NCERT Class X Chapter 1: Real Numbers Exercise 1.2 Question 3(iii)

NCERT Class X Chapter 1: Real Numbers

Question:

Prove that \(6 + \sqrt{2}\) is irrational.

Given:

  • The number \(6 + \sqrt{2}\)

To Find:

  • Prove that \(6 + \sqrt{2}\) is irrational.

Formula:

  • If the sum of a rational number and an irrational number is rational, then the irrational number would also be rational, which is a contradiction.

Solution:

Step 1: Assume, for contradiction, that \(6 + \sqrt{2}\) is a rational number.

Step 2: Let \(6 + \sqrt{2} = \frac{p}{q}\), where \(p\) and \(q\) are integers, \(q \ne 0\), and \(\gcd(p, q) = 1\).

$$ 6 + \sqrt{2} = \frac{p}{q} $$

Step 3: Subtract 6 from both sides to isolate \(\sqrt{2}\).

$$ 6 + \sqrt{2} - 6 = \frac{p}{q} - 6 $$ $$ \sqrt{2} = \frac{p}{q} - 6 $$

Step 4: Write 6 as \(\frac{6q}{q}\) to get a common denominator.

$$ \sqrt{2} = \frac{p}{q} - \frac{6q}{q} $$

Step 5: Combine the fractions.

$$ \sqrt{2} = \frac{p - 6q}{q} $$

Step 6: Since \(p\) and \(q\) are integers, \(\frac{p - 6q}{q}\) is rational. This implies \(\sqrt{2}\) is rational.

Step 7: But it is known that \(\sqrt{2}\) is irrational. This is a contradiction.

Step 8: Therefore, our assumption is false. So, \(6 + \sqrt{2}\) is irrational.

Result:

Therefore, \(6 + \sqrt{2}\) is irrational.

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