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

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

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

Let be the Number are $4\sqrt{3}  and  2\sqrt{3}$
Difference of Number  $4\sqrt{3} - 2\sqrt{3} = 2\sqrt{3}$
Which is a irrational number

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 quotient is an irrational number.

  1. $\sqrt{15},\sqrt{5}$
  2. $\sqrt{45},\sqrt{5}$
  3. $\sqrt{20},\sqrt{5}$
  4. $\sqrt{80},\sqrt{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let be the Number are $\sqrt{15}  and  \sqrt{5}$
Quotient of Numbers  $\frac{\sqrt{15}}{\sqrt{5}} = \sqrt{\frac{15}{5}} = \sqrt{3} $
Which is a irrational number

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 an irrational number.

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

Let be the Number are $2\sqrt{5}  and  3\sqrt{5}$
Sum of Number  $2\sqrt{5} + 3\sqrt{5} = 5\sqrt{5}$
Which is a irrational number

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 quotient is a rational number.

  1. $\sqrt{5},\sqrt{2}$
  2. $\sqrt{8},\sqrt{2}$
  3. $\sqrt{3},\sqrt{2}$
  4. $\sqrt{7},\sqrt{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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 product is a rational number.

  1. $\sqrt{8},\sqrt{2}$
  2. $\sqrt{5},\sqrt{2}$
  3. $2+\sqrt{8},\sqrt{2}$
  4. $\sqrt{8},2+\sqrt{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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 product is an irrational number.

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

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

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

  1. rational

  2. whole

  3. integer

  4. irrational
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

$6+\sqrt{2}$ is a rational number.

  1. True

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

Let's assume that $6+\sqrt2$ is rational..... 

then 

$6+\sqrt2 = p/q $

$\sqrt2 =( p-6q)/(q) $ 

now take $p-6q$ to be P and $q$ to be Q........where P and Q are integers 

which means, $\sqrt2= P/Q$...... 

But this contradicts the fact that $\sqrt2$ is rational 

So our assumption is wrong and $6+\sqrt2$ is irrational.

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Let x and y be rational and irrational numbers, respectively, then x + y necessarily an irrational number.


State True or False.

  1. True

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

Yes.

Let x $= 21, y =\sqrt{2}$ be a rational number
Now $x+y=21 +\sqrt{2}=21+1.4142....=22.4142....$ , which is non-terminating and non-recurring. Hence x+y is irrational.

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Identify the irrational number(s) between $2\sqrt{3}$ and $3\sqrt{3}$

  1. $\sqrt{19}$
  2. $\sqrt{29}$
  3. $\cfrac { 4\sqrt { 3 } }{ \sqrt { 3 } } $
  4. $\sqrt{17}$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation
$2\sqrt{3}=\sqrt{12}$
$3\sqrt{3}=\sqrt{27}$
$\therefore \sqrt{176}\sqrt{19}$ are irrational no between them $\sqrt{29}$ lie out of it.
As $\dfrac{4\sqrt{3}}{\sqrt{3}}=4$ (Rational)