Introduction
In Class 9, you studied rational and irrational numbers. Together, these form the real numbers. In this chapter, we explore two powerful ideas about real numbers:
- Euclid’s Division Lemma - a method for finding the HCF (Highest Common Factor) of two positive integers.
- The Fundamental Theorem of Arithmetic - every composite number can be uniquely expressed as a product of primes.
Using these, we will also understand why numbers like √2 are irrational, and why the decimal expansions of certain rational numbers terminate.
1. Euclid’s Division Lemma
Lemma: Given any two positive integers a and b, there exist unique integers q and r satisfying:
a = bq + r, where 0 ≤ r < b
Here q is the quotient and r is the remainder.
Example
Let a = 455 and b = 42:
- 455 = 42 × 10 + 35
- So q = 10 and r = 35 (and 0 ≤ 35 < 42 ✓)
Finding HCF (Euclid’s Algorithm)
To find the HCF of two numbers:
- Divide the larger by the smaller.
- If the remainder is 0, the divisor is the HCF.
- Otherwise, the divisor becomes the new a and the remainder becomes the new b. Repeat step 1.
Example: Find the HCF of 455 and 42.
| Step | Equation | Divisor | Remainder |
|---|---|---|---|
| 1 | 455 = 42 × 10 + 35 | 42 | 35 |
| 2 | 42 = 35 × 1 + 7 | 35 | 7 |
| 3 | 35 = 7 × 5 + 0 | 7 | 0 |
HCF = 7
2. The Fundamental Theorem of Arithmetic
Every composite number can be expressed as a product of primes, and this factorization is unique - apart from the order of the factors.
Examples
- 12 = 2 × 2 × 3 = 2² × 3
- 60 = 2 × 2 × 3 × 5 = 2² × 3 × 5
- 156 = 2² × 3 × 13
Finding HCF and LCM
For two numbers - say 6 and 20:
- 6 = 2 × 3
- 20 = 2² × 5
| 2 | 3 | 5 | |
|---|---|---|---|
| 6 | 1 | 1 | 0 |
| 20 | 2 | 0 | 1 |
- HCF: take the smaller power of each prime → 2¹ × 3⁰ × 5⁰ = 2
- LCM: take the larger power of each prime → 2² × 3¹ × 5¹ = 60
Note: HCF × LCM = a × b → 2 × 60 = 120 = 6 × 20 ✓
3. Irrational Numbers
A number that cannot be written as p/q (where p, q are integers and q ≠ 0) is called an irrational number.
Prove that √2 is irrational
Proof by contradiction:
Assume √2 is rational. Then √2 = p/q, where p and q share no common factor (i.e., p/q is in simplest form).
- Squaring: 2 = p²/q²
- So: p² = 2q²
- This means p² is even ⟹ p is even.
- So p = 2k for some integer k.
- Then: (2k)² = 2q² ⟹ 4k² = 2q² ⟹ q² = 2k² ⟹ q² is even ⟹ q is even.
But now both p and q are divisible by 2 - contradicting our assumption that p/q was in simplest form.
Therefore √2 cannot be rational. √2 is irrational. ∎
4. Decimal Expansions of Rational Numbers
A rational number p/q (in simplest form) has a terminating decimal expansion if and only if the prime factorization of q contains only 2’s and 5’s (no other primes).
Examples
- 7/8 = 7/2³ → terminating (= 0.875)
- 13/3125 = 13/5⁵ → terminating
- 1/3 → 0.3333… (non-terminating, repeating - because 3 is neither 2 nor 5)
- 1/6 = 1/(2 × 3) → non-terminating, repeating
Key Takeaways
- Euclid’s algorithm finds the HCF.
- Fundamental Theorem: every composite number has a unique prime factorization.
- p × q = HCF(p, q) × LCM(p, q)
- √2, √3, √5, π - all irrational.
- p/q terminates ⟺ q’s prime factorization contains only 2’s and/or 5’s.
Practice Questions
- Use Euclid’s algorithm to find the HCF of 867 and 255.
- Find the LCM and HCF of 12, 15, and 21 using prime factorization.
- Prove that √5 is irrational.
- Without dividing, decide which of the following have a terminating decimal expansion: (a) 13/3125 (b) 17/8 (c) 64/455 (d) 15/1600
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