Tag: rational and irrational numbers

Questions Related to rational and irrational numbers

Give an example of two irrational numbers, whose product is a rational number.

maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

  1. $\sqrt{8},\sqrt{2}$

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

  3. $2+\sqrt{8},\sqrt{2}$

  4. $\sqrt{8},2+\sqrt{2}$

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Correct Option: A
Explanation:

Let be the Number are $\sqrt{8}  and  \sqrt{2}$
Product of Numbers  $\sqrt{8}\times \sqrt{2} = \sqrt{16} = 4$
Which is a rational number

Give an example of two irrational numbers, whose product is an irrational number.

maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

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

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

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

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

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Correct Option: D
Explanation:

Let be the Number are $\sqrt{2}  and  \sqrt{3}$
Product of Numbers  $\sqrt{2}\times \sqrt{3} = \sqrt{6} $
Which is a irrational number

$\displaystyle log _{4}18$ is 

maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

  1. an irrational number

  2. a rational number

  3. natural number

  4. whole number

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Correct Option: A
Explanation:

$  Log AB = log A + log B $
Also, $ log a^b = b log a $

So, $ log _{4} 18 = \frac { log 2 \times 3^2}{log 4}  = \frac { log 2 \times 3^2}{log 2^2}  = \frac { log 2}{2log 2} + \frac {2 log 3}{2 log 2}  = \frac {1}{2} +  \frac {log 3}{log 2}   $

As both $ log 2 $ and $ log 3 $ are irrational numbers, $ log _{x} 18 $ is an irrational number too. 

Number of integers lying between $1 $ to $102$  which are divisible by all $\displaystyle \sqrt{2},\sqrt{3},\sqrt{6}, $ is 

maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

  1. $16$

  2. $17$

  3. $15$

  4. $0$

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Correct Option: D
Explanation:

For a number to be divisible by $\sqrt { 2 } $, it must be an irrational number. An integer is not an irrational,

so  there are no  numbers between  $ 1$ to  $102$ which are divisible by all  $\sqrt{2},\sqrt{3},\sqrt{6}$.

Simplify by combining similar terms :$\displaystyle 3\sqrt{147}-\frac{7}{3}\sqrt{\frac{1}{3}}+7\sqrt{\frac{1}{3}}$

maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

  1. $\displaystyle \frac{189}{3\sqrt{3}}$

  2. $\displaystyle \frac{175}{3\sqrt{3}}$

  3. $\displaystyle \frac{208\sqrt{3}}{3}$

  4. $\displaystyle \frac{203}{3\sqrt{3}}$

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Correct Option: D

Which of the following is an irrational number?

maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

  1. $\sqrt {41616}$

  2. $23.232323.....$

  3. $\dfrac {(1+\sqrt 3)^3-(1-\sqrt 3)^3}{\sqrt 3}$

  4. $23.10100100010000....$

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Correct Option: D
Explanation:

Irrational numbers are non-repeating and non-terminating numbers.

$\sqrt {5}$ is a\an ......... number.

maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

  1. rational

  2. whole

  3. integer

  4. irrational

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Correct Option: D
Explanation:

$\sqrt {5} = \dfrac {a}{b}$

$b\sqrt {5} = a$ $(a$ and $b$ are co-prime i.e. they have no common factors$) ...(1)$ 
$5b^{2} = a^{2}$ (squaring both sides)
Therefore $5$ divides $a^{2}$
As per Fundamental Theorem of Arithmetic, $5$ divides $a.$
Let's take it as $a = 5c,$
$5b^{2} = 25 c^{2}$
$b^{2} = 5c^{2}$
As per Fundamental Theorem of Arithmetic, $5$ divides $a.$
So $a$ and $b$ have $5$ as a common factor but $a$ and $b$ have only $1$ common factor $1$ from equation $(1),$ so it is not rational.
So, we conclude that $\sqrt {5}$ is irrational.
Therefore, $D$ is the correct answer.

How many irrational numbers are there between $2$ and $6$?

maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

  1. $1$

  2. $3$

  3. $4$

  4. $10$

  5. Infinitely many

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Correct Option: E
Explanation:

There can be infinite number of irrational numbers between any two rational numbers. Hence the answer is infinitely many.

$\sqrt{21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}}=$...........

maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

  1. $\sqrt{5}-2+2\sqrt{3}$

  2. $\sqrt{5}-\sqrt{4}-\sqrt{12}$

  3. $-\sqrt{5}+\sqrt{4}+\sqrt{12}$

  4. $-\sqrt{5}-\sqrt{4}+\sqrt{12}$

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Correct Option: C

State whether the following statements are true or false. 
$\sqrt {n}$ is not irrational if n is a perfect square

maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

  1. True

  2. False

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Correct Option: A
Explanation:

False ,

$\sqrt{4}=2$ where 2 is a rational number.Here n is perfect square the  $\sqrt{n}$ is rational number 
$\sqrt{5}=2.236..$ is not rational  number But it is irrational number . here n is not a perfect square the  $\sqrt{n}$ is  irrational  number
So $\sqrt{n}$ is not irrational number if n is perfect square