CBSE Class 10 Mathematics Real Numbers Notes Set 17

Euclid's Division Lemma and Algorithm

Euclid's division lemma: for positive integers a and b there are unique integers q and r with a = bq + r and 0 ≤ r < b.
Euclid's algorithm finds HCF(a, b) for a > b: divide a by b; if the remainder is not zero, divide the divisor by the remainder, and repeat. The last non-zero remainder is the HCF.
Every positive integer is of the form 2q or 2q + 1; every odd integer is 4q + 1 or 4q + 3; the square of any integer has the form 3m or 3m + 1, and also 4m or 4m + 1.

Fundamental Theorem of Arithmetic

Every composite number can be written as a product of primes, and this factorisation is unique apart from the order of the factors.
HCF is the product of the smallest power of each common prime factor; LCM is the product of the greatest power of every prime factor that appears.
For two positive integers, HCF × LCM = product of the numbers. This does not hold for three numbers.

Irrational Numbers

If a prime p divides a², then p divides a. This is used to prove that √2, √3, √5 and similar numbers are irrational.
Sum, difference, product or quotient of a non-zero rational and an irrational number is irrational. The sum or product of two irrationals may be rational, for example √2 × √2 = 2.

Decimal Expansions

A rational number p/q in lowest terms has a terminating decimal expansion exactly when q = 2ᵐ5ⁿ for some non-negative integers m and n. The number of decimal places is the larger of m and n.
If q has any other prime factor, the expansion is non-terminating and repeating.
Irrational numbers have non-terminating, non-repeating expansions.

Worked Examples

Example 1. Find the HCF of 867 and 255 by Euclid's algorithm.
Solution: 867 = 255 × 3 + 102, 255 = 102 × 2 + 51, 102 = 51 × 2 + 0. HCF = 51.

Example 2. Find the HCF and LCM of 336 and 54, and verify HCF × LCM = product.
Solution: 336 = 2⁴ × 3 × 7 and 54 = 2 × 3³. HCF = 2 × 3 = 6, LCM = 2⁴ × 3³ × 7 = 3024. Product 6 × 3024 = 18144 = 336 × 54.

Example 3. Show that the square of any positive integer is of the form 3m or 3m + 1.
Solution: Take n = 3q, 3q + 1 or 3q + 2. Squares: 9q² = 3(3q²), 9q² + 6q + 1 = 3(3q² + 2q) + 1, 9q² + 12q + 4 = 3(3q² + 4q + 1) + 1.

Example 4. Prove that 3 + 2√5 is irrational.
Solution: Suppose it equals a rational r. Then √5 = (r − 3)/2 is rational, which contradicts the irrationality of √5.

Example 5. Without long division, state whether 129/(2² × 5⁷ × 7⁵) terminates.
Solution: The denominator contains the prime 7 and 129 = 3 × 43 has no factor 7 to cancel it, so the expansion is non-terminating and repeating.