Introduction of Irrational Numbers
An irrational number is a real number that cannot be expressed as a ratio of integers. example, √ 2. The real numbers which are not rational i.e. which cannot be expressed in the form of p/q, where p and q are integers and q ≠ 0 are known as irrational numbers. The decimal expansion of an irrational number is neither terminating nor recurring.
For Example √ 2, √ 3, √ pi, etc.
The above numbers are not a rational number as 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.
For Example, 2, 5/11, -5.12, 0.31 are all examples of rational numbers because the decimal expansion of an irrational number either terminates or repeats.
From the study of irrational numbers, if p is a prime number, then √ p is irrational. From this, it can be said that √ 2, √ 3, √ pi, etc. are all 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, p = 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 × p2 × p3……….. × pn …..(1)
Where, p1, p2, p3, ……, pn represent all the prime factors of a.
Squaring both the sides of equation (1),
a2 = ( p1 × p2 × 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: 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.
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