NCERT Class X Chapter 1: Real Numbers Example 5
NCERT Class X Chapter 1: Real Numbers Example 5
Question:
Prove that \( \sqrt{3} \) is irrational.
Given:
We need to prove that \( \sqrt{3} \) is an irrational number.
To Find:
Show that \( \sqrt{3} \) cannot be expressed as a ratio of two integers.
Formula:
No specific formula is required. We use proof by contradiction and properties of rational and irrational numbers.
Solution:
Step 1: Assume, for contradiction, that \( \sqrt{3} \) is rational. That is,
$$ \sqrt{3} = \frac{a}{b} $$where \( a \) and \( b \) are coprime integers and \( b \neq 0 \).
Step 2: Square both sides to remove the square root:
$$ (\sqrt{3})^2 = \left( \frac{a}{b} \right)^2 $$So,
$$ 3 = \frac{a^2}{b^2} $$Step 3: Multiply both sides by \( b^2 \) to clear the denominator:
$$ 3b^2 = a^2 $$Step 4: Since 3 divides \( a^2 \), 3 must also divide \( a \) (because 3 is prime).
Let \( a = 3k \) for some integer \( k \).
Step 5: Substitute \( a = 3k \) into the equation:
$$ 3b^2 = (3k)^2 = 9k^2 $$Step 6: Divide both sides by 3:
$$ b^2 = 3k^2 $$This means 3 divides \( b^2 \), so 3 divides \( b \).
Step 7: Thus, both \( a \) and \( b \) are divisible by 3.
This contradicts our assumption that \( a \) and \( b \) are coprime (have no common factor other than 1).
Step 8: Therefore, our assumption is false. So, \( \sqrt{3} \) is irrational.
Result:
Hence, \( \sqrt{3} \) is irrational.
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