In order to know about irrational numbers, first let us learn about rational numbers. Rational numbers are the numbers that can be written in fractional form where both the numerator and denominator are integers and the denominator is not equal to zero. Irrational numbers can also be said as a number which can be represented in the fractional form as $\frac{P}{Q}$ where Q is not equal to 0 and P, Q are both Integers.

A number is said to be an irrational number, if we are not able to express it as an integer's quotient. $\frac{p}{q}$, q $\neq$ 0. Examples of irrational numbers are $\sqrt{5}$, $\sqrt{13}$, $\sqrt{11}$, etc... The irrational numbers definition is as follows:

Real number which are not Rational numbers are called as Irrational Numbers.

The irrational number list consists of numbers that cannot be expressed in the form of a ratio. Irrational numbers present from 0 – 50 are as follows:


List of
Irrational Numbers
Present Form of Irrational Numbers
$\sqrt{2}$ 1.4142135623730950488016887242097
$\sqrt{3}$ 1.7320508075688772935274463415059
$\sqrt{5}$ 2.2360679774997896964091736687313
$\sqrt{7}$ 2.6457513110645905905016157536393
$\sqrt{11}$ 3.3166247903553998491149327366707
$\sqrt{13}$ 3.6055512754639892931192212674705
$\sqrt{17}$ 4.1231056256176605498214098559741
$\sqrt{19}$ 4.3588989435406735522369819838596
$\sqrt{23}$ 4.7958315233127195415974380641627
$\sqrt{29}$ 5.3851648071345040312507104915403
$\sqrt{31}$ 5.5677643628300219221194712989185
$\sqrt{37}$ 6.0827625302982196889996842452021
$\sqrt{41}$ 6.4031242374328486864882176746218
$\sqrt{43}$ 6.557438524302000652344109997636
$\sqrt{47}$ 6.8556546004010441249358714490848

Listed below are a few facts related to irrational numbers

a) Negative of an irrational number is also negative.
If m is an irrational number then -m is also irrational.

p is irrational and - p is also irrational

b) The sum of an irrational number and a rational number is Irrational.
If m is an irrational number and n is a rational number then (m + n) is irrational

$\sqrt2$ is an irrational number and 3 is a rational number. Their sum $\sqrt2$ + 3 is irrational number.

c) The product of an irrational number and a non-zero rational number is Irrational.
If m is an irrational number and a is any non-zero rational number then a • m is Irrational.

$\sqrt2$ is irrational number and 3 is a rational number. their product 3$\sqrt2$ is irrational number.

d) We can find an irrational number between any two distinct positive rational numbers. If m and n are distinct positive rational numbers then there will be an irrational number $\sqrt(mn)$ between m and n if mn is not a perfect square.
Consider two distinct positive rational numbers 2 and 3. The product of 2 and 3 is 6 which is not a perfect square number and so there exists an irrational number $\sqrt6$ between 2 and 3.

Here are some examples based on irrational numbers

Solved Examples

Question 1:

Prove that $\sqrt{5}$ is an irrational number.


Solution:

Let us prove the required result by contradiction.

Step 1: Assume $\sqrt{5}$ is a rational number. Hence

$\frac{p}{q}$ where p, q are integers and q $\neq$ 0

Also p, q have no common factors that is $\frac{p}{q}$ is in its simplest form. The only common factor between p and q is 1.

Step 2: Square on both sides

($\sqrt{5}$)2 = ($\frac{p}{q}$)2

5 = $\frac{p^2}{q^2}$

5q2 = p2

This means p2 is a multiple of 5 which in turn implies that p is also a multiple of 5. Let us assume p = 5a

Step 3: Substitute p = 5a in 5q2 = p2

5q2 = p2

5q2 = (5a)2

5q2 = 25a2

q2 = 5a2

This means that q2 is a multiple of 5 which in turn implies q is also a multiple of 5.

If both p and q are multiples of 5 that means there is at least one common factor between p and q.

This contradicts our assumption that there is no common factor between p and q except 1.

Hence $\sqrt{5}$ is an irrational number.



Question 2:

Is (3 + $\sqrt{3}$)(3 - $\sqrt{3}$) an irrational number?


Solution:

(3 + $\sqrt{3}$)(3 - $\sqrt{3}$) = 32 - ($\sqrt{3}$)2

Using the identity (a+b)(a-b) = a2 - b2

32 - ($\sqrt3$)2 = 9 - 3 = 6

6 is a rational number.

Hence(3 + $\sqrt{3}$)(3 - $\sqrt{3}$) is a rational number. It is not an irrational number.