Irrational Numbers & Decimal Expansions of Real Numbers | Advance Learner Course: Mathematics (Maths) Class 9 PDF Download

Introduction

Irrational numbers are the real numbers that cannot be represented as a simple fraction. It cannot be expressed in the form of a ratio, such as p/q, where p and q are integers, q≠0. It is a contradiction of rational numbers.Irrational numbers are expressed usually in the form of R\Q, where the backward slash symbol denotes ‘set minus’. it can also be expressed as R – Q, which states the difference between a set of real numbers and a set of rational numbers.
The calculations based on these numbers are a bit complicated. For example, √5, √11, √21, etc., are irrational. If such numbers are used in arithmetic operations, then first we need to evaluate the values under root. These values could be sometimes recurring also. Now let us find out its definition, lists of irrational numbers, how to find them, etc., in this article.

Irrational Numbers Definition

An irrational number is a real number that cannot be expressed as a ratio of integers, for example, √ 2 is an irrational number. Again, the decimal expansion of an irrational number is neither terminating nor recurring.
Irrational Meaning: The meaning of irrational is not having a ratio or no ratio can be written for that number. That means the number which cannot be expressed other than by means of roots. In other words, we can say that irrational numbers cannot be represented as the ratio of two integers.

How do you know a number is irrational?

The real numbers which cannot be expressed in the form of p/q, where p and q are integers and q ≠ 0 are known as irrational numbers. For example  √ 2 and √ 3 etc. are irrational. Whereas any number which can be represented in the form of p/q, such that, p and q are integers and q ≠ 0 is known as a rational number.

Is Pi an irrational number?

Pi (π) is an irrational number because it is non-terminating. The approximate value of pi is 22/7. Also, the value of π is 3.14159 26535 89793 23846 264…

Symbol 

Generally, the symbol used to represent the irrational symbol is “P”.  Since the irrational numbers are defined negatively, the set of real numbers (R) that are not the rational number (Q), is called an irrational number. The symbol P is often used because of the association with the real and rational number. (i.e) because of the alphabetic sequence P, Q, R. But mostly, it is represented using the set difference of the real minus rationals, in a way R- Q or R\Q.

Properties 

Since irrational numbers are the subsets of the real numbers, irrational numbers will obey all the properties of the real number system. The following are the properties of irrational numbers:

• The addition of an irrational number and a rational number gives an irrational number. For example, let us assume that x is an irrational number, y is a rational number and the addition of both the numbers x +y gives a rational number z.
• Multiplication of any irrational number with any nonzero rational number results in an irrational number. Let us assume that if xy=z is rational, then x =z/y is rational, contradicting the assumption that x is irrational. Thus, the product xy must be irrational.
• The least common multiple (LCM) of any two irrational numbers may or may not exist.
• The addition or the multiplication of two irrational numbers may be rational; for example, √2. √2 = 2. Here, √2 is an irrational number. If it is multiplied twice, then the final product obtained is a rational number. (i.e) 2.
• The set of irrational numbers is not closed under the multiplication process, unlike the set of rational numbers.

List of Irrational Numbers 

The famous irrational numbers consist of Pi, Euler’s number, Golden ratio. Many square roots and cube roots numbers are also irrational, but not all of them. For example, √3 is an irrational number but √4 is a rational number. Because 4 is a perfect square, such as 4 = 2 x 2 and √4 = 2, which is a rational number.  It should be noted that there are infinite irrational numbers between any two real numbers. For example, say 1 and 2, there are infinitely many irrational numbers between 1 and 2. Now, let us have a look at the values of famous irrational numbers.

Are Irrational Numbers Real Numbers? 

In Mathematics, all the irrational numbers are considered as real numbers, which should not be rational numbers. It means that irrational numbers cannot be expressed as the ratio of two numbers. The irrational numbers can be expressed in the form of non-terminating fractions and in different ways. For example, the square roots which are not perfect squares will always result in an irrational number.

Sum and Product of Two Irrational Numbers

Now, let us discuss the sum and the product of the irrational numbers.

Product of Two Irrational Numbers

Statement: The product of two irrational numbers is sometimes rational or irrational

For example, √2 is an irrational number, but when √2  is multiplied by √2, we get the result 2, which is a rational number.

(i.e.,) √2 x √2 = 2 
We know that π is also an irrational number, but if π is multiplied by π, the result is π², which is also an irrational number.
(i.e..) π x π = π²
It should be noted that while multiplying the two irrational numbers, it may result in an irrational number or a rational number. 
Sum of Two Irrational Numbers
Statement: The sum of two irrational numbers is sometimes rational or irrational.
Like the product of two irrational numbers, the sum of two irrational numbers will also result in a rational or irrational number.
For example, if we add two irrational numbers, say 3√2+ 4√3, a sum is an irrational number.
But, let us consider another example, (3+4√2) + (-4√2 ), the sum is 3, which is a rational number.
So, we should be very careful while adding and multiplying two irrational numbers, because it might result in an irrational number or a rational number.

Irrational Number Proof
The following theorem is used to prove the above statement
Theorem: Given p is a prime number and a² is divisible by p, (where a is any positive integer), then it can be concluded that p also divides a.
Proof: Using the Fundamental Theorem of Arithmetic, the positive integer can be expressed in the form of the product of its primes as:
a = p₁ × p₂ × p_3………..  × p_{n …..(1)}
Where, p_1, p_2, p₃, _……, p_n represent all the prime factors of a.
Squaring both the sides of equation (1),
a² = ( p₁ × p₂ × p_3………..  × p_{n) (} p₁ × p₂ × p₃………..  × p_n)
⇒a² = (p₁)² × (p₂)² × (^p₃ )^{2_………..}× (p_n)²
According to the Fundamental Theorem of Arithmetic, the prime factorization of a natural number is unique, except for the order of its factors.
The only prime factors of a² are p₁, p_2, p_{3………..,} p_n. If p is a prime number and a factor of a², then p is one of  p₁, p_{2 ,} p_{3………..,} p_n. So, p will also be a factor of a.
Hence, if a²  is divisible by p, then p also divides a.
Now, using this theorem, we can prove that √ 2 is irrational.

How to Find an Irrational Number?
Let us find the irrational numbers between 2 and 3.
We know, square root of 4 is 2; √4 =2
and the square root of 9 is 3; √9 = 3
Therefore, the number of irrational numbers between 2 and 3 are √5, √6, √7, and √8, as these are not perfect squares and cannot be simplified further. Similarly, you can also find the irrational numbers, between any other two perfect square numbers.
Another case:
Let us assume a case of √2. Now, how can we find if √2 is an irrational number?
Suppose, √2 is a rational number. Then, by the definition of rational numbers, it can be written that,
√ 2 =p/q    …….(1)
Where p and q are co-prime integers and q ≠ 0 (Co-prime numbers are those numbers whose common factor is 1).
Squaring both the sides of equation (1), we have
2 = p²/q²
⇒ p² = 2 q ²    ………. (2)
From the theorem stated above, if 2 is a prime factor of p², then 2 is also a prime factor of p.
So, p = 2 × c, where c is an integer.
Substituting this value of p in equation (3), we have
(2c)² = 2 q ²
⇒ q² = 2c ² 
This implies that 2 is a prime factor of q² also. Again from the theorem, it can be said that 2 is also a prime factor of q.
According to the initial assumption, p and q are co-primes but the result obtained above contradicts this assumption as p and q have 2 as a common prime factor other than 1. This contradiction arose due to the incorrect assumption that √2  is rational.

So, root 2 is irrational.
Similarly, we can justify the statement discussed in the beginning that if p is a prime number, then √ p  is an irrational number. Similarly, it can be proved that for any prime number p,√ p is irrational.

Problems and Solutions

Question 1: Which of the following are Rational Numbers or Irrational Numbers?
2, -.45678…, 6.5, √ 3, √ 2
Solution: Rational Numbers – 2, 6.5 as these have terminating decimals.
Irrational Numbers – -.45678…, √ 3, √ 2 as these have a non-terminating non-repeating decimal expansion.

Question 2: Check if below numbers are rational or irrational. 
2, 5/11, -5.12, 0.31 
Solution: Since the decimal expansion of a rational number either terminates or repeats. So, 2, 5/11, -5.12, 0.31 are all rational numbers.

Decimal Expansions of Rational Numbers

Real numbers are simply the combination of rational and irrational numbers, in the number system. In general, all the arithmetic operations can be performed on these numbers and they can be represented in the number line, also. So in this article let’s discuss some rational and irrational numbers and their proof.

Rational Numbers

A number of the form p/q, where p and q are integers and q ≠ 0 are called rational numbers.
Examples: 1) All natural numbers are rational,
1, 2, 3, 4, 5…….. all are rational numbers.
2) Whole numbers are rational.
0,1, 2, 3, 4, 5, 6,,,,,, all are rational.
3) All integers are rational numbers.
-4.-3,-2,-1, 0, 1, 2, 3, 4, 5,,,,,,,, all are rational numbers.

Irrational Numbers

The numbers which when expressed in decimal form are expressible as non-terminating and non-repeating decimals are known as irrational numbers.
Examples: 
1) If m is a positive integer which is not a perfect square, then √m is irrational.
√2 ,√3, √5, √6, √7, √8, √10,….. etc., all are irrational.
2) If m is a positive integer which is not a perfect cube , then 3√m is irrational.
3√2,  3√3,  3√4,…..  etc., all are irrational.
3) Every Non Repeating and Non Terminating Decimals are Irrational Numbers.
0.1010010001……  is an non-terminating and non repeating decimal. So it is irrational number.
0.232232223…….. is irrational.
0.13113111311113…… is irrational.

Nature of the Decimal Expansions of Rational Numbers

• Theorem 1: Let x be a rational number whose simplest form is p/q, where p and q are integers and q ≠ 0. Then, x is a terminating decimal only when q is of the form (2^m x 5ⁿ) for some non-negative integers m and n.
• Theorem 2: Let x be a rational number whose simplest form is p/q, where p and q are integers and q ≠ 0. Then, x is a nonterminating repeating decimal, if q ≠ (2^m x 5ⁿ).
• Theorem 2: Let x be a rational number whose simplest form is p/q, where p and q are integers and q = 2^m x 5ⁿ then x has a decimal expansion which terminates.

Proof 1: √2 is irrational
Let √2 be a rational number and let its simplest form is p/q.
Then, p and q are integers having no common factor other than 1, and q ≠ 0.
Now √2 = p/q
⇒ 2 = p²/q²    (on squaring both sides)
⇒ 2q² = p²    ……..(i)
⇒ 2 divides p²   (2 divides 2q² )
⇒ 2 divides p   (2 is prime and divides p² ⇒ 2 divides p)
Let p = 2r for some integer r.
putting p = 2r in eqn (i)
2q² = 4r² 
⇒ q²= 2r²
⇒ 2 divides q²   (2 divides 2r² )
⇒ 2 divides p    (2 is prime and divides q² ⇒ 2 divides q)
Thus 2 is common factor of p and q. But, this contradict the fact that a and b have common factor other than 1. The contradiction arises by assuming that √2 is rational. So, √2 is irrational.

Proof 2: Square roots of prime numbers are irrational
Let p be a prime number and if possible, let √p be rational.
Let its simplest form be √p=m/n, where m and n are integers having n no common factor other than 1, and      
n ≠0.
Then, √p = m/n
⇒ p = m²/n²         [on squaring both sides]
⇒ pn² = m² ……..(i)
⇒ p divides m²  (p divides pn²)
⇒ p divides m    (p is prime and p divides m² ⇒ p divides m)
Let m = pq for some integer q.
Putting m = pq in eqn (i), we get:
pn² = p²q² 
⇒ n² = pq²
⇒ p divides n² [ p divides pq²]
⇒ p divides n [p is prime and p divides n² = p divides n].
Thus, p is a common factor of m and n. But, this contradicts the fact that m and n have no common factor other than 1. The contradiction arises by assuming that /p is rational. Hence, p is irrational.

Proof 3: 2 + √3 is irrational.
If possible, let (2 + √3) be rational. Then, (2 + √3) is rational, 2 is rational 
⇒  {( 2 + √3) – 2} is rational          [difference of rationales is rational]
⇒ √3 is rational. This contradicts the fact that √3 is irrational. 
The contradiction arises by assuming that (2 + √3) is irrational.
Hence, (2 + √3) is irrational.

Proof 4: √2 + √3 is irrational.
Let us suppose that (√2 + √3 ) is rational. 
Let (√2 + √3) = a, where a is rational. 
Then, √2 = (a – √3 )    ………….(i)
On squaring both sides of (i), we get: 
2 = a² + 3 – 2a√3 ⇒  2a√3 = a² + 1 
Hence, √3 = (a² +1)/2a  
This is impossible, as the right hand side is rational, while √3  is irrational. This is a contradiction. 
Since the contradiction arises by assuming that (√2 + √3) is rational, hence (√2 + √3) is irrational.

Identifying Terminating Decimals

To Check Whether a Given Rational Number is a Terminating or Repeating Decimal Let x be a rational number whose simplest form is p/q, where p and q are integers and q ≠ 0. Then,
(i) x is a terminating decimal only when q is of the form (2^m x 5ⁿ) for some non-negative integers m and n.  
(ii) x is a nonterminating repeating decimal if q ≠ (2^m x 5ⁿ). 
Examples:
(i) 33/50
Now, 50 = (2×5²)  and 2 and 5  is not a factor of 33.
So, 33/ 50 is in its simplest form.
Also, 50 = (2×5²) = (2^m × 5ⁿ) where m = 1 and n = 2.
53/343 is a terminating  decimal.
33/50 = 0.66.

(ii) 41/1000
Now, 1000 = (2³x5³) = and 2 and 5  is not a factor of 41.
So, 41/ 1000 is in its simplest form.
Also, 1000 = (2³x2³) = (2^m × 5ⁿ) where m = 3 and n = 3.
4 /1000 is a terminating  decimal.
41/1000 = 0.041
(iii) 53/343
Now,  343 = 7³ and 7  is not a factor of 53.
So, 53/ 343 is in its simplest form.
Also, 343 =7³ ≠ (2^m × 5ⁿ) .
53 /343 is a non-terminating repeating decimal.