Tag: existence of irrational numbers

Questions Related to existence of irrational numbers

State whether the following statements are true or false. Justify your answers.
Every real number need not be a rational number

  1. True

  2. False


Correct Option: A
Explanation:

Real number are all numbers on number line 

And  a rational number is any number that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non-zero denominator q. Since q may be equal to 1, every integer is a rational number.
And other numbers are not rational number are called irrational number.
Then every real number need not be a rational is true
Eg:  $\sqrt{3},\sqrt{2},\pi $

State whether the following statement is true or false:
All real numbers are irrational

  1. True

  2. False


Correct Option: B
Explanation:

False,

The set of real numbers is made by combining the set of rational numbers and the set of irrational numbers. The real numbers include natural numbers or counting numbers, whole numbers, integers, rational numbers (fractions and repeating or terminating decimals), and irrational numbers.
Real number includes number like $\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{7}....$ which are not irrational numbers.

So the statement, all real numbers are irrational is false.

Classify the following numbers as rational or irrational:  $\displaystyle \frac{\sqrt{12}}{\sqrt{75}}$

  1. Rational

  2. Irrational

  3. Can't be determined

  4. None of these


Correct Option: A
Explanation:

$\dfrac { \sqrt { 12 }  }{ \sqrt { 75 }  } =\dfrac { 2\sqrt { 3 }  }{ 5\sqrt { 3 }  } =\dfrac { 2 }{ 3 } $ which is a rational number 

Hence, the correct answer will be option A

Which of the following number is different from others?

  1. $\sqrt 7$

  2. $\sqrt 6$

  3. $\sqrt {25}$

  4. $\sqrt{10}$


Correct Option: C
Explanation:

$\sqrt{7}$ is an irrational number

$\sqrt{6}$ is an irrational number
$\sqrt{10}$ is an irrational number
$\sqrt{25}=5$ is different from others because others are irrational number but $\sqrt{25}$ is a rational number
Hence, option C is correct.

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

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

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

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

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


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?

  1. $\pi$

  2. $\sqrt{9}$

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

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


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 ${x}^{2}$

  1. It is rational

  2. It can be irrational.

  3. It can be rational.

  4. Both $B$ and $C$


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.

  1. True

  2. False


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.

  1. True

  2. False


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 :

  1. Irrational number

  2. Rational number

  3. Whole number

  4. Natural number


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.