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← CBSE Class 10

Real Numbers

Board: CBSE Class: 10

Chapter Overview

Real numbers include rational and irrational numbers. Euclid's division lemma states: for positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 <= r < b. This lemma is used to compute HCF of two numbers. The Fundamental Theorem of Arithmetic states that every composite number can be expressed uniquely as a product of primes, up to order. This theorem is used to find LCM and HCF of positive integers. The chapter also covers the irrationality of numbers like √(2), √(3), √(5) using proof by contradiction, and the decimal expansions of rational numbers (terminating and non-terminating repeating).

Topics Covered

  • Euclid Division Lemma
  • Euclid Algorithm for HCF
  • Fundamental Theorem of Arithmetic
  • Prime Factorization Method
  • LCM and HCF
  • Irrational Numbers Proof
  • Decimal Expansions
  • Rational vs Irrational

Key Formulas

a = bq + r, 0 <= r < b
HCF x LCM = a x b
LCM = product of highest powers of primes
HCF = product of lowest powers of common primes

Real-World Applications

Applications: Computer algorithms (GCD), cryptography (prime factorization), number theory foundations.

Study Tips

Tip: Practice Euclid division algorithm step by step
Tip: Master prime factorization method for HCF/LCM
Tip: Learn proof of irrationality by contradiction

NCERT Exercise Solutions

Exercise 1.1

  1. Use Euclid's division algorithm to find the HCF of 135 and 225.
    Show Solution
    Apply Euclid's division algorithm:
    225 = 135 x 1 + 90
    135 = 90 x 1 + 45
    90 = 45 x 2 + 0

    Since remainder is 0, the divisor at this step is 45.
    Therefore, HCF(135, 225) = 45.
  2. Use Euclid's division algorithm to find the HCF of 196 and 38220.
    Show Solution
    Apply Euclid's division algorithm:
    38220 = 196 x 195 + 0

    Since remainder is 0, the divisor is 196.
    Therefore, HCF(196, 38220) = 196.
  3. Use Euclid's division algorithm to find the HCF of 867 and 255.
    Show Solution
    Apply Euclid's division algorithm:
    867 = 255 x 3 + 102
    255 = 102 x 2 + 51
    102 = 51 x 2 + 0

    Since remainder is 0, the divisor is 51.
    Therefore, HCF(867, 255) = 51.
  4. Show that any positive odd integer is of the form 6q + 1, 6q + 3, or 6q + 5, where q is some integer.
    Show Solution
    Let a be any positive integer and b = 6. By Euclid's division lemma:
    a = 6q + r, where 0 <= r < 6
    So r can be 0, 1, 2, 3, 4, or 5.

    If r = 0: a = 6q (even)
    If r = 1: a = 6q + 1 (odd)
    If r = 2: a = 6q + 2 (even)
    If r = 3: a = 6q + 3 (odd)
    If r = 4: a = 6q + 4 (even)
    If r = 5: a = 6q + 5 (odd)

    Therefore, any odd integer is of the form 6q + 1, 6q + 3, or 6q + 5.
  5. Use Euclid's division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
    Show Solution
    Let a be any positive integer and b = 3. By Euclid's lemma:
    a = 3q + r, where r = 0, 1, or 2.

    Case 1: a = 3q
    a2 = (3q)2 = 9q2 = 3(3q2) = 3m where m = 3q2

    Case 2: a = 3q + 1
    a2 = (3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 = 3m + 1 where m = 3q2 + 2q

    Case 3: a = 3q + 2
    a2 = (3q + 2)2 = 9q2 + 12q + 4 = 9q2 + 12q + 3 + 1 = 3(3q2 + 4q + 1) + 1 = 3m + 1

    Therefore, square of any integer is of the form 3m or 3m + 1.

Exercise 1.2

  1. Express each number as a product of its prime factors:
    (i) 140
    (ii) 156
    (iii) 3825
    (iv) 5005
    (v) 7429
    Show Solution
    (i) 140 = 2 x 2 x 5 x 7 = 22 x 5 x 7
    (ii) 156 = 2 x 2 x 3 x 13 = 22 x 3 x 13
    (iii) 3825 = 3 x 3 x 5 x 5 x 17 = 32 x 52 x 17
    (iv) 5005 = 5 x 7 x 11 x 13
    (v) 7429 = 17 x 19 x 23
  2. Find the LCM and HCF of the following pairs of integers and verify that LCM x HCF = product of the two numbers:
    (i) 26 and 91
    (ii) 510 and 92
    (iii) 336 and 54
    Show Solution
    (i) 26 = 2 x 13, 91 = 7 x 13
    HCF = 13, LCM = 2 x 7 x 13 = 182
    Verification: LCM x HCF = 182 x 13 = 2366, Product = 26 x 91 = 2366. Verified.

    (ii) 510 = 2 x 3 x 5 x 17, 92 = 22 x 23
    HCF = 2, LCM = 22 x 3 x 5 x 17 x 23 = 23460
    Verification: LCM x HCF = 23460 x 2 = 46920, Product = 510 x 92 = 46920. Verified.

    (iii) 336 = 24 x 3 x 7, 54 = 2 x 33
    HCF = 2 x 3 = 6, LCM = 24 x 33 x 7 = 3024
    Verification: LCM x HCF = 3024 x 6 = 18144, Product = 336 x 54 = 18144. Verified.
  3. Find the LCM and HCF of the following integers by prime factorization:
    (i) 12, 15 and 21
    (ii) 17, 23 and 29
    (iii) 8, 9 and 25
    Show Solution
    (i) 12 = 22 x 3, 15 = 3 x 5, 21 = 3 x 7
    HCF = 3, LCM = 22 x 3 x 5 x 7 = 420

    (ii) 17, 23, 29 are all prime numbers.
    HCF = 1, LCM = 17 x 23 x 29 = 11339

    (iii) 8 = 23, 9 = 32, 25 = 52
    HCF = 1 (no common prime factors), LCM = 23 x 32 x 52 = 1800
  4. Given that HCF(306, 657) = 9, find LCM(306, 657).
    Show Solution
    We know: LCM x HCF = Product of numbers
    LCM x 9 = 306 x 657
    LCM = (306 x 657) / 9 = 306 x 73 = 22338

    Verification: LCM = 22338, HCF = 9, Product = 306 x 657 = 201042
    LCM x HCF = 22338 x 9 = 201042. Correct.
  5. Check whether 6n can end with the digit 0 for any natural number n.
    Show Solution
    If 6n ends with 0, it must be divisible by 5 (since any number ending with 0 is divisible by 5).

    6n = 2n x 3n
    Its prime factors are only 2 and 3. There is no factor 5.

    For a number to end with 0, it must have prime factors 2 and 5.
    Since 6n has no factor 5, it cannot end with 0 for any natural number n.
  6. Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.
    Show Solution
    7 x 11 x 13 + 13 = 13(7 x 11 + 1) = 13(77 + 1) = 13 x 78
    = 13 x 2 x 3 x 13 = 2 x 3 x 132
    Since it has factors other than 1 and itself, it is composite.

    7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5(7 x 6 x 4 x 3 x 2 x 1 + 1)
    = 5(1008 + 1) = 5 x 1009
    Since it has factors 5 and 1009 (other than 1 and itself), it is composite.
  7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round, while Ravi takes 12 minutes. If they both start at the same time and go in the same direction, after how many minutes will they meet again at the starting point?
    Show Solution
    Sonia takes 18 minutes per round, Ravi takes 12 minutes.
    They will meet at the starting point after LCM(18, 12) minutes.

    18 = 2 x 32
    12 = 22 x 3
    LCM = 22 x 32 = 36

    Therefore, they will meet again at the starting point after 36 minutes.

Exercise 1.3

  1. Prove that √(5) is irrational.
    Show Solution
    Assume √(5) is rational. Then √(5) = p/q where p, q are coprime integers, q != 0.
    Squaring both sides: 5 = p2/q2 => p2 = 5q2
    So 5 divides p2, hence 5 divides p.
    Let p = 5k. Then (5k)2 = 5q2 => 25k2 = 5q2 => 5k2 = q2
    So 5 divides q2, hence 5 divides q.
    Thus p and q have common factor 5, contradicting our assumption that p and q are coprime.
    Therefore, √(5) is irrational.
  2. Prove that 3 + 2 √(5) is irrational.
    Show Solution
    Assume 3 + 2 √(5) is rational. Then 3 + 2 √(5) = a/b where a, b are integers, b != 0.
    => 2 √(5) = a/b - 3 = (a - 3b)/b
    => √(5) = (a - 3b)/(2b)

    Since a, b are integers, RHS is rational. This implies √(5) is rational.
    But we have proved √(5) is irrational (Exercise 1.3 Q1).
    Contradiction. Therefore, 3 + 2 √(5) is irrational.
  3. Prove that the following are irrationals:
    (i) 1/√(2)
    (ii) 7 √(5)
    (iii) 6 + √(2)
    Show Solution
    (i) Assume 1/√(2) is rational. Let 1/√(2) = p/q, where p, q are coprime.
    => √(2) = q/p, which is rational.
    But we know √(2) is irrational. Contradiction.
    Therefore, 1/√(2) is irrational.

    (ii) Assume 7 √(5) is rational. Let 7 √(5) = p/q, where p, q are coprime.
    => √(5) = p/(7q), which is rational.
    But we know √(5) is irrational. Contradiction.
    Therefore, 7 √(5) is irrational.

    (iii) Assume 6 + √(2) is rational. Let 6 + √(2) = p/q, where p, q are integers.
    => √(2) = p/q - 6 = (p - 6q)/q, which is rational.
    But we know √(2) is irrational. Contradiction.
    Therefore, 6 + √(2) is irrational.

Exercise 1.4

  1. Without actually performing long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
    (i) 13/3125
    (ii) 17/8
    (iii) 64/455
    (iv) 15/1600
    (v) 29/343
    (vi) 23/23 x 52
    (vii) 129/22 x 57 x 75
    (viii) 6/15
    (ix) 35/50
    (x) 77/210
    Show Solution
    A rational number p/q has terminating decimal expansion if q is of the form 2n x 5m.

    (i) 3125 = 55 = 20 x 55 => Terminating
    (ii) 8 = 23 = 23 x 50 => Terminating
    (iii) 455 = 5 x 7 x 13 => Not of form 2n x 5m => Non-terminating repeating
    (iv) 1600 = 26 x 52 => Terminating
    (v) 343 = 73 => Not of form 2n x 5m => Non-terminating repeating
    (vi) 23 x 52 => Terminating
    (vii) 22 x 57 x 75 => Has factor 7 => Non-terminating repeating
    (viii) 6/15 = 2/5, denominator = 5 = 20 x 51 => Terminating
    (ix) 35/50 = 7/10, denominator = 10 = 2 x 5 => Terminating
    (x) 77/210 = 11/30, denominator = 30 = 2 x 3 x 5 => Has factor 3 => Non-terminating repeating
  2. Write the decimal expansion of the following rational numbers without doing long division and say what kind of decimal expansion they have:
    (i) 13/3125
    (ii) 17/8
    (iii) 64/455
    (iv) 15/1600
    (v) 29/343
    (vi) 23/23 x 52
    (vii) 129/22 x 57 x 75
    (viii) 6/15
    (ix) 35/50
    (x) 77/210
    Show Solution
    Only terminating decimals can be written without long division by converting denominator to 10n:

    (i) 13/3125 = 13/55 = (13 x 25)/(105) = 416/100000 = 0.00416 (Terminating)
    (ii) 17/8 = 17/23 = (17 x 53)/103 = 2125/1000 = 2.125 (Terminating)
    (iv) 15/1600 = 15/26 x 52 = 15/(24 x 102) = (15 x 54)/106 = 9375/1000000 = 0.009375 (Terminating)
    (vi) 23/(23 x 52) = 23/(8 x 25) = 23/200 = (23 x 53)/103 = 2875/100000 = 0.02875 (Terminating)
    (viii) 6/15 = 2/5 = (2 x 2)/10 = 4/10 = 0.4 (Terminating)
    (ix) 35/50 = 7/10 = 0.7 (Terminating)

    (iii), (v), (vii), (x) have non-terminating repeating decimal expansions.
  3. The following real numbers have decimal expansions as given below. Decide whether they are rational or not. If rational, express them in p/q form:
    (i) 43.123456789
    (ii) 0.120120012000120000...
    (iii) 43.123456789 with bar on 123456789
    Show Solution
    (i) 43.123456789 = 43123456789/109
    It is a terminating decimal, so it is rational.
    In p/q form: 43123456789/1000000000 (can be reduced by dividing by common factors).

    (ii) 0.120120012000120000...
    This decimal does not have a repeating pattern (the number of zeros keeps increasing).
    Therefore, it is irrational.

    (iii) 43.123456789 (with bar, meaning 123456789 repeats)
    This is a non-terminating repeating decimal, so it is rational.
    Let x = 43.123456789123456789...
    Using the standard method for recurring decimals:
    x = 43 + 0.123456789123456789...
    Let y = 0.123456789123456789...
    109 y = 123456789.123456789...
    109 y - y = 123456789
    => 999999999y = 123456789
    => y = 123456789/999999999
    => x = 43 + 123456789/999999999
    => x = (43 x 999999999 + 123456789)/999999999
    => x = 42999999957 + 123456789/999999999 = 43123456746/999999999

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