Real Numbers – Class 10 Mathematics (Lecture 3)

Theorem 1.2: Let \(p\) be a prime number. If \(p\) divides \(a^2\), then \(p\) divides \(a\), where \(a\) is a positive integer.

Example: 2 divides \(36 = 6^2\), so 2 divides 6 also.

This theorem is fundamental in proving irrationality of square roots of prime numbers.

Proof by Contradiction:

1. Assume \(\sqrt{5}\) is rational: \(\sqrt{5} = \frac{p}{q}\) where \(p, q\) are integers, \(q \neq 0\), HCF(\(p, q\)) = 1
2. Squaring: \(5 = \frac{p^2}{q^2}\) ⇒ \(p^2 = 5q^2\)
3. So 5 divides \(p^2\). By Theorem 1.2, 5 divides \(p\) ⇒ \(p = 5m\)
4. Substitute: \((5m)^2 = 5q^2\) ⇒ \(25m^2 = 5q^2\) ⇒ \(q^2 = 5m^2\)
5. So 5 divides \(q^2\), thus 5 divides \(q\)
6. But if 5 divides both \(p\) and \(q\), then HCF(\(p, q\)) ≥ 5, contradicting HCF = 1
7. Therefore, \(\sqrt{5}\) is irrational

Proof:

1. Assume \(\sqrt{3} + \sqrt{5}\) is rational = \(\frac{p}{q}\)
2. Squaring: \((\sqrt{3} + \sqrt{5})^2 = \frac{p^2}{q^2}\)
3. \(3 + 5 + 2\sqrt{15} = \frac{p^2}{q^2}\) ⇒ \(8 + 2\sqrt{15} = \frac{p^2}{q^2}\)
4. \(2\sqrt{15} = \frac{p^2}{q^2} – 8 = \frac{p^2 – 8q^2}{q^2}\)
5. \(\sqrt{15} = \frac{p^2 – 8q^2}{2q^2}\)
6. LHS is irrational (\(\sqrt{15}\)), RHS is rational (since \(p, q\) are integers)
7. Contradiction! Therefore, \(\sqrt{3} + \sqrt{5}\) is irrational

Proof:

1. Assume \(5 – \frac{3}{7}\sqrt{3}\) is rational = \(\frac{p}{q}\)
2. Rearranging: \(\frac{3}{7}\sqrt{3} = 5 – \frac{p}{q}\)
3. \(\frac{3}{7}\sqrt{3} = \frac{5q – p}{q}\)
4. \(\sqrt{3} = \frac{7(5q – p)}{3q}\)
5. LHS is irrational (\(\sqrt{3}\)), RHS is rational (since \(p, q\) are integers)
6. Contradiction! Therefore, \(5 – \frac{3}{7}\sqrt{3}\) is irrational

Terminating Decimal Expansion: Decimal that ends after a finite number of digits.

Example: \(\frac{15}{2} = 7.5\) (ends after 1 decimal place)

Non-terminating Recurring Decimal Expansion: Decimal that never ends but has repeating digits.

Example: \(\frac{1}{3} = 0.3333…\) (repeating 3)

Non-terminating Non-recurring Decimal Expansion: Decimal that never ends and has no repeating pattern (irrational numbers).

Example: \(\sqrt{2} = 1.41421356…\) (no repeating pattern)

Solution:

Simplify: \(\frac{395}{10500} = \frac{79}{2100}\) (dividing numerator and denominator by 5)

Prime factorize denominator: \(2100 = 2^2 \times 3 \times 5^2 \times 7\)

Rule: A rational number \(\frac{p}{q}\) in simplest form has terminating decimal expansion if and only if \(q\) is of the form \(2^n \times 5^m\) where \(n, m\) are non-negative integers.

Here, denominator has prime factors 3 and 7 besides 2 and 5.

Answer: Non-terminating repeating decimal expansion.

Solution:

For terminating decimal expansion, denominator must have only prime factors 2 and/or 5.

Current denominator: \(2^3 \cdot 5^2 \cdot 3^p \cdot 7^q\)

To make it terminating, we must eliminate factors 3 and 7 by making \(p = 0\) and \(q = 0\).

So the fraction becomes: \(\frac{10^5}{2^3 \cdot 5^2}\)

Answer: Least possible values are \(p = 0\) and \(q = 0\).

Solution:

Even integers are multiples of 2.

If \(m\) is any integer, then \(2m\) is always even.

Examples: If \(m = 1\), \(2m = 2\) (even)

If \(m = 2\), \(2m = 4\) (even)

If \(m = 3\), \(2m = 6\) (even)

Answer: C) \(2m\)

Solution:

Odd integers are one more than even integers.

If \(q\) is any integer, then \(2q\) is even, and \(2q+1\) is odd.

Examples: If \(q = 1\), \(2q+1 = 3\) (odd)

If \(q = 2\), \(2q+1 = 5\) (odd)

If \(q = 3\), \(2q+1 = 7\) (odd)

Answer: D) \(2q+1\)

Solution:

Test with examples:

If \(n = 1\) (odd): \(2^1 – 1 = 1\) (not divisible by 8)

If \(n = 2\) (even): \(2^2 – 1 = 3\) (not divisible by 8)

If \(n = 3\) (odd): \(2^3 – 1 = 7\) (not divisible by 8)

If \(n = 4\) (even): \(2^4 – 1 = 15\) (not divisible by 8)

Actually, \(2^n – 1\) is divisible by 8 when \(n\) is a multiple of 3.

Let’s check: If \(n = 3\): \(2^3 – 1 = 7\) (not divisible)

If \(n = 6\): \(2^6 – 1 = 63\) (not divisible by 8)

Correction: The question might be incomplete or have different context.

Note: \(2^n – 1\) is divisible by 3 when \(n\) is even.

Solution:

Find HCF(65, 117):

\(65 = 5 \times 13\)

\(117 = 3^2 \times 13\)

HCF = 13

Given: \(65m – 117 = 13\)

\(65m = 13 + 117 = 130\)

\(m = \frac{130}{65} = 2\)

Answer: B) 2

Solution:

Let the number be \(d\).

When dividing 70, remainder is 5 ⇒ \(70 – 5 = 65\) is divisible by \(d\)

When dividing 125, remainder is 8 ⇒ \(125 – 8 = 117\) is divisible by \(d\)

So \(d\) divides both 65 and 117.

We need largest such \(d\) = HCF(65, 117)

\(65 = 5 \times 13\)

\(117 = 3^2 \times 13\)

HCF = 13

Answer: A) 13