Mathematics · Quantitative Aptitude

Real Number System

277 Questions

The real number system encompasses rational and irrational numbers, forming the basis of arithmetic and number theory. It is a key area in the quantitative aptitude and mathematics sections of school level and competitive exams. Practice these questions to master the properties and identification of different types of numbers.

Identifying irrational numbersOrdering rational numbersProperties of square rootsPerfect number propertiesRational and irrational products

Real Number System Questions

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

$\pi$ is _______

  1. a rational number

  2. an integer

  3. an irrational number

  4. a whole number
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Sometimes we use $π = 22/7$ which is a popular approximation

$π = 3.14159265358...$

$22/7 = 3.142857142857...$

But $π$ and $22/7$ are close but not accurate.

Rational Numbers - $P/Q$ when $Q$ is not equal to $0$.

Let $x = 33.33333…. $——-(1)

$10x = 33.333333….. $——-(2)

Equation $(2) - (1)$

$9x = 30$

$x = 30/9$ which is in form of $P/Q$ and $x = 33.3333…$

The digit $‘3′$ is repeating itself and that’s why it can be written as $100/3.$

When it’s π, the value is $3.14159265358...$ The order of digits will not repeat itself in it but in $22/7 = 3.142857142857….$ you can see that $142857…$ is repeating itself that’s why $22/7$ is rational but $π$ is irrational.

So, option C is correct.
Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

Which of the following number is irrational ?

  1. $\sqrt{16}-4$
  2. $(3-\sqrt{3}) (3+\sqrt{3})$
  3. $\sqrt{5}+3$
  4. $-\sqrt{25}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In the given options $\sqrt { 16 }$ and $\sqrt { 25 } $ are irrational numbers. Their real values are 4 and 5 respectively. So, option A and C are incorrect.

Option B can solved and its real value becomes 6. So it is also a rational number.
In option C, $\sqrt { 5 }$ is a irrational number. So, option C is a irrational number. 
So, correct answer is option C. 

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

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

  1. non $-$ terminating
  2. non $-$ terminating and repeating
  3. non $-$ terminating and non $-$ repeating

  4. terminating
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Irrational numbers have decimal expansions that neither terminate nor repeating

So the correct answer is option C.

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

Which of the following is an irrational number ?

  1. $\sqrt{23}$
  2. $\sqrt{225}$
  3. $0.3796$
  4. $7.478$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In the given options, 

$\sqrt { 225 }$ = 15. So, it is not a irrational number,
Option C and D are terminating decimals. So, they are also rational numbers.
$\sqrt{23}$ is a irrational number. 
So, option A is correct answer.  

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

$\pi$ is a(n) ________ while $\dfrac{22}{7}$ is rational.

  1. Integer

  2. Whole Number

  3. <p>Rational Number
    </p>

  4. <p>Irrational Number
    </p>
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The value $\dfrac{22}7$ is a rational number, as it can be expressed in the form $\dfrac pq$. 

We consider it as an approximate value of $\pi$ because $\pi$ is close to $\dfrac{22}7$. 
But actually its value is $3.14159....$, which is neither terminating nor repeating. 
Thus, $\pi $ is irrational, but $\dfrac{22}7$ is rational.

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

$3+2\sqrt{5}$ a rational number.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let's assume that $3+2\sqrt5$ is rational..... 

then 

$3+2\sqrt5 = p/q $

$\sqrt5 =( p-3q)/(2q) $ 

now take $p-3q$ to be P and $2q$ to be Q........where P and Q are integers 

which means, $\sqrt5= P/Q$...... 

But this contradicts the fact that $\sqrt5$ is rational 

So our assumption is wrong and $3+2\sqrt5$ is irrational.

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

$\sqrt { 2 } ,\sqrt { 3 }$ are

  1. Whole numbers

  2. Rational numbers

  3. Irrational numbers

  4. Integers
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A rational number is any number that can be expressed as a fraction $\dfrac pq$ of two integers with $q$ not equal to zero.
As in the case of $\sqrt2$ and $\sqrt3$, it cannot be expressed as a fraction $\dfrac pq$.

Hence, option $A$ is the correct answer.

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

If $p$ is prime, then $\sqrt{p}$ is irrational and if $a, b$ are two odd prime numbers, then $a^2 -b^2$ is composite. As per the above passage mark the correct answer to the following question.
$\sqrt{7}$ is:

  1. a rational number

  2. an irrational number

  3. not a real number

  4. terminating decimal
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The basic definition for a rational number is that it can be represented in the form of $p/q$, where p and q are integers and q is a non-zero integer. Here, $\sqrt7$ is not a perfect square and thus cannot be expressed in the form of $p/q$, thus it is an irrational number.

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

Consider the given statements:
I. All surds are irrational numbers.
II. All irrationals numbers are surds.
Which of the following is true.

  1. Only I

  2. Only II

  3. Both I and II

  4. Neither I nor II
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A surd, by its very definition is an irrational number.

However, not every irrational number can be expressed as a surd.
Hence, only statement 1 is true.

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

If $a\neq 1$ and $ln{ a }^{ 2 }+{ \left( ln{ a }^{ 2 } \right)  }^{ 2 }+{ \left( ln{ a }^{ 2 } \right)  }^{ 3 }+........=3\left( lna+{ \left( ln{ a } \right)  }^{ 2 }+{ \left( ln{ a } \right)  }^{ 3 }+{ \left( ln{ a } \right)  }^{ 4 }+...... \right)$ then $a$ is

  1. $an\ irrational\ number$
  2. $a\ transcendental\ number$
  3. $an\ algeberaic\ number$
  4. $a\ surd$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, for $a\ne 1$,

$ln{ a }^{ 2 }+{ \left( ln{ a }^{ 2 } \right)  }^{ 2 }+{ \left( ln{ a }^{ 2 } \right)  }^{ 3 }+........=3\left( lna+{ \left( ln{ a } \right)  }^{ 2 }+{ \left( ln{ a } \right)  }^{ 3 }+{ \left( ln{ a } \right)  }^{ 4 }+...... \right)$
or, $\dfrac{\ln a^2}{1-\ln a^2}=3\times \dfrac{\ln a}{1-\ln a}$
or, $\dfrac{2\ln a}{1-2\ln a}=3\times \dfrac{\ln a}{1-\ln a}$
or, $2(1-\ln a)=3(1-2\ln a)$ [Since $a\ne 1\Rightarrow \ln a \ne 0$ ]
or, $4\ln a =1$
or, $a=\sqrt[4]{e}$.
So clearly $a$ is an irrational number.

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

Simplify the following expressions.
Classify the following numbers as rational or irrational.

  1. $\left( 5+\sqrt { 7 } \right) \left( 2+\sqrt { 5 } \right)$
  2. $\left( 5+\sqrt { 5 } \right) \left( 5-\sqrt { 5 } \right)$
  3. ${ \left( \sqrt { 3 } +\sqrt { 7 } \right) }^{ 2 }$
  4. $\left( \sqrt { 11 } -\sqrt { 7 } \right) \left( \sqrt { 11 } +\sqrt { 7 } \right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$A:$

$\left( {{\rm{5}} + \sqrt {\rm{7}} } \right)\left( {{\rm{2}} + \sqrt {\rm{5}} } \right)$  

$=10+5\sqrt5+2\sqrt7+\sqrt{35}$

Now, $10$ is rational and $\sqrt5,\sqrt7$ are non terminating , non repeating is an irrational 

and we know that $rational + irrational = irrational$ 

Therefore,  $\left( {{\rm{5}} + \sqrt {\rm{7}} } \right)\left( {{\rm{2}} + \sqrt {\rm{5}} } \right)$  is  irrational 


$B:$
$\left( {{\rm{5}} + \sqrt {\rm{5}} } \right)\left( {5 - \sqrt {\rm{5}} } \right)$

$={{\rm{5}}^2} + {\left( {\sqrt {\rm{5}} } \right)^2} = 25 - 5$

$=5$, which is rational 

So, $\left( {{\rm{5}} + \sqrt {\rm{5}} } \right)\left( {5 - \sqrt {\rm{5}} } \right)$
Is rational number.


$C:$
${\left( {\sqrt {\rm{3}}  + \sqrt {\rm{7}} } \right)^{\rm{2}}}$

$={\left( {\sqrt {\rm{3}} } \right)^2} + {\left( {\sqrt {\rm{7}} } \right)^2} + 2\sqrt {\rm{3}} \sqrt 7 $

$={\left( {\sqrt {\rm{3}} } \right)^2} + {\left( {\sqrt {\rm{7}} } \right)^2} + 2\sqrt {{\rm{21}}} =3 + 7 + 2\sqrt {{\rm{21}}} =10+2\sqrt{21}$
and $10$ and $\sqrt{21}$ are both rational.

Therefore, ${\left( {\sqrt {\rm{3}}  + \sqrt {\rm{7}} } \right)^{\rm{2}}}$ is rational.


$D:$
$\left( {{\rm{11}} - \sqrt {\rm{7}} } \right)\left( {{\rm{11 + }}\sqrt {\rm{7}} } \right)$

$={\left( {{\rm{11}}} \right)^2} - {\left( {\sqrt {\rm{7}} } \right)^2}$

$=11-7=4$, which is rational.

Therefore $\left( {{\rm{11}} - \sqrt {\rm{7}} } \right)\left( {{\rm{11 + }}\sqrt {\rm{7}} } \right)$ is rational.

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

Which of the following numbers are an irrational number. 

  1. $2- \sqrt 5$
  2. $\left( {3 + \sqrt {23} } \right) - \left( {\sqrt {23} } \right)$
  3. $\frac{1}{\sqrt 2}$
  4. $2\pi $
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation

$A$ is a irrational number as it cannot be expressed of the form $\cfrac{p}{q}$

$B$ is a rational number. As it can be expressed in the form of $\cfrac{3}{1}$
$C$ is a irrational number as it cannot be expressed of the form $\cfrac{p}{q}$
$D$ is a irrational number as it cannot be expressed of the form $\cfrac{p}{q}$ of two integers