NCERT Class X Chapter 1: Real Numbers Exercise 1.2 Question 1

NCERT Class X Chapter 1: Real Numbers

Question:

Prove that $$\sqrt{5}$$ is irrational.

Given:

We need to prove that $$\sqrt{5}$$ is an irrational number.

To Find:

Show that $$\sqrt{5}$$ is irrational.

Formula:

  • Proof by Contradiction: Assume the opposite of what is to be proved and show that it leads to a contradiction.
  • Prime Divisibility Property: If a prime number \( p \) divides \( a^2 \), then \( p \) divides \( a \).
  • Coprime Integers: Two integers \( a \) and \( b \) are coprime if their greatest common divisor (GCD) is 1.

Solution:

Step 1: Assume, for contradiction, that $$\sqrt{5}$$ is rational. Then it can be written as $$\sqrt{5} = \frac{a}{b}$$, where \( a \) and \( b \) are coprime integers and \( b \ne 0 \).

$$ \sqrt{5} = \frac{a}{b} $$

Step 2: Square both sides to remove the square root.

$$ 5 = \frac{a^2}{b^2} $$

Step 3: Multiply both sides by \( b^2 \) to clear the denominator.

$$ 5b^2 = a^2 $$

Step 4: Since 5 divides \( a^2 \), and 5 is prime, 5 must also divide \( a \).

Let \( a = 5m \) for some integer \( m \).

Step 5: Substitute \( a = 5m \) into the equation \( 5b^2 = a^2 \).

$$ 5b^2 = (5m)^2 $$ $$ 5b^2 = 25m^2 $$ $$ b^2 = 5m^2 $$

Step 6: Since 5 divides \( b^2 \), and 5 is prime, 5 must also divide \( b \).

Step 7: Thus, both \( a \) and \( b \) are divisible by 5. This contradicts the assumption that \( a \) and \( b \) are coprime.

Step 8: Therefore, our initial assumption is false. Hence, $$\sqrt{5}$$ is irrational.

Result:

$$\sqrt{5}$$ is an irrational number.

© Kaliyuga Ekalavya. All rights reserved.

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