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

State True or False.

$\sqrt{5}-2$ is an irrational number.

  1. True

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

$\ { (\sqrt { 5 } -2) }\ \sqrt { 5 } =2.2360679775........\ \ \sqrt { 5 } is\quad an\quad irrational\quad number,\quad since\quad its\quad decimal\quad representaion\quad is\quad non\quad terminating\quad non\quad repeating.\ Subtraction\quad of\quad rational\quad with\quad irrational\quad is\quad irrational.\ Hence,\quad { (\sqrt { 5 } -2) }\quad is\quad an\quad irrational\quad number.\ \quad $

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

State True or False.

$-\displaystyle\frac{2}{5}\sqrt{8}$ is an irrational number.

  1. True

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

$\ { \frac { -2 }{ 5 } \sqrt { 8 }  }={ \frac { -4 }{ 5 } \sqrt { 2 }  }=-0.8*\sqrt { 2 } \ \sqrt { 2 } =1.41421356237........\ \ \sqrt { 2 } is\quad an\quad irrational\quad number,\quad since\quad its\quad decimal\quad representaion\quad is\quad non\quad terminating\quad non\quad repeating.\ Multiplication\quad of\quad rational\quad with\quad irrational\quad is\quad irrational.\ Hence,\quad { (\frac { -2 }{ 5 } \sqrt { 8 } ) }\quad is\quad an\quad irrational\quad number.\ \quad $

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

 $\sqrt3$ is 

  1. rational number

  2. irrational number

  3. natural number

  4. None
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $\sqrt3$ is a rational number
$\therefore \sqrt3 = \displaystyle \frac{a}{b}$ [Where a & b are co-primes]
$a^2=3b^2$ .......(i)
$\Rightarrow$ 3 divides $a^2$
$\Rightarrow$ 3 also divides a
$\Rightarrow$ a=3c
[Where c is any non-zero positive integer]
$\Rightarrow a^2 = 9c^2$
From equation (i)
$3b^2=9c^2$
$\Rightarrow b^2 = 3c^2  \Rightarrow$ 3 divides $b^2$
$\Rightarrow$ 3 also divides b
So, 3 is a common factor of a and b.
Our assumption is wrong, because a and b are not co - primes.
It means $\sqrt3$ is an irrational number.

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

 $\sqrt2 + \sqrt3$ is 

  1. irrational

  2. rational

  3. natural

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\cfrac{m}{n} = \sqrt{2} + \sqrt{3} $
Square both sides:
$\cfrac{m^2}{ n^2} = 5 + 2\sqrt{6} $

"Solve" for $\sqrt{6}$
$\sqrt{6} = \cfrac{\left(m^{2} - 5n^{2}\right)}{\left(2n^{2}\right)} $
so if  $\sqrt{2} + \sqrt{3} $ is  rational,  then  so  is $ \sqrt{6}$
Let a and b be the integers with gcd(a,b) = 1 such that
$\cfrac{a}{b} = \sqrt{6}$
Square both sides and multiply by $b^2$:
$a^2 = 6b^2 $
Now, the right side is divisible by 2, so $a^2$ is divisible by 2, which
then implies that a is divisible by 2 (since 2 is prime).
Therefore we  can write a=2k for some integer k:
$4k^{2} = \left(2k \right)^{2} = 6b^{2} $
Divide by 2:
$2k^{2} = 3b^{2} $
Now the left side is divisible by 2, so $3b^{2}$ is divisible by 2, from which it follows that b is divisible by 2.
However, this would mean that 2 divides gcd(a,b) = 1. Contradiction.
$\therefore  \sqrt{6} $ is  irrational
,  and  $\therefore \sqrt{2} + \sqrt{3} $ is  also irrational.


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

Every surd is

  1. a natural number

  2. an irrational number

  3. a whole number

  4. a rational number
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
When a number cannot be simplified further to remove a square root then it is a surd.  
A surd is an irrational number.

For. eg: square root of 2 cannot be simplified. thus it is a surd.

By definition, a surd is an irrational root of a rational number. So we know that surds are always irrational and they are always roots.

For eg, $\sqrt2$ is a surd since 2 is rational and $\sqrt 2$ is irrational.

Similarly, the cube root of 9 is also a surd since 9 is rational and the cube root of 9 is irrational.

On the other hand, $\sqrtπ$ is not a surd even though $\sqrtπ$ is irrational because π is not rational.

Thus, to answer the question, every surd is an irrational number, though an irrational number may or may not be a surd.

The answer is Option B
Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

Irrational number is defined as 

  1. a real number that cannot be made by dividing two integers.

  2. a real number that can be made by dividing two integer.

  3. a number that can be made derived after multiplying two integers.

  4. a real number that can be written as whole number.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
An irrational is any real number that cannot be expressed as a ratio of integers.

Therefore, $A$ is the correct answer.

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. $\dfrac{11}{2}$
  2. $\sqrt{16}$
  3. $\sqrt{9}$
  4. $\sqrt{11}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
An irrational is any real number that cannot be expressed as a ratio of integers.
Option $A$ is a rational number.
Option $B$ and $C$ are $\sqrt{16}$ and $\sqrt{9}$, i.e. $4$ and $3$ respectively.
$D$ cannot be expressed as a ratio of integers.
$D$ is the correct answer.
Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

$m$ is not a perfect square, then $\sqrt {m}$ is 

  1. an irrational number

  2. a composite number

  3. a rational number
  4. None of these as $m$ is not on a number line
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\sqrt {m}$ is irrational when it is not being a perfect square.
Example $\sqrt3$ which is an irrational number.

Therefore, $A$ is the correct answer.
Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

How many of the following four numbers are rational?
$\sqrt{3}+\sqrt{3}, \sqrt{3}-\sqrt{3}, \sqrt{3} \times \sqrt{3}, \sqrt{3} / \sqrt{3}$

  1. One

  2. Two

  3. Three

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

$\sqrt { 3 } +\sqrt { 3 } =2\sqrt { 3 } \quad irrational\quad number\ \sqrt { 3 } -\sqrt { 3 } =0\quad rational\quad number\ \sqrt { 3 } \times \sqrt { 3 } =3\quad rational\quad number\ \frac { \sqrt { 3 }  }{ \sqrt { 3 }  } =1\quad rational\quad number$

Now it is clear that there are three rational number so correct answer will be option C