Mathematics

Example: Insert a rational and an irrational number between 2 and 3.

Solutions: If a and b are two positive rational numbers such that ab is not a perfect square of a rational number, then √ab is an irrational number lying between a and b. Also, if a,b are rational numbers, then is a rational number between them. Therefore, A rational number between 2 and 3 is  = 2.5.  
An irrational number between 2 and 3 is = = √6

Example: Find two irrational numbers between 2 and 2.5.

Solutions: If a and b are two distinct positive rational numbers such that ab is not a perfect square of a rational number, then √ab is an irrational number lying between a and b. Thereofre, Irrational number between 2 and 2.5 is
= = √5. 
Similarly, irrational number between 2 and is √5 is   
So, required numbers are √5 and .

Example: Find two irrational numbers lying between √2 and √3.

Solution: We know that, if a and b are two distinct positive irrational numbers, then √ab is an irrational number lying between a and b. Therefore, Irrational number between √2 and √3 is equal to  = 6^1/4.

Irrational number between √2 and  is equal to 2^1/4 × 6^1/8.
Hence required irrational number are 6^1/4 and 2^1/4 × 6^1/8.

Example: Find two irrational numbers between 0.12 and 0.13.

Solution: Let a = 0.12 and b = 0.13. Clearly, a and b are rational numbers such that a < b.
We observe that the number a and b have a 1 in the first place of decimal. But in the second place of decimal a has a 2 and b has 3. So, we consider the numbers
c = 0.1201001000100001 ...... & d = 0.12101001000100001.......
Clearly, c and d are irrational numbers such that a < c < d < b.

To understand the concept, use the the Theorem :

Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.
Proof : Let the prime factorisation of a be as follows :
a = p₁p₂…..p_n, where p₁,p₂,….. p_n are primes, not necessarily distinct.
Therefore,
a² = (p₁p₂…..p_n) (p₁p₂…..p_n) = p₁²p₂²…..p_n²
Now, we are given that p divides a². Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a². However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a² are p₁p₂…..p_n. So p is one of p₁, p₂,……, p_n.
Now, since a = p₁ p₂ …… p_n, p divides a.
We are now ready to give a proof that is irrational.
The proof is based on a technique called ‘proof by contradiction’.