Real Numbers – Class 10 Mathematics (Lecture 6)

Proof by Contradiction:

1. Assume \(3\sqrt{2}\) is rational.
2. Then \(3\sqrt{2} = \frac{p}{q}\), where \(p, q\) are integers, \(q \neq 0\), HCF(\(p, q\)) = 1.
3. \(\sqrt{2} = \frac{p}{3q}\)
4. LHS (\(\sqrt{2}\)) is irrational.
5. RHS (\(\frac{p}{3q}\)) is rational (since \(p, 3q\) are integers).
6. This is a contradiction because an irrational number cannot equal a rational number.
7. Therefore, \(3\sqrt{2}\) is irrational.

Alternative approach: Since \(\sqrt{2}\) is irrational and 3 is rational, the product of a rational (non-zero) and an irrational is always irrational.

Proof by Contradiction:

1. Assume \(2 – 3\sqrt{5}\) is rational = \(\frac{p}{q}\), where \(p, q\) are integers, \(q \neq 0\), HCF(\(p, q\)) = 1.
2. Rearrange: \(2 – \frac{p}{q} = 3\sqrt{5}\)
3. \(\frac{2q – p}{q} = 3\sqrt{5}\)
4. \(\sqrt{5} = \frac{2q – p}{3q}\)
5. LHS (\(\sqrt{5}\)) is irrational.
6. RHS (\(\frac{2q – p}{3q}\)) is rational (since \(2q-p\) and \(3q\) are integers).
7. This is a contradiction.
8. Therefore, \(2 – 3\sqrt{5}\) is irrational.

Proof by Contradiction:

1. Assume \(\sqrt{3} + \sqrt{5}\) is rational = \(\frac{p}{q}\), where \(p, q\) are integers, \(q \neq 0\), HCF(\(p, q\)) = 1.
2. Squaring both sides: \((\sqrt{3} + \sqrt{5})^2 = \frac{p^2}{q^2}\)
3. \(3 + 5 + 2\sqrt{15} = \frac{p^2}{q^2}\)
4. \(8 + 2\sqrt{15} = \frac{p^2}{q^2}\)
5. \(2\sqrt{15} = \frac{p^2}{q^2} – 8 = \frac{p^2 – 8q^2}{q^2}\)
6. \(\sqrt{15} = \frac{p^2 – 8q^2}{2q^2}\)
7. LHS (\(\sqrt{15}\)) is irrational.
8. RHS (\(\frac{p^2 – 8q^2}{2q^2}\)) is rational (since \(p^2-8q^2\) and \(2q^2\) are integers).
9. This is a contradiction.
10. Therefore, \(\sqrt{3} + \sqrt{5}\) is irrational.

Proof by Contradiction:

1. Assume \(\sqrt{p} + \sqrt{q}\) is rational = \(\frac{a}{b}\), where \(a, b\) are integers, \(b \neq 0\), HCF(\(a, b\)) = 1.
2. Squaring both sides: \((\sqrt{p} + \sqrt{q})^2 = \frac{a^2}{b^2}\)
3. \(p + q + 2\sqrt{pq} = \frac{a^2}{b^2}\)
4. \(2\sqrt{pq} = \frac{a^2}{b^2} – (p+q) = \frac{a^2 – (p+q)b^2}{b^2}\)
5. \(\sqrt{pq} = \frac{a^2 – (p+q)b^2}{2b^2}\)
6. LHS (\(\sqrt{pq}\)) is irrational because:
   • \(p\) and \(q\) are primes
   • \(pq\) is not a perfect square (unless \(p = q\), but even then \(\sqrt{p}\) is irrational for prime \(p\))
7. RHS is rational (since \(a^2-(p+q)b^2\) and \(2b^2\) are integers).
8. This is a contradiction.
9. Therefore, \(\sqrt{p} + \sqrt{q}\) is irrational.

Note: This proof works for any distinct primes \(p\) and \(q\). If \(p = q\), then \(\sqrt{p} + \sqrt{q} = 2\sqrt{p}\), which is also irrational.