Real Numbers – Class 10 Mathematics (Lecture 4)

Proof by Contradiction:

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

Proof by Contradiction:

1. Assume \(\sqrt{7}\) is rational: \(\sqrt{7} = \frac{p}{q}\) where \(p, q\) are integers, \(q \neq 0\), HCF(\(p, q\)) = 1
2. Squaring: \(7 = \frac{p^2}{q^2}\) ⇒ \(p^2 = 7q^2\)
3. So 7 divides \(p^2\). By Theorem 1.2, 7 divides \(p\) ⇒ \(p = 7m\)
4. Substitute: \((7m)^2 = 7q^2\) ⇒ \(49m^2 = 7q^2\) ⇒ \(q^2 = 7m^2\)
5. So 7 divides \(q^2\), thus 7 divides \(q\)
6. But if 7 divides both \(p\) and \(q\), then HCF(\(p, q\)) ≥ 7, contradicting HCF = 1
7. Therefore, \(\sqrt{7}\) is irrational

Solution:

Multiply numerator and denominator by the conjugate \((2 + \sqrt{3})\):

\(\frac{1}{2 – \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{2 + \sqrt{3}}{(2)^2 – (\sqrt{3})^2}\)

\(= \frac{2 + \sqrt{3}}{4 – 3} = \frac{2 + \sqrt{3}}{1} = 2 + \sqrt{3}\)

Answer: \(2 + \sqrt{3}\)

Laws of Exponents:

1. \(a^m \times a^n = a^{m+n}\)
2. \(\frac{a^m}{a^n} = a^{m-n}\)
3. \((a^m)^n = a^{m \times n}\)
4. \(a^m \times b^m = (ab)^m\)
5. \(\frac{a^m}{b^m} = \left(\frac{a}{b}\right)^m\)
6. \(a^0 = 1\) (for \(a \neq 0\))
7. \(a^{-n} = \frac{1}{a^n}\)

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:

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

Solution:

Check denominator: \(1250 = 2 \times 5^4\)

A rational number \(\frac{p}{q}\) in simplest form has terminating decimal expansion if \(q = 2^m \times 5^n\).

Here, \(1250 = 2^1 \times 5^4\)

The number of decimal places = maximum of \(m\) and \(n\) = max(1, 4) = 4

Answer: D) Four decimal places

Verification: \(\frac{14587}{1250} = 11.6696\) (terminates after 4 decimal places)

Solution:

Smallest prime number = 2

Smallest composite number = 4

HCF(2, 4) = 2

Answer: C) 2

Solution:

Two numbers are called co-prime if their HCF is 1.

Examples: (8, 15) have HCF = 1, so they are co-prime

(5, 9) have HCF = 1, so they are co-prime

Note: Twin primes are a special case of co-primes where both numbers are prime and differ by 2 (e.g., 3 and 5).

Answer: B) Co-primes

Solution:

Using the relationship: \(a \times b = \text{HCF}(a,b) \times \text{LCM}(a,b)\)

\(a \times b = 2 \times 27 = 54\)

Answer: C) 54

Example: Numbers could be 2 and 27, or 6 and 9, etc.

Check: HCF(6, 9) = 3 (not 2), so actual numbers could be different but product remains 54.

Solution:

1. \(\sqrt{a} \times \sqrt{b} = \sqrt{a \times b} = \sqrt{ab}\)

2. \((\sqrt{a} + \sqrt{b})(\sqrt{a} – \sqrt{b})\)

Using identity: \((x + y)(x – y) = x^2 – y^2\)

\(= (\sqrt{a})^2 – (\sqrt{b})^2 = a – b\)

Answers:

1. \(\sqrt{ab}\)

2. \(a – b\)