Real Numbers: Lemma, Fundamental Theorem & Decimals Master Guide Class 10

In the architecture of mathematics, numbers are the foundational atoms upon which all algebra, calculus, and geometry are constructed. In CBSE Class 10 Mathematics, Chapter 1 (Real Numbers) builds a bridge between elementary arithmetic and higher number theory: the divisibility structure of integers formalized by Euclid's Division Lemma, the prime factorization uniqueness guaranteed by the Fundamental Theorem of Arithmetic, and the characterization of rational numbers through their terminating decimal expansions.

In this master guide, we synthesize the theoretical theorems, divisibility proofs, and decimal criteria that frequently appear on the board examination.

What You Will Learn

• Statement and applications of Euclid's Division Lemma: $a = bq + r$ ($0 \le r < b$)
• Proving positive integer forms (e.g., $3q, 3q+1, 3q+2$ and odd integer forms $4q+1, 4q+3$)
• The Fundamental Theorem of Arithmetic and unique prime factorization
• The HCF and LCM product relationship: $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$
• Why numbers of the form $6^n$ or $4^n$ can never end with the digit $0$
• The $2^n 5^m$ Criterion for terminating decimal expansions of rational numbers
• Converting rational numbers to decimals without long division

1. Euclid's Division Lemma and Divisibility Forms

Theorem Statement

Given two positive integers $a$ and $b$, there exist unique integers $q$ and $r$ satisfying: $\mathbf{a = bq + r, \quad \text{where } 0 \le r < b}$

• $a$: Dividend, $b$: Divisor, $q$: Quotient, $r$: Remainder

Solved Example: Proving Every Odd Integer is of Form $4q+1$ or $4q+3$

Problem: Show that any positive odd integer is of the form $4q + 1$ or $4q + 3$, where $q$ is some integer.

Proof:

1. Let $a$ be any positive integer, and take divisor $b = 4$.
2. By Euclid's Division Lemma: $a = 4q + r, \quad \text{where } 0 \le r < 4$
3. Therefore, the possible remainders are $r = 0, 1, 2,$ or $3$.
4. This gives four possible forms for $a$:
   • If $r = 0 \implies a = 4q = 2(2q)$ (Even, divisible by 2).
   • If $r = 1 \implies a = 4q + 1 = 2(2q) + 1$ (Odd, not divisible by 2).
   • If $r = 2 \implies a = 4q + 2 = 2(2q + 1)$ (Even, divisible by 2).
   • If $r = 3 \implies a = 4q + 3 = 2(2q + 1) + 1$ (Odd, not divisible by 2).
5. <u>Since any positive integer can be either even or odd, and $4q$ and $4q + 2$ are even, any positive ODD integer must be of the form $4q + 1$ or $4q + 3$!</u> Hence Proved.

2. The Fundamental Theorem of Arithmetic

Theorem Statement

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

Finding HCF and LCM via Prime Factorization:

• HCF (Highest Common Factor): Product of the smallest power of each common prime factor.
• LCM (Lowest Common Multiple): Product of the greatest power of each prime factor involved.

The HCF-LCM Product Property: For any two positive integers $a$ and $b$: $\mathbf{\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b}$ (Note: This formula holds strictly for TWO numbers; it does NOT hold for three numbers $a, b, c$!)

3. Why Can $6^n$ Never End with the Digit Zero? (CBSE High-Frequency Question)

Problem: Check whether $6^n$ can end with the digit $0$ for any natural number $n$.

The Rigorous Proof:

1. For any number to end with the digit $0$, it must be divisible by $10$.
2. Since $10 = 2 \times 5$, any number ending with $0$ must have BOTH $2$ and $5$ as prime factors.
3. Now, consider the prime factorization of $6^n$: $6^n = (2 \times 3)^n = \mathbf{2^n \times 3^n}$
4. The only prime factors in the expansion of $6^n$ are $2$ and $3$.
5. By the uniqueness of the Fundamental Theorem of Arithmetic, there are no other prime factors in the factorization of $6^n$.
6. <u>Because the prime factor $5$ does not occur in the prime factorization of $6^n$, $6^n$ can NEVER end with the digit $0$ for any natural number $n$!</u>

4. Decimal Expansions of Rational Numbers: The $2^n 5^m$ Rule

Let $x = \frac{p}{q}$ be a rational number in simplest co-prime form ($\text{HCF}(p, q) = 1$):

                        Decimal Expansion of Rational Number p/q
                                           |
       +-----------------------------------+-----------------------------------+
       |                                                                       |
TERMINATING DECIMAL EXPANSION                                           NON-TERMINATING REPEATING
Prime factorization of denominator q is of form:                        Prime factorization of denominator q
$$\mathbf{q = 2^n 	imes 5^m}$$                                         contains prime factors OTHER THAN 2 or 5
(where n, m are non-negative integers)                                  (e.g., factors of 3, 7, 11, 13)

Converting to Decimals Without Long Division:

Problem: Find the decimal expansion of $\frac{13}{3125}$ without actual division.

Solution:

1. Factorize the denominator: $3125 = 5^5$.
2. The denominator is of the form $2^0 \times 5^5$ (Terminating decimal).
3. To convert to a power of $10$, multiply numerator and denominator by $2^5$ ($32$): $\frac{13}{3125} = \frac{13}{5^5} = \frac{13 \times 2^5}{5^5 \times 2^5} = \frac{13 \times 32}{(5 \times 2)^5} = \frac{416}{10^5} = \frac{416}{100000} = \mathbf{0.00416}$
4. <u>The decimal expansion terminates after exactly 5 decimal places!</u>

5. Summary and Examination Tips

Exam Tip: In decimal expansion questions, ALWAYS reduce the fraction to simplest co-prime form FIRST! For example, in $\frac{6}{15} = \frac{2}{5}$, the denominator has only factor $5$, making it terminating ($0.4$). If you factorized $15 = 3 \times 5$ without cancelling $3$, you would incorrectly conclude it is non-terminating!

Common Mistake: Applying $\text{HCF} \times \text{LCM} = a \times b \times c$ for three numbers. For three numbers, this formula is mathematically FALSE!