Irrational Numbers

Introduction of Irrational Numbers

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. 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. 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.

Rational Numbers and Irrational Numbers

Proof that root 2 is irrational.

Let us assume that 2 is a rational number.

Going 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, = 2 × c, where c is any 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.

Since according to 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.

Irrational Number theorem

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 a can be expressed in the form of the product of its primes as:

a = p1 × p× p3………..  × pn …..(1)

Where, p1, p2p3, ……, pn represent all the prime factors of a.

Squaring both the sides of equation (1),

a2 = ( p1 × p× p3………..  × pn) ( p1 × p2 × p3………..  × pn)

⇒a2 = (p1)2 × (p2)2 × (p3 )2………..× (pn)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 , p2 , p3……….., pn. If p is a prime number and a factor of a2, then p is one of  p1 , p2 , p3……….., pn . 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.

Solved Examples of Irrational Numbers

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 are terminating numbers.

Irrational Numbers – -.45678…,  3,  2 as these are non-terminating numbers.

Question 2: Check if below numbers are rational or irrational. 

2, 5/11, -5.12, 0.31 

Solution: Since decimal expansion of an irrational number either terminates or repeats. So, 2, 5/11, -5.12, 0.31 are all rational numbers.

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