Proving that a number is irrational, especially for Class 10 mathematics, typically involves using proof by contradiction. Here's how to prove that numbers like √2, √3, or √5 are irrational. Remember that the real numbers which cannot be expressed in the form of p/q, where p and q are integers and q ≠ 0 are known as irrational numbers (as per the reference).
Understanding Rational and Irrational Numbers
Before diving into the proof, let's clarify the difference between rational and irrational numbers.
- Rational Numbers: These can be expressed in the form p/q, where p and q are integers, and q is not zero. Examples include 1/2, -3/4, 5, etc.
- Irrational Numbers: These cannot be expressed in the form p/q. Examples include √2, π, and e.
Proving √2 is Irrational (A Common Example)
The most common example used in Class 10 to illustrate the proof of irrationality is demonstrating that √2 is irrational. Here's the proof by contradiction:
- Assume the opposite: Suppose, for the sake of contradiction, that √2 is rational.
- Express as a fraction: This means we can write √2 = a/b, where a and b are integers, and b ≠ 0. We also assume that a/b is in its simplest form, meaning a and b have no common factors other than 1 (they are co-prime).
- Rearrange the equation: Squaring both sides of the equation √2 = a/b, we get 2 = a²/b².
- Further manipulation: Multiplying both sides by b², we get 2b² = a².
- Deduction: This implies that a² is even (since it's equal to 2 times an integer).
- Implication for 'a': If a² is even, then 'a' must also be even. (The square of an odd number is always odd.)
- Express 'a' in terms of another integer: Since 'a' is even, we can write a = 2k, where k is some integer.
- Substitute: Substituting a = 2k into the equation 2b² = a², we get 2b² = (2k)², which simplifies to 2b² = 4k².
- Simplify: Dividing both sides by 2, we get b² = 2k².
- Deduction about 'b': This implies that b² is also even, and therefore, 'b' must also be even.
- Contradiction: We've now shown that both 'a' and 'b' are even. This means they have a common factor of 2. However, we initially assumed that a and b were co-prime (had no common factors other than 1). This is a contradiction.
- Conclusion: Since our initial assumption leads to a contradiction, the assumption must be false. Therefore, √2 is irrational.
Generalizing the Proof for √p (where p is a prime number)
The same logic can be extended to prove the irrationality of the square root of any prime number (like √3, √5, √7, etc.). The key steps remain the same:
- Assume √p is rational.
- Express √p as a/b, where a and b are co-prime integers.
- Square both sides: p = a²/b².
- Rearrange: pb² = a². This means a² is divisible by p.
- Therefore, a is also divisible by p (a = pk for some integer k).
- Substitute: pb² = (pk)² = p²k².
- Simplify: b² = pk². This means b² is divisible by p, and therefore, b is also divisible by p.
- Contradiction: Both a and b are divisible by p, contradicting the assumption that a and b are co-prime.
- Conclusion: √p is irrational.
Tips for Class 10 Students
- Understand the Logic: Focus on why the proof works, not just memorizing the steps.
- Practice: Work through several examples (√2, √3, √5) until you're comfortable.
- State Assumptions Clearly: Be explicit about your assumptions, especially the "co-prime" assumption.
- Show Each Step: Don't skip steps in your proof, as clarity is important.
