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 one of the following is not true?

  1. $\sqrt{2}$ is an irrational number
  2. If a is a rational number and $\sqrt{b}$ is an irrational number then $a\sqrt{b}$ is irrational number

  3. Every surd is an irrational number

  4. The square root of every positive integer is always irrational
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

(a) All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction. An irrational number has endless non-repeating digits to the right of the decimal point. Here are some irrational numbers:


$π = 3.141592…$
$\sqrt {2} = 1.414213…$

Therefore, $\sqrt {2}$ is an irrational number.

(b) Let us take a rational number $a=\dfrac {2}{1}$ and an irrational number $b=\sqrt {2}$, then their product can be determined as:

$a\times b=2\times \sqrt { 2 } =2\sqrt { 2 }$ which is also an irrational number.

Therefore, if $a$ is a rational number and $\sqrt {b}$ is an irrational number than $a\sqrt {b}$ is an irrational number.

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

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

Surds are numbers left in root form $\sqrt {}$ to express its exact value. It has an infinite number of non-recurring decimals. 

Therefore, every surd is an irrational number.

(d) Let us take a positive integer $4$, now square root of $4$ will be:

$\sqrt {4}=2$ which is not an irrational number 

Hence, the square root of every positive integer is not always irrational.

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

Which one of the following is not true?

  1. When x is not a perfect square, $\sqrt{x}$ is an irrational number
  2. The index form of $\sqrt[m]{x^n}$ is $x^{\frac{n}{m}}$
  3. The radical form of $\left(x^{\frac{1}{n}}\right)^{\frac{1}{m}}$ is $\sqrt[m]{x^n}$

  4. Every real number is an irrational number
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$(D)\,\, Real = Rational + Irrational$

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

If $ x = ( 2 + \sqrt3)^n , n \epsilon N $ and $ f = x - [x],$ then $ \dfrac {f^2}{1-f} $ is :

  1. An irrational number

  2. A non-integer rational number

  3. An odd number

  4. An even number
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let x = (2 + sqrt(3))^n. Let y = (2 - sqrt(3))^n. Since 0 < 2 - sqrt(3) < 1, y is between 0 and 1. x + y is an integer, so f = x - [x] = 1 - y. Then f^2 / (1 - f) = (1 - y)^2 / y. This simplifies to an even integer.

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

The product of two irrational numbers is 

  1. Always irrational

  2. Always rational

  3. Can be both rational and irrational

  4. always an integer
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $p=\sqrt 3$ and $q=\sqrt 3$ be two irrational numbers 
$pq=\sqrt 3\times \sqrt 3=3$
which is rational
Now let $p=\sqrt 3$ and $q=\sqrt 2$
$pq=\sqrt 3\times \sqrt 2=\sqrt 6$
which is an irrational number
So the product can be both rational and irrational .
Option $C$ is correct.

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

State whether True or False :


All the following numbers are irrationals.
(i) $\dfrac { 2 }{ \sqrt { 7 }  } $ (ii) $\dfrac { 3 }{ 2\sqrt { 5 }  }$ (iii) $4+\sqrt { 2 } $ (iv) $5\sqrt { 2 } $

  1. True

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

In all of the above questions $\sqrt7,\sqrt5,\sqrt2$ are a $irrational$ numbers


And $\text{the addition, subtraction, division and product between  rational and irrational gives  an irrational number}$

So that all of the above are irrational numbers.

hence option A is correct.