NCERT Class X Chapter 1: Real Numbers Exercise 1.2 Question 2

NCERT Class X Chapter 1: Real Numbers

Question:

Prove that \(3 + 2\sqrt{5}\) is irrational.

Given:

The number \(3 + 2\sqrt{5}\).

To Find:

To prove that \(3 + 2\sqrt{5}\) is irrational.

Formula:

We will use the method of contradiction to prove irrationality.

Definitions:

  • Rational Number: A number that can be written as \(\frac{a}{b}\), where \(a\) and \(b\) are integers and \(b \neq 0\).
  • Irrational Number: A number that cannot be written as a fraction of two integers. Its decimal expansion is non-terminating and non-repeating.

Solution:

Step 1: Assume, to the contrary, that \(3 + 2\sqrt{5}\) is rational.

Step 2: Let \(3 + 2\sqrt{5} = \frac{a}{b}\), where \(a\) and \(b\) are integers, \(b \neq 0\).

Step 3: Subtract 3 from both sides to isolate the term with \(\sqrt{5}\).

$$ 3 + 2\sqrt{5} - 3 = \frac{a}{b} - 3 $$ $$ 2\sqrt{5} = \frac{a}{b} - 3 $$

Step 4: Express the right side with a common denominator.

$$ \frac{a}{b} - 3 = \frac{a - 3b}{b} $$ So, $$ 2\sqrt{5} = \frac{a - 3b}{b} $$

Step 5: Divide both sides by 2 to solve for \(\sqrt{5}\).

$$ \sqrt{5} = \frac{a - 3b}{2b} $$

Step 6: Since \(a\) and \(b\) are integers, \(\frac{a - 3b}{2b}\) is rational. This implies \(\sqrt{5}\) is rational, which is a contradiction because \(\sqrt{5}\) is irrational.

Step 7: Therefore, our initial assumption is false. Hence, \(3 + 2\sqrt{5}\) is irrational.

Result:

Therefore, \(3 + 2\sqrt{5}\) is irrational.

© Kaliyuga Ekalavya. All rights reserved.

Comments

Post a Comment