NCERT Class X Chapter 1: Real Numbers Example 6
NCERT Class X Chapter 1: Real Numbers Example 6
Question:
Show that \(5 - \sqrt{3}\) is irrational.
Given:
Assume, for contradiction, that \(5 - \sqrt{3}\) is rational.
To Find:
Prove that \(5 - \sqrt{3}\) is irrational.
Formula:
Proof by contradiction using properties of rational and irrational numbers.
Definitions:
- Rational number: A number that can be expressed as \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \ne 0\).
- Irrational number: A number that cannot be expressed as \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \ne 0\).
Solution:
Step 1: Assume that \(5 - \sqrt{3}\) is rational.
Let \(5 - \sqrt{3} = \frac{a}{b}\), where \(a\) and \(b\) are integers, \(b \ne 0\).
Step 2: Rearranging the equation to isolate \(\sqrt{3}\).
$$ 5 - \sqrt{3} = \frac{a}{b} $$Add \(\sqrt{3}\) to both sides:
$$ 5 = \frac{a}{b} + \sqrt{3} $$Subtract \(\frac{a}{b}\) from both sides:
$$ 5 - \frac{a}{b} = \sqrt{3} $$Step 3: Express the left side as a single fraction.
$$ 5 = \frac{5b}{b} $$ $$ 5 - \frac{a}{b} = \frac{5b - a}{b} $$So,
$$ \sqrt{3} = \frac{5b - a}{b} $$Step 4: Analyze the result.
\(\frac{5b - a}{b}\) is a rational number (since \(a\) and \(b\) are integers, \(b \ne 0\)).
This means \(\sqrt{3}\) is rational.
Step 5: Contradiction.
But it is known that \(\sqrt{3}\) is irrational.
This is a contradiction.
Step 6: Conclusion.
Therefore, our assumption is false.
Hence, \(5 - \sqrt{3}\) is irrational.
Result:
Thus, \(5 - \sqrt{3}\) is irrational.
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