Revisiting Irrational Numbers and Proofs of Irrationality for CBSE Class 10

In ancient Greece, the followers of Pythagoras believed that the entire physical universe was governed by whole numbers and their clean ratios—what we call rational numbers. However, when the Pythagorean mathematician Hippasus proved that the diagonal of a unit square ($\sqrt{2}$) could never be expressed as the ratio of two integers, legend says the Pythagoreans were so scandalized that they drowned him at sea. The existence of irrational numbers shook the foundations of mathematics.

In CBSE Class 10 Mathematics, Chapter 1 (Real Numbers) concludes with one of the most intellectually elegant deductive proofs in algebra: proving the irrationality of numbers using Proof by Contradiction (reductio ad absurdum). This is a guaranteed 3-mark question on every CBSE board exam paper.

What You Will Learn

• Definition of Rational vs. Irrational Numbers
• Theorem 1.3: If $p$ is a prime number and $p$ divides $a^2$, then $p$ divides $a$
• The classical Proof by Contradiction method
• Rigorous step-by-step proofs that $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{5}$ are irrational
• Proving the irrationality of composite expressions: $a + b\sqrt{p}$ (e.g., $3 + 2\sqrt{5}$, $5 - \sqrt{3}$)
• Proving expressions of the form $\frac{1}{\sqrt{2}}$ and $7\sqrt{5}$
• The exact presentation template required to score full marks in board exams

1. Rational vs. Irrational Numbers: A Quick Recall

1. Rational Numbers ($\mathbb{Q}$): Any real number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q \ne 0$, and $p$ and $q$ are co-prime (they share no common factors other than $1$).
   • Their decimal expansions are either terminating (e.g., $0.75$) or non-terminating repeating (e.g., $0.333\dots$).
2. Irrational Numbers: Real numbers that cannot be written in the form $\frac{p}{q}$.
   • Their decimal expansions are non-terminating and non-recurring (e.g., $\sqrt{2} = 1.41421356\dots$, $\pi = 3.14159265\dots$).

2. The Fundamental Lemma (Theorem 1.3)

Before proving irrationality, we establish the foundational lemma derived from the Fundamental Theorem of Arithmetic:

Theorem Statement

Let $p$ be a prime number. If $p$ divides $a^2$ (where $a$ is a positive integer), then $p$ divides $a$.

Proof Outline:

• By the Fundamental Theorem of Arithmetic, $a$ can be factored into primes: $a = p_1 p_2 p_3 \dots p_k$.
• Squaring both sides: $a^2 = (p_1 p_2 p_3 \dots p_k)^2 = p_1^2 p_2^2 \dots p_k^2$.
• Since $p$ divides $a^2$, $p$ must be one of the prime factors of $a^2$.
• Because the prime factors of $a^2$ are identical to the prime factors of $a$, $p$ must be one of the prime factors of $a$.
• <u>Therefore, if prime $p$ divides $a^2$, then $p$ MUST divide $a$!</u>

3. Proof of Irrationality of $\sqrt{5}$ (The Gold Standard CBSE Proof)

Theorem: Prove that $\mathbf{\sqrt{5}}$ is an irrational number.

Step-by-Step Proof by Contradiction:

Step 1: State the Opposite Assumption

Let us assume, to the contrary, that $\sqrt{5}$ is a rational number. Therefore, we can find two co-prime integers $a$ and $b$ ($b \ne 0$) such that: $\sqrt{5} = \frac{a}{b}$ where $a$ and $b$ are co-prime (i.e., $\text{HCF}(a, b) = 1$).

Step 2: Rearrange and Square Both Sides

$b\sqrt{5} = a$ Squaring both sides: $5b^2 = a^2 \quad \text{--- (Equation 1)}$

Step 3: Deduce that 5 Divides $a$

Equation (1) shows that $5$ divides $5b^2$, which means $5$ divides $a^2$. By Theorem 1.3, since $5$ is a prime number and divides $a^2$:

$5$ divides $a$.

Step 4: Substitute $a = 5c$

Since $5$ divides $a$, we can write $a = 5c$ for some integer $c$. Substitute $a = 5c$ into Equation (1): $5b^2 = (5c)^2$ $5b^2 = 25c^2$ Dividing both sides by $5$: $b^2 = 5c^2 \quad \text{--- (Equation 2)}$

Step 5: Deduce that 5 Divides $b$

Equation (2) shows that $5$ divides $5c^2$, which means $5$ divides $b^2$. By Theorem 1.3, since $5$ is a prime number:

$5$ divides $b$.

Step 6: Identify the Contradiction

From Steps 3 and 5, we have established that:

• $5$ divides $a$
• $5$ divides $b$ This means that $5$ is a common factor of both $a$ and $b$.

The Conclusion: <u>This contradicts the fundamental fact that $a$ and $b$ are co-prime (having no common factor other than 1)! This contradiction has arisen because of our incorrect assumption that $\sqrt{5}$ is rational. Hence, we conclude that $\mathbf{\sqrt{5}}$ is irrational.</u>

(Note: The proofs for $\sqrt{2}$ and $\sqrt{3}$ are completely identical—simply replace $5$ with $2$ or $3$ throughout the proof!)

4. Proving Irrationality of Composite Expressions ($a + b\sqrt{p}$)

When a question asks to prove that an expression like $3 + 2\sqrt{5}$ or $5 - \sqrt{3}$ is irrational, you do NOT need to re-prove that $\sqrt{5}$ is irrational from scratch (unless explicitly asked). You can use the fact that $\sqrt{5}$ is known to be irrational!

Solved Example: Prove that $3 + 2\sqrt{5}$ is Irrational

Problem: Prove that $3 + 2\sqrt{5}$ is irrational, given that $\sqrt{5}$ is an irrational number.

Solution:

1. Let us assume, to the contrary, that $3 + 2\sqrt{5}$ is a rational number.
2. Therefore, we can find co-prime integers $a$ and $b$ ($b \ne 0$) such that: $3 + 2\sqrt{5} = \frac{a}{b}$
3. Rearrange the equation to isolate the radical term $\sqrt{5}$ on the left-hand side: $2\sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b}$ $\mathbf{\sqrt{5} = \frac{a - 3b}{2b}}$
4. Analyze the Right-Hand Side:
   • Since $a$ and $b$ are integers, $(a - 3b)$ and $2b$ are also integers.
   • Therefore, the fraction $\frac{a - 3b}{2b}$ is a rational number.
5. Analyze the Contradiction:
   • If the RHS is rational, the LHS must also be rational, which implies that $\sqrt{5}$ is a rational number.
   • But this contradicts the established fact that $\sqrt{5}$ is irrational!
6. <u>This contradiction has arisen because of our incorrect assumption that $3 + 2\sqrt{5}$ is rational. Hence, $3 + 2\sqrt{5}$ is an irrational number.</u>

5. Summary and Examination Tips

Exam Tip: In composite proofs, NEVER forget to state that "since $a$ and $b$ are integers, rac{a-3b}{2b} is rational". Board marking schemes reserve 1 mark specifically for this line of mathematical justification!

Common Mistake: Forgetting to state that $a$ and $b$ are co-prime in the opening assumption of the $\sqrt{5}$ proof. The entire proof hinges on contradicting co-primality; omitting the word "co-prime" loses half a mark!