REAL NUMBERS

Real numbers \((R)\) encompass all numbers that can be represented on a number line. They are the union of rational and irrational numbers.

• Classification of Real 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 \(\frac{p}{q}\), where \(p, q\) are integers and \(q \neq 0\). Their decimal expansion is either terminating or non-terminating repeating.
• Examples: \(\frac{1}{2} = 0.5\) (terminating), \(\frac{1}{3} = 0.333...\) (non-terminating repeating).
• Irrational Numbers \((I)\): Numbers that cannot be expressed in the form \(\frac{p}{q}\). Their decimal expansion is non-terminating and non-repeating.
• Examples: \(\sqrt{2}, \sqrt{3}, \pi, e\).

• Properties of Real Numbers:
• Commutative Property:
• Addition: \(a + b = b + a\)
• Multiplication: \(a \times b = b \times a\)
• Associative Property:
• Addition: \((a + b) + c = a + (b + c)\)
• Multiplication: \((a \times b) \times c = a \times (b \times c)\)
• Distributive Property:
• \(a(b + c) = ab + ac\)

• Order of Operations (BODMAS/PEMDAS):

1. Brackets (Parentheses)
2. Orders (Exponents/Indices)
3. Division and Multiplication (from left to right)
4. Addition and Subtraction (from left to right)

Euclid's Division Lemma

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 \(0 \le r < b\)

• Meaning: This lemma states that any positive integer \(a\) can be divided by another positive integer \(b\) in such a way that it leaves a remainder \(r\) that is less than \(b\).
• Application: Primarily used to find the HCF (Highest Common Factor) of two positive integers.

Euclid's Division Algorithm

This is a technique to compute the HCF of two given positive integers, say \(c\) and \(d\), where \(c > d\).

Steps:

1. 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. If \(r = 0\), then \(d\) is the HCF of \(c\) and \(d\).
3. If \(r \neq 0\), then make \(d\) the new dividend and \(r\) the new divisor. Apply the division lemma again.
4. Continue this process until the remainder is zero. The divisor at this stage will be the HCF of the two original numbers.

• Why it works: The HCF of \((c, d)\) is the same as the HCF of \((d, r)\). This property allows us to reduce the problem to smaller numbers until the remainder is zero.

• Example: Find HCF of 135 and 225.

1. \(225 = 135 \times 1 + 90\) (Here \(r = 90 \neq 0\))
2. \(135 = 90 \times 1 + 45\) (Here \(r = 45 \neq 0\))
3. \(90 = 45 \times 2 + 0\) (Here \(r = 0\))

The divisor at this stage is 45. Therefore, HCF(225, 135) = 45.

Statement

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

• Composite Number: A positive integer that has at least one divisor other than 1 and itself.
• Examples: 4, 6, 8, 9, 10...
• Prime Number: A positive integer greater than 1 that has no positive divisors other than 1 and itself.
• Examples: 2, 3, 5, 7, 11...

Applications

1. Finding HCF and LCM:

• HCF (Highest Common Factor): Product of the smallest power of each common prime factor in the numbers.
• LCM (Least Common Multiple): Product of the greatest power of each prime factor (common or not common) in the numbers.

• Relationship between HCF and LCM: For any two positive integers \(a\) and \(b\),

$$HCF(a, b) \times LCM(a, b) = a \times b$$ This relation holds true only for two numbers. For three or more numbers, it does not necessarily hold.

1. Proving Irrationality: This theorem forms the basis for proving numbers like \(\sqrt{2}\), \(\sqrt{3}\) irrational (covered in t4).

1. Determining Decimal Expansions: Used to predict whether a rational number has a terminating or non-terminating repeating decimal expansion (covered in t5).

• Example: Find HCF and LCM of 6 and 20 using prime factorization.
• Prime factorization of \(6 = 2^1 \times 3^1\)
• Prime factorization of \(20 = 2^2 \times 5^1\)
• HCF(6, 20): Common prime factor is 2. Smallest power of 2 is \(2^1\). So, HCF = 2.
• LCM(6, 20): Prime factors involved are 2, 3, 5. Greatest powers are \(2^2, 3^1, 5^1\). So, LCM = \(2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60\).
• Verification: \(HCF \times LCM = 2 \times 60 = 120\). Also, \(a \times b = 6 \times 20 = 120\). The relation holds.

Recall: Irrational Numbers

Numbers that cannot be expressed in the form \(\frac{p}{q}\), where \(p, q\) are integers and \(q \neq 0\). Their decimal expansion is non-terminating and non-repeating.

Theorem 1.3 (NCERT)

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

• Proof Idea: If \(p\) divides \(a^2\), then \(p\) is a prime factor of \(a^2\). By the Fundamental Theorem of Arithmetic, the prime factors of \(a^2\) are the same as the prime factors of \(a\). Therefore, \(p\) must be a prime factor of \(a\), which means \(p\) divides \(a\).

Proof of Irrationality (by Contradiction)

This method assumes the opposite of what we want to prove and then shows that this assumption leads to a contradiction, thus proving the original statement.

• General Steps for proving \(\sqrt{p}\) is irrational (where \(p\) is a prime number):

1. Assume the contrary: Assume \(\sqrt{p}\) is rational.
2. If \(\sqrt{p}\) is rational, then it can be written as \(\frac{a}{b}\), where \(a\) and \(b\) are coprime integers (i.e., HCF(a, b) = 1) and \(b \neq 0\).
3. Square both sides: \(p = \frac{a^2}{b^2}\) \(\Rightarrow pb^2 = a^2\).
4. This implies that \(p\) divides \(a^2\). By Theorem 1.3, if \(p\) divides \(a^2\), then \(p\) divides \(a\).
5. Since \(p\) divides \(a\), we can write \(a = pk\) for some integer \(k\).
6. Substitute \(a = pk\) into \(pb^2 = a^2\): \(pb^2 = (pk)^2 \Rightarrow pb^2 = p^2k^2 \Rightarrow b^2 = pk^2\).
7. This implies that \(p\) divides \(b^2\). By Theorem 1.3, if \(p\) divides \(b^2\), then \(p\) divides \(b\).
8. Contradiction: We have shown that \(p\) divides both \(a\) and \(b\). This contradicts our initial assumption that \(a\) and \(b\) are coprime (have no common factors other than 1). Therefore, our initial assumption that \(\sqrt{p}\) is rational must be false.
9. Conclusion: Hence, \(\sqrt{p}\) is irrational.

• Extension: This method can be used to prove the irrationality of numbers like \(2 + \sqrt{3}\), \(5 - \sqrt{2}\), \(\frac{1}{\sqrt{2}}\) etc. The key is to isolate the irrational part and then apply the contradiction method.

Recall: Rational Numbers

Numbers of the form \(\frac{p}{q}\), where \(p, q\) are integers, \(q \neq 0\), and HCF(p, q) = 1 (coprime).

Terminating vs. Non-terminating Repeating Decimals

• Terminating Decimal Expansion: A decimal expansion that ends after a finite number of digits.
• Example: \(\frac{1}{2} = 0.5\), \(\frac{3}{8} = 0.375\)
• Non-terminating Repeating (or Recurring) Decimal Expansion: A decimal expansion that continues indefinitely, with a repeating block of digits.
• Example: \(\frac{1}{3} = 0.333...\), \(\frac{2}{7} = 0.285714285714...\)

Theorem 1.5 (NCERT): Condition for Terminating Decimal Expansion

Let \(x = \frac{p}{q}\) be a rational number, where \(p\) and \(q\) are coprime, such that the prime factorization of the denominator \(q\) is of the form \(2^n 5^m\), where \(n\) and \(m\) are non-negative integers. Then \(x\) has a terminating decimal expansion.

• Explanation: If the denominator \(q\) only has prime factors of 2 and/or 5, we can multiply the numerator and denominator by appropriate powers of 2 or 5 to make the denominator a power of 10 (e.g., \(10^k\)). This directly results in a terminating decimal.
• Example: \(\frac{3}{8} = \frac{3}{2^3} = \frac{3 \times 5^3}{2^3 \times 5^3} = \frac{375}{1000} = 0.375\)

Theorem 1.6 (NCERT): Condition for Non-terminating Repeating Decimal Expansion

Let \(x = \frac{p}{q}\) be a rational number, where \(p\) and \(q\) are coprime, such that the prime factorization of the denominator \(q\) is not of the form \(2^n 5^m\), where \(n\) and \(m\) are non-negative integers. Then \(x\) has a non-terminating repeating decimal expansion.

• Explanation: If the denominator \(q\) has any prime factor other than 2 or 5, it's impossible to convert the denominator into a power of 10. Hence, the decimal expansion will be non-terminating and repeating.
• Example: \(\frac{1}{7}\). Denominator 7 is not of the form \(2^n 5^m\). So, \(\frac{1}{7}\) will have a non-terminating repeating decimal expansion.

Steps to determine decimal expansion without actual division:

1. Simplify the rational number: Ensure the fraction \(\frac{p}{q}\) is in its simplest form (i.e., \(p\) and \(q\) are coprime).
2. Prime factorize the denominator \(q\): Find the prime factors of \(q\).
3. Check the factors:

• If the prime factors of \(q\) are only 2s and/or 5s, then the decimal expansion is terminating.
• If the prime factors of \(q\) include any other prime number (like 3, 7, 11, etc.) besides 2 and 5, then the decimal expansion is non-terminating repeating.