Introduction
Real numbers include rational numbers, which can be written as p/q where p and q are integers and q ≠ 0, and irrational numbers, which cannot. For CBSE 2026-27, this chapter focuses on the Fundamental Theorem of Arithmetic and algebraic proofs that numbers such as √2, √3 and √5 are irrational.
The Fundamental Theorem of Arithmetic
You already know that any composite number can be broken down into a product of primes (for example, 30 = 2 × 3 × 5). This chapter states that fact formally, and adds an important detail: this factorisation is unique.
Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
Express 140 as a product of its prime factors.
140 = 2 × 70 = 2 × 2 × 35 = 2 × 2 × 5 × 7 = 22 × 5 × 7
No matter which pair of factors you start with, you will always arrive at the same set of prime factors. This is what "unique factorisation" means.
Proving a Number is Irrational
A number is irrational if it cannot be written in the form p/q, where p and q are integers and q ≠ 0. In this section, we prove that certain square roots are irrational. This is one of the most important proof-based questions in this chapter, and it appears frequently in board exams.
√2 is irrational.
Proof. We prove this by contradiction: we assume the opposite of what we want to prove, and show that this assumption leads to something impossible.
Let us assume, to the contrary, that √2 is rational. Then we can write
√2 = p/q
where p and q are coprime integers (that is, HCF(p, q) = 1) and q ≠ 0. This is always possible, since any rational number can be reduced to its simplest form.
Squaring both sides:
2 = p2/q2 ⟹ p2 = 2q2 ...(1)
From (1), p2 is divisible by 2. It follows that p itself must be divisible by 2: if a prime number divides the square of an integer, it must also divide the integer itself, by the Fundamental Theorem of Arithmetic. So we can write p = 2c for some integer c.
Substituting p = 2c back into (1):
(2c)2 = 2q2 ⟹ 4c2 = 2q2 ⟹ q2 = 2c2
This shows q2 is also divisible by 2, and so q must be divisible by 2 as well.
But this means both p and q are divisible by 2, which contradicts our assumption that p and q are coprime (HCF(p, q) = 1). This contradiction has arisen because of our incorrect assumption that √2 is rational.
So, √2 is irrational.
The same method (assume the number is rational, write it in the form p/q with p and q coprime, and show that this forces both p and q to share a common factor) can be used to prove that √3, √5, and in general √p for any prime number p, are irrational.
Prove that 3 + 2√5 is irrational, given that √5 is irrational.
Assume, to the contrary, that 3 + 2√5 is rational. Then we can write
3 + 2√5 = p/q
for some coprime integers p and q, with q ≠ 0. Rearranging,
2√5 = p/q − 3 = (p − 3q)/q ⟹ √5 = (p − 3q)/2q
Since p and q are integers, the right-hand side (p − 3q)/2q is a rational number. This would mean √5 is rational, but this contradicts the known fact that √5 is irrational.
This contradiction shows our assumption was wrong. So, 3 + 2√5 is irrational.
This same reasoning applies generally: the sum of a rational number and an irrational number is always irrational, and the product of a nonzero rational number and an irrational number is always irrational.
Summary
- Fundamental Theorem of Arithmetic: every composite number can be factorised into primes in exactly one way, apart from the order of the factors.
- √2, √3, √5, and √p for any prime p, are irrational (proved by contradiction).
- The sum of a rational number and an irrational number is irrational. The product of a non-zero rational number and an irrational number is also irrational.
Practice Questions
- Express 156 as a product of its prime factors.
- Prove that √3 is irrational.
- Prove that 5 − √2 is irrational.