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Real Numbers Notes: Class 10 Maths Chapter 1

Class 10 Maths Chapter 1 Real Numbers covers Euclid's Division Lemma, the Fundamental Theorem of Arithmetic, HCF and LCM by prime factorisation, and proofs of irrationality of certain numbers.

Overview of Topics & Subtopics (Click to Jump)

1. Euclid's Division Lemma

Short Note:

• For any two positive integers a and b, there exist unique whole numbers q and r such that a = bq + r, where 0 ≤ r < b.

• Euclid's Division Algorithm is a technique based on this lemma to compute the HCF of two positive integers by repeated division.

2. The Fundamental Theorem of Arithmetic

Short Note:

• 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.

• This theorem is used to find the HCF and LCM of two or more positive integers using prime factorisation: HCF = product of the smallest power of each common factor; LCM = product of the greatest power of each factor involved.

• For any two positive integers a and b: HCF(a, b) × LCM(a, b) = a × b.

3. Revisiting Irrational Numbers

Short Note:

• A number is irrational if it cannot be written in the form p/q, where p and q are integers and q ≠ 0.

• Using the Fundamental Theorem of Arithmetic, it can be proved that √2, √3, √5, etc. are irrational, typically using proof by contradiction.

NCERT Textbook Exercise Questions

Use Euclid's division algorithm to find the HCF of 135 and 225.
Answer: 225 = 135 × 1 + 90; 135 = 90 × 1 + 45; 90 = 45 × 2 + 0. Since the remainder is now 0, the HCF is the last non-zero divisor, i.e., 45.
Express 156 as a product of its prime factors.
Answer: 156 = 2 × 78 = 2 × 2 × 39 = 2² × 3 × 13. So 156 = 2² × 3 × 13.
Find the LCM and HCF of 6 and 20 by the prime factorisation method.
Answer: 6 = 2 × 3; 20 = 2² × 5. HCF = 2 (smallest power of common factor). LCM = 2² × 3 × 5 = 60. Check: HCF × LCM = 2 × 60 = 120 = 6 × 20 ✓
Prove that √5 is irrational.
Answer: Assume, for contradiction, √5 = p/q in lowest terms. Then p² = 5q², so 5 divides p², hence 5 divides p. Let p = 5m; substituting gives 5q² = 25m², so q² = 5m², meaning 5 divides q as well. This contradicts p and q having no common factor, so √5 is irrational.
Show that 5 − √3 is irrational.
Answer: If 5 − √3 were rational, then √3 = 5 − (5 − √3) would also be rational (difference of two rationals is rational). But √3 is known to be irrational, a contradiction. Hence 5 − √3 is irrational.
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Class 10 › Mathematics › Real Numbers
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Topics in this chapter: Euclid's Division LemmaThe Fundamental Theorem of ArithmeticRevisiting Irrational Numbers