Real Numbers

Real Numbers (\(R\)) form the complete set of numbers that can be represented on a number line. They include both Rational and Irrational numbers.

• Natural Numbers (\(N\)): Counting numbers. \(\{1, 2, 3, ...\}\)
• Whole Numbers (\(W\)): Natural numbers including zero. \(\{0, 1, 2, 3, ...\}\)
• Integers (\(Z\)): Whole numbers and their negatives. \(\{..., -3, -2, -1, 0, 1, 2, 3, ...\}\)
• Rational Numbers (\(Q\)): Numbers that can be expressed in the form \(p/q\), where \(p, q\) are integers and \(q \neq 0\).
• Decimal expansions are either terminating or non-terminating repeating.
• Examples: \(1/2 = 0.5\), \(1/3 = 0.333...\), \(5 = 5/1\), \(-7/4 = -1.75\).
• Irrational Numbers (\(I\)): Numbers that cannot be expressed in the form \(p/q\).
• Decimal expansions are non-terminating non-repeating.
• Examples: \(\sqrt{2}, \sqrt{3}, \pi, 0.101101110...\).

Relationships between Number Systems

\(N \subset W \subset Z \subset Q \subset R\)

• Every Natural number is a Whole number.
• Every Whole number is an Integer.
• Every Integer is a Rational number.
• Every Rational number is a Real number.
• Irrational numbers are a subset of Real numbers, but distinct from Rational numbers. \(Q \cap I = \emptyset\) and \(Q \cup I = R\).

Euclid's Division Lemma (EDL)

For any two given positive integers \(a\) and \(b\), there exist unique whole numbers \(q\) (quotient) and \(r\) (remainder) such that:

\(a = bq + r\), where \(0 \le r < b\).

• Dividend (a) = Divisor (b) \(\times\) Quotient (q) + Remainder (r)
• This lemma is the basis for the Euclid's Division Algorithm.

Euclid's Division Algorithm (EDA)

EDA is a technique to compute the Highest Common Factor (HCF) of two given positive integers.

Steps to find HCF of two positive integers \(c\) and \(d\) (where \(c > d\)):

1. Apply EDL: Apply Euclid's Division Lemma to \(c\) and \(d\) to find whole numbers \(q\) and \(r\) such that \(c = dq + r\), where \(0 \le r < d\).
2. Check Remainder:

• If \(r = 0\), then \(d\) is the HCF of \(c\) and \(d\).
• If \(r \neq 0\), then make the divisor \(d\) the new dividend and the remainder \(r\) the new divisor.

1. Repeat: Continue the process until the remainder is zero. The divisor at this stage will be the HCF.

Applications of EDA

• Finding HCF of large numbers.
• Proving properties of numbers (e.g., squares of positive integers are of the form \(3m\) or \(3m+1\)).

Statement

Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

• Composite Number: A natural number greater than 1 that is not prime (i.e., has more than two factors).
• Examples: \(4, 6, 8, 9, 10, 12, ...\)
• Prime Number: A natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself.
• Examples: \(2, 3, 5, 7, 11, 13, ...\)

Prime Factorization

The process of expressing a composite number as a product of its prime factors.

• Example: \(12 = 2 \times 2 \times 3 = 2^2 \times 3\)
• The order of prime factors doesn't matter: \(2 \times 3 \times 2\) is the same as \(2 \times 2 \times 3\).

Applications of FTA: Finding HCF and LCM

Given two positive integers, say \(a\) and \(b\), their HCF and LCM can be found using their prime factorizations.

Let \(a = p_1^{x_1} p_2^{x_2} ... p_n^{x_n}\) and \(b = p_1^{y_1} p_2^{y_2} ... p_n^{y_n}\), where \(p_i\) are prime numbers and \(x_i, y_i \ge 0\).

1. HCF (Highest Common Factor):

• Product of the smallest power of each common prime factor in the numbers.
• \(HCF(a, b) = p_1^{\min(x_1, y_1)} p_2^{\min(x_2, y_2)} ... p_n^{\min(x_n, y_n)}\)

1. LCM (Least Common Multiple):

• Product of the greatest power of each prime factor involved in the numbers.
• \(LCM(a, b) = p_1^{\max(x_1, y_1)} p_2^{\max(x_2, y_2)} ... p_n^{\max(x_n, y_n)}\)

Important Relation between HCF and LCM

For any two positive integers \(a\) and \(b\):

\(HCF(a, b) \times LCM(a, b) = a \times b\)

• This relation is useful for verifying calculations or finding one value if the other two are known.
• Note: This relation holds true only for two numbers. It does not necessarily hold for three or more numbers.

The method used to prove irrationality is usually Proof by Contradiction.

Key Theorem

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

• This theorem is crucial for proving irrationality of square roots of prime numbers.

General Steps for Proving \(\sqrt{p}\) is Irrational (where \(p\) is a prime number)

1. Assume the contrary: Assume \(\sqrt{p}\) is rational.
2. Express as a fraction: If \(\sqrt{p}\) is rational, then it can be written as \(\sqrt{p} = a/b\), where \(a\) and \(b\) are coprime integers (i.e., their HCF is 1) and \(b \neq 0\).
3. Square both sides: \(p = a^2/b^2 \implies a^2 = pb^2\).
4. Deduce divisibility: Since \(a^2 = pb^2\), it means \(p\) divides \(a^2\). By the theorem above, if \(p\) divides \(a^2\), then \(p\) must divide \(a\).
5. Substitute and deduce again: If \(p\) divides \(a\), then we can write \(a = pk\) for some integer \(k\). Substitute this into the equation from step 3: \((pk)^2 = pb^2 \implies p^2k^2 = pb^2 \implies pk^2 = b^2\).
6. Deduce divisibility for \(b\): Since \(pk^2 = b^2\), it means \(p\) divides \(b^2\). Again, by the theorem, if \(p\) divides \(b^2\), then \(p\) must divide \(b\).
7. Contradiction: From steps 4 and 6, we have shown that \(p\) divides both \(a\) and \(b\). This contradicts our initial assumption in step 2 that \(a\) and \(b\) are coprime (HCF is 1). Our assumption was that \(a\) and \(b\) have no common factors other than 1, but we found a common factor \(p\).
8. Conclusion: Since our assumption led to a contradiction, the assumption must be false. Therefore, \(\sqrt{p}\) is irrational.

Proving Irrationality of Expressions

Similar methods are used for expressions like \(a + b\sqrt{p}\) or \(a\sqrt{p}\).

• Example: To prove \(3 + 2\sqrt{5}\) is irrational.

1. Assume \(3 + 2\sqrt{5}\) is rational. So, \(3 + 2\sqrt{5} = a/b\) (where \(a, b\) are coprime integers, \(b \neq 0\)).
2. Rearrange to isolate \(\sqrt{5}\): \(2\sqrt{5} = a/b - 3 \implies 2\sqrt{5} = (a - 3b)/b \implies \sqrt{5} = (a - 3b)/(2b)\).
3. Contradiction: Since \(a, b, 2, 3\) are integers, \((a - 3b)/(2b)\) is a rational number. This implies \(\sqrt{5}\) is rational, which contradicts the known fact that \(\sqrt{5}\) is irrational.
4. Conclusion: Therefore, \(3 + 2\sqrt{5}\) is irrational.

We know that rational numbers have either terminating or non-terminating repeating decimal expansions.

Terminating Decimal Expansions

A rational number \(x = p/q\) (where \(p\) and \(q\) are coprime) has a terminating decimal expansion if and only if the prime factorization of the denominator \(q\) is of the form \(2^n 5^m\), where \(n\) and \(m\) are non-negative integers.

• Condition: Denominator \(q\) must only have prime factors 2 or 5 (or both).
• Example: \(3/8 = 3/2^3\). Denominator is \(2^3\) (form \(2^n 5^m\) where \(m=0\)). So, it's terminating (\(0.375\)).
• Example: \(7/20 = 7/(2^2 \times 5)\). Denominator is \(2^2 \times 5^1\). So, it's terminating (\(0.35\)).

Non-terminating Repeating Decimal Expansions

A rational number \(x = p/q\) (where \(p\) and \(q\) are coprime) has a non-terminating repeating decimal expansion if and only if the prime factorization of the denominator \(q\) is not of the form \(2^n 5^m\).

• Condition: Denominator \(q\) has prime factors other than 2 or 5.
• Example: \(1/3\). Denominator is \(3\). It has a prime factor other than 2 or 5. So, it's non-terminating repeating (\(0.333...\)).
• Example: \(5/12 = 5/(2^2 \times 3)\). Denominator is \(2^2 \times 3\). It has a prime factor \(3\) other than 2 or 5. So, it's non-terminating repeating (\(0.41666...\)).

Determining Decimal Expansion without Actual Division

To determine if \(p/q\) has a terminating decimal expansion:

1. Simplify: Express the rational number \(p/q\) in its simplest form (i.e., \(p\) and \(q\) are coprime).
2. Prime Factorize Denominator: Find the prime factorization of \(q\).
3. Check Factors: If the prime factors of \(q\) are only 2s or 5s (or both), then it's terminating. Otherwise, it's non-terminating repeating.