Tag: rational and irrational numbers

Questions Related to rational and irrational numbers

Which of the following are irrational numbers?
(i) $\sqrt{2+\sqrt{3}}$
(ii) $\sqrt{4+\sqrt{25}}$
(iii) $\sqrt[3]{5+\sqrt{7}}$
(iv) $\sqrt{8-\sqrt[3]{8}}$.

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

  1. (ii), (iii) and (iv)

  2. (i), (ii) and (iv)

  3. (i), (ii) and (iii)

  4. (i), (iii) and (iv)

Show answer
Correct Option: D
Explanation:

Option (i)

$\sqrt3$ is irrational, so (i) is irrational.

Option (ii)
$\sqrt{25} = 5$, so we get $\sqrt{4+5} = \sqrt9 = 3$ which is rational.

Option (iii)
$\sqrt7$ is irrational, so (iii) is irrational.

Option (iv)
$\sqrt[3]{8} = 2$, so we get $\sqrt{8-2} = \sqrt6$ which is irrational.

$\therefore$ (i),(iii) and (iv) are irrational.

Which one of the following is an irrational number?

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

  1. $\pi$

  2. $\sqrt{9}$

  3. $\displaystyle\frac{1}{4}$

  4. $\displaystyle\frac{1}{5}$

Show answer
Correct Option: A
Explanation:

A number having non-terminating and non-recurring decimal expansion is  a Irrational Number


A number having non-terminating and recurring decimal expansion is  a Rational Number

now looking at the options

$\pi$  is an irrational number 

$\pi = 3.1415926535897932384626433832............$


the number has non-terminating decimal expansion and non-recurring.

$\sqrt9 = 3$  is a rational number

$\dfrac14 = 0.25$ is a rational number

$\dfrac15 = 0.2$ is a rational number

So option $A $ is correct

Let $x$ be an irrational number then what can be said about ${x}^{2}$

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

  1. It is rational

  2. It can be irrational.

  3. It can be rational.

  4. Both $B$ and $C$

Show answer
Correct Option: D
Explanation:

$x$ is any irrational number 
Let $x=\sqrt [ 4 ]{ 3 } $
$\Rightarrow { x }^{ 2 }=\sqrt { 3 } $
which is irrational so option $B$ is correct.
Now let $x=\sqrt 3$
$\Rightarrow {x}^{2}=3$
which is rational so option $C$ is correct.
So correct answer is $D$

State the following statement is true or false.

7.4848..is an irrational number.

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

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

$7.4848...=7.\bar{48}\Rightarrow $Rational  number

State true or false:
There are numbers which cannot be written in the form $\frac{p}{q}$, where $q\neq 0$  and both p, q are integers.

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Any irrational number can not be written as $\dfrac{p}{q}$.

The product of a non-zero rational number with an irrational number is always :

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

  1. Irrational number

  2. Rational number

  3. Whole number

  4. Natural number

Show answer
Correct Option: A
Explanation:

By definition, an irrational number in decimal form goes on forever without repeating (a non-repeating, non-terminating decimal). By definition, a rational number in decimal form either terminates or repeats. 

By multiplying a non repeating non terminating number to repeating or terminating/repeating number, the result will always be a non terminating non repeating number. 
So, option A is correct. 

Which is not an Irrational number?

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

  1. $5-\sqrt{3}$

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

  3. $4+\sqrt{2}$

  4. $6+\sqrt{9}$

Show answer
Correct Option: D
Explanation:

We know that sum of two irrational number or one rational and one irrational number will be irrational number. Option A, B , C stisfies this criteria but option D have two rational number i.e. $6 + \sqrt { 9 }$ = $6+ 3=9$

So correct answer is option D

$\left ( 2+\sqrt{5} \right )\left ( 2+\sqrt{5} \right )$ expression is :

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

  1. A rational number

  2. A whole number

  3. An irrational number

  4. A natural number

Show answer
Correct Option: C
Explanation:

${ (2+\sqrt { 5 } ) }^{ 2 }\ =4+5+4\sqrt { 5 } \ =9+4\sqrt { 5 } $

In the above equation $4\sqrt { 5 } $ is irrational number so $9+4\sqrt { 5 } $ will also be irrational number 
So correct answer is option C.

A pair of irrational numbers whose product is a rational number is:

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

  1. $\sqrt{16}, \sqrt{4}$

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

  3. $\sqrt{3}, \sqrt{27}$

  4. $\sqrt{36}, \sqrt{2}$

Show answer
Correct Option: C
Explanation:

In the given options,

for option $(A)$ $\sqrt { 16 } \& \sqrt { 4 }$ are not irrational numbers i.e. their real values are 4 & 2 respectively. It cannot be correct answer. 

Now multiplying other options, we get

$(B):\sqrt { 5 } \times \sqrt { 2 } =\sqrt { 10 } $

$(C)\sqrt { 27 } \times \sqrt { 3 } =\sqrt { 81 } =9$

$(D) \sqrt { 36 } \times \sqrt { 2 } =\sqrt { 72 } $

So, correct answer is option C.

A number is an irrational if and only if its decimal representation is :

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

  1. non terminating

  2. non terminating and repeating

  3. non terminating and non repeating

  4. terminating

Show answer
Correct Option: C
Explanation:

According to definition of irrational number, If written in decimal notation, an irrational number would have an infinite number of digits to the right of the decimal point, without repetition.

So, correct answer is option C.