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 numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

Which of the following is always true 

  1. $irrational + irrational =irrational $
  2. $\dfrac{rational }{rational }=rational $
  3. $\dfrac{integer }{integer}=integer$

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

Counter-example for A: $(\sqrt{2}) + (4-\sqrt{2}) = 4$

Counter-example for C: $\dfrac{1}{2}=0.5$

Proof for B:

Let $q _1, q _2$ be two rational numbers such that $q _2\neq0$.


As they are rational, they can be written as $a/b, c/d$ respectively for some integers $a, b, c, d$. $(b,c,d\neq0)$

$\dfrac{q _1}{q _2}=\dfrac{a/b}{c/d}=\dfrac{ad}{bc}$

Since, $a,b,c,d$ were integers, even $ad$ and $bc(\neq0)$ are integers and therefore, the above expression is rational.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

If the product of two irrational numbers is rational, then which of the following can be concluded?

  1. The ratio of the greater and the smaller numbers is an integer.

  2. The sum of the numbers must be rational.

  3. The excess of the greater irrational number over the irrational number must be rational.

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

If a and b are irrational and ab is rational, let a = x*sqrt(k) and b = y*sqrt(k). The ratio a/b = x/y, which is rational. The options provided are limited, but A is the most mathematically sound conclusion.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

State whether the following statement is true or not:
$\left( 3+\sqrt { 5 }  \right) $ is an irrational number. 

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let us suppose $3+\sqrt 5$ is rational.

$=>3+\sqrt 5$ is in the form of $\dfrac pq$ where $p$ and $q$ are integers and $q\neq0$

$=>\sqrt5=\dfrac pq-3$

​$=>\sqrt5=\dfrac{p-3q}{q}$

as $p, q$ and $3$ are integers $\dfrac{p-3q}{q}$ is a rational number.

$=>\sqrt 5$ is a rational number.

But we know that $\sqrt 5$ is an irrational number.

So this is a contradiction.

This contradiction has arisen because of our wrong assumption that $3+\sqrt 5$ is a rational number.

Hence $3+ \sqrt5$  is an irrational number.
Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

A rational number equivalent to  $ \displaystyle \frac{-5}{-3}  $ is -

  1. $ \displaystyle \frac{25}{15} $
  2. $ \displaystyle \frac{-15}{25} $
  3. $ \displaystyle \frac{-25}{15} $

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

 $ \displaystyle  \because  \frac{-5}{-3} $= $ \displaystyle  \frac{-5}{-3} $X $ \displaystyle  \frac{-5}{-5} $= $ \displaystyle  \frac{25}{15} $

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

Every irrational number is

  1. a surd

  2. a prime number

  3. not a surd

  4. none
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

An irrational number is a real number that cannot be represented as a ratio or a simple fraction.


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 C.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

For three irrational numbers $p,q$ and $r$ then $p.(q+r)$ can be 

  1. A rational number

  2. An irrational number

  3. An integer

  4. All of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$p,q$ and $r$ are all irrational 
Let $p=q=r=\sqrt2$
$p(q+r)=p.q+p.r$
$\Rightarrow \sqrt { 2 } (\sqrt { 2 } +\sqrt { 2 } )=\sqrt { 2 } .\sqrt { 2 } +\sqrt { 2 } .\sqrt { 2 } =2+2=4$
which is rational as well as integer
Let us take another case in which $p=\sqrt2$ and $q=r=\sqrt3$
$\Rightarrow \sqrt { 2 } (\sqrt { 3 } +\sqrt { 3 } )=\sqrt { 2 } .\sqrt { 3 } +\sqrt { 2 } .\sqrt { 3 } =\sqrt { 6 } +\sqrt { 6 } =2\sqrt { 6 } $
which is an irrational number.
So on applying distributive property on three irrational numbers we can get an integer,a rational as well as an irrational number .
So option $D$ is correct.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

Which one of the following statements is not correct?

  1. If $a$ is a rational number and $b$ is irrational, then $a+b$ is irrational.

  2. The product of non-zero rational number with an irrational number is always irrational.

  3. The addition of any two rational numbers can be an integer.

  4. The division of any two integers is an integer.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Sum of rational and an irrational number is rational (i.e., need not to be irrational)

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

$\sqrt{5}\left{(\sqrt{5}+1)^{50}-(\sqrt{5}-1)^{50}\right}$ is?

  1. An irrational number
  2. $0$

  3. A natural number

  4. A prime number
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This expression involves powers of (sqrt(5)+1) and (sqrt(5)-1). Using the binomial expansion, the irrational parts cancel out, leaving an irrational result due to the leading sqrt(5).