Mathematics · Quantitative Aptitude

Real Number System

261 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 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

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

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

If $p$ and $q$ are two distinct irrational numbers, then which of the following is always is an irrational number

  1. $\dfrac{p}{q}$
  2. $pq$
  3. $(p+q)^2$
  4. $\dfrac{p^2q+qp}{pq}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As, given $p$ and $q$ are two distinct irrational numbers.


Let $p=2+\sqrt 3$ and $q=2-\sqrt 3$

Then,

Option $A$
$\dfrac{p}{q}=\dfrac{2+\sqrt3}{2-\sqrt 3}$
$\dfrac{p}{q}=\dfrac{4+3+4\sqrt3}{4-3}=7+4\sqrt 3$

Option $B$
$pq=(2+\sqrt3)(2-\sqrt 3)=4-3=1$


Option $C$
$(p+q)^2=(2+\sqrt3+2-\sqrt 3)^2=4^2=16$

Option $D$
$\dfrac{p^2q+pq}{pq}=p+1$ is always an irrational number, because sum of rational and irrational is always irrational.

Hence, this is irrational.

Hence, this is the answer.

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

$\sqrt 7 $ is irrational.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Lets assume that √7 is rational number. ie √7 = p/q.
suppose p/q have common factor then
we divide by the common factor to get √7 = a/b were a and b are co-prime number.
that is a and b have no common factor.
√7 =  a/b co- prime number
√7 = a/b
a = √7b
squaring
a² = 7b²                                   ....(i)
a² is divisible by 7
a = 7c
substituting values in eq (i)
(7c)² = 7b²
49c² = 7b²
7c² = b²
b² = 7c²
b² is divisible by 7
that is a and b have at least one common factor 7. 
√7 is irrational
Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers
Say true or false:
$87, 54, 0, -13, -4.7, \sqrt{5}, 2{1}{7}, \sqrt{15}, -{8}{7}, 3\sqrt{2}, 4.807, 0.002, \sqrt{16}$ and $2+\sqrt{3}.$ are rational numbers
 

  1. True

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

$\sqrt { 5 } ,\quad \sqrt { 15 } ,\quad 3\sqrt { 2 } ,\quad 2\quad +\sqrt { 3 } $ are irrational numbers as they cannot be expressed as a ratio.

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

Say True or False
$3+2\sqrt 5$ is an irrational number

  1. True

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

Let us assume, to the contrary, that $3+2\sqrt{5}$ is rational.


That is, we can find coprime integers $a$ and $b$ $(b0)$ such that $3+2\sqrt{5}=\dfrac{a}{b}$.

Therefore, $\dfrac{a}{b} - 3=2\sqrt{5}$

$\dfrac{a-3b}{b}=2\sqrt{5}$

$\dfrac{a-3b}{2b}=\sqrt{5}$

$\dfrac{a}{2b}-\frac{3}{2}=\sqrt{5}$

Since $a$ and $b$ are integers, we get $\dfrac{a}{2b}-\dfrac{3}{2}$ is rational, and so $\dfrac{a-3b}{2b}=\sqrt{5}$ is rational.

But this contradicts the fact that $\sqrt{5}$ is irrational.

This contradiction has arisen because of our incorrect assumption that $3+2\sqrt{5}$ is rational.
So, we conclude that  $3+2\sqrt{5}$ is irrational.

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

The number $\displaystyle\frac{3-\sqrt{3}}{3+\sqrt{3}}$ is 

  1. Rational

  2. Irrational

  3. Both

  4. Can't say
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$Here,\quad we\quad will\quad carry\quad out\quad rationalization.\quad \ \frac { 3-\sqrt { 3 }  }{ 3+\sqrt { 3 }  } =\frac { 3-\sqrt { 3 }  }{ 3+\sqrt { 3 }  } x\frac { 3-\sqrt { 3 }  }{ 3-\sqrt { 3 }  } =\frac { { (3-\sqrt { 3 } ) }^{ 2 } }{ (3+\sqrt { 3) } (3-\sqrt { 3 } ) } =\frac { 9+3-6\sqrt { 3 }  }{ 9-3 } =\frac { 12-6\sqrt { 3 }  }{ 6 } =\frac { 2-\sqrt { 3 }  }{ 1 } \ Since\quad \sqrt { 3 } is\quad irrational\quad number\quad and\quad subtraction\quad of\quad rational\quad and\quad irrational\quad is\quad irrational.\ The\quad given\quad expression\quad is\quad irrational.\ \quad $

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

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

  1. $4 +\sqrt{5},-\sqrt{5}$
  2. $4 +\sqrt{5},\sqrt{5}$
  3. $4 -\sqrt{5},-\sqrt{5}$
  4. $ 2+\sqrt{5},2+\sqrt{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let be the Number are $\sqrt{5}  and  -\sqrt{5}$
Sum of Number  $\left(\sqrt{5}\right) + \left(-\sqrt{5}\right)$
$\sqrt{5}-\sqrt{5} = 0$
Which is a rational number