line - class-XI

Description: representation of irrational numbers on number line
Number of Questions: 18
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Tags: real numbers basic algebra number system maths number systems real numbers (rational and irrational numbers)
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Which of the following numbers lie between $1$ and $3$?

  1. $\dfrac13$

  2. $\sqrt{2}$

  3. $\sqrt{10}$

  4. $\dfrac{8}3$


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.

  1. an integer

  2. an irrational number

  3. a natural number

  4. a rational number


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.

  1. $1,2,3,4$

  2. $2,4,1,3$

  3. $3,2,4,1$

  4. $3,2,1,4$


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 $8$?

  1. $\sqrt{49}$

  2. $\sqrt{19}$

  3. $\sqrt{47}$

  4. $\sqrt{62}$


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 $b+a = \, ?$

  1. $4$

  2. $5$

  3. $7$

  4. None of these


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?

  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


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 ____.

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

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

  3. Both are equal

  4. Cannot comare


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 :

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

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

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

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


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?

  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


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:

  1. $1$ and $1$

  2. $1$ and $2$

  3. $2$ and $0$

  4. $-1$ and $-1$


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).

Use ______________ to represent an irrational number on number line.

  1. Isosceles-angle theorem

  2. Scalene angle theorem

  3. Right-angled theorem

  4. None of the above


Correct Option: C
Explanation:
Using the Pythagoras Theorem, we can represent some irrational numbers, which are surds, on a number line.
Since it involves Pythagoras Theorem, we get to use Right Angle Theorem.

Hence, to represent an irrational number, we generally use right angled theorem.

$D$ is a real number with non terminating digits $a _1$ and $a _2$ after the decimal point. Let $D = 0, a _1 a _2 a _1 a _2 ........ $  with $a _1 & a _2$ both not zero which of the following when multiplied by $D$ will necessarily give an integer ?

  1. $99$

  2. $18$

  3. $125$

  4. $75$


Correct Option: A
Explanation:

its straight question
give $D=0.abababab$ $(say - 1)$ 
Multiply both sides by $100$ $i.e.$ 
$100D = ab.abababab$ $(say - 2)$
now subtract $1$ from $2 .$ That gives
$99D = ab => D = ab/99$ hence it should be multiplied by $99k$ to get an integer ab$.$

The number $5\sqrt{34}$ lies between

  1. $29$ and $30$

  2. $30$ and $31$

  3. $31$ and $32$

  4. $32$ and $33$


Correct Option: A
Explanation:
$5\sqrt {34}\ \le \ 5\sqrt {36} \ = \ 5\times 6=30$
less than $30$
i.e., $=29$ to $30$


Can $\sqrt { 3 } -3$ be represented on the number line. 

  1. True

  2. False


Correct Option: A
Explanation:

Yes, 


Number line consists of all real numbers i.e., Both rational and Irrational Numbers

Since $\sqrt 3-3$ is a real number It can be represented on number line 

Give an example of two irrational numbers whose difference is an irrational number.

  1. $\sqrt{3},-\sqrt{3}$

  2. $\sqrt{5,}-\sqrt{5}$

  3. $4\sqrt{3},-2\sqrt{3}$

  4. None of the above


Correct Option: C
Explanation:

$4\sqrt{3},2\sqrt{3}$ are the irrational numbers and thier difference,


$4\sqrt{3}-2\sqrt{3}=2\sqrt 3$ is also an irrational number.

Which is the wrong step that shows $\displaystyle 5-\sqrt{3}$ is irrational?
(I) Contradiction : Assume that $\displaystyle 5-\sqrt{3}$ is rational
(II) Find coprime a & b $\displaystyle \left ( b\neq 0 \right )$ such that $\displaystyle 5-\sqrt{3}=\frac{a}{b},\therefore 5-\frac{a}{b}=\sqrt{3}$
Rearranging above equation $\displaystyle \sqrt{3}=5-\frac{a}{b}=\frac{5b-a}{b}$
(III) Since a & b are integers we get $\displaystyle 5-\frac{a}{b}$ is irrational and so $\displaystyle \sqrt{3}$ is irrational
(IV) But this contradicts the fact that $\displaystyle \sqrt{3}$ is irrational Hence $\displaystyle 5-\sqrt{3}$ is irrational

  1. Both I and II

  2. Only III

  3. Only II

  4. Both II and III


Correct Option: B

Which of the following irrational numbers lie between $4$ and $7$?

  1. $\sqrt{25}$

  2. $\sqrt{19}$

  3. $\sqrt{47}$

  4. $\sqrt{50}$


Correct Option: B,C
Explanation:

$4^{2} = 16$

$5^{2} = 25$
$6^{2} = 36$
$7^{2} = 49$


$\Rightarrow \sqrt19$ and $\sqrt47$ are irrational numbers which lie between $4$ and $7$


$\sqrt25 = 5$ which is a rational number

The ascending order of the surds $\sqrt[3]{2}, \sqrt[6]{3}, \sqrt[9]{4}$ is 

  1. $\sqrt[9]{4}, \sqrt[6]{3}, \sqrt[3]{2}$

  2. $\sqrt[9]{4}, \sqrt[3]{2}, \sqrt[6]{3}$

  3. $\sqrt[3]{2}, \sqrt[6]{3}, \sqrt[9]{4}$

  4. $\sqrt[6]{3}, \sqrt[9]{4}, \sqrt[3]{2}$


Correct Option: A
Explanation:

Lets keep in mind the following common log values:

$log(2)=0.30$
$log(3)=0.47$
Lets compare these three surds by their log values.
Reason: Since $log(x) $ is an increasing function  when $x>1$ so we can compare these values by comparing their log values.

$log(\sqrt[3]{2})=log(2^{\frac{1}3})=\left(\dfrac{1}3\right)log(2)=\left(\dfrac{1}3\right) \times 0.3=0.100$

$log(\sqrt[6]{3})=log(3^{\frac{1}6})=\left(\dfrac{1}6\right)log(3)=\left(\dfrac{1}6\right) \times 0.47=0.078$

$log(\sqrt[9]{4})=log(\sqrt[9]{2^2})=log(2^{\frac{2}9})=\left(\dfrac{2}9\right)log(2)=\left(\dfrac{2}9\right) \times 0.3=0.060$

By looking at the log values,it is clear that,

$\sqrt[9]{4} < \sqrt[6]{3} < \sqrt[3]{2}$

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