Proof of Irrationality | Real Numbers | Class 10 Mathematics

01 · FOUNDATIONAL THEOREM

Theorem 1.2 (NCERT)

Before proving irrationality, we use an essential theorem based on prime factorisation:

Let p be a prime number. If p divides a² (where a is a positive integer), then p divides a.

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Proof by Contradiction Technique

To prove a number is irrational, we assume the opposite: that it is rational (can be written as a/b where a, b are coprime integers). We then logically derive a contradiction!

02 · PROOF OF √2

Step-by-Step Proof that √2 is Irrational

PROOF 1

Prove that √2 is Irrational

Step 1: Assumption (Contradiction)

Let us assume to the contrary that √2 is rational. Then there exist coprime integers a and b (b ≠ 0) such that:

√2 = a / b ⟹ a = b√2

Step 2: Squaring both sides

a² = 2b² — (Equation 1)

This means 2 divides a². By Theorem 1.2, 2 divides a.

So, we can write a = 2c for some integer c.

Step 3: Substitute a = 2c in Equation 1

(2c)² = 2b² ⟹ 4c² = 2b² ⟹ b² = 2c²

This means 2 divides b², so 2 divides b.

Step 4: Contradiction & Conclusion

From above, 2 is a common factor of both a and b. But this contradicts our initial assumption that a and b are coprime (have no common factor other than 1).

Hence, our assumption was wrong. √2 is Irrational. (Proved)

03 · SUM & PRODUCT PROOFS

Proving Numbers like 3 + 2√5 are Irrational

PROOF 2

Prove that 3 + 2√5 is Irrational (given √5 is irrational)

1. Assume 3 + 2√5 is rational = a/b (where a, b are coprime integers, b ≠ 0):

3 + 2√5 = a / b
2√5 = (a / b) - 3 = (a - 3b) / b
√5 = (a - 3b) / (2b)

2. Since a and b are integers, (a - 3b)/(2b) is rational. This implies √5 is rational.

3. But this contradicts the fact that √5 is irrational!

Hence, 3 + 2√5 is Irrational. (Proved)

04 · QUICK SUMMARY

Key Takeaways

If prime p divides a², then p divides a.

Always use proof by contradiction assuming p/q coprime form.

Sum or product of a rational and an irrational number is always irrational.