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🔢 Chapter 3: The World of Numbers • End-of-Chapter Exercises

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Exercise Set 3.1 4Exercise Set 3.2 4Exercise Set 3.3 8Exercise Set 3.4 5Exercise Set 3.5 5End-of-Chapter Exercises 14All Exercises 40
Q1

End-of-Chapter Exercises

1. Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal, whichever the case may be, by the process of long division:

(i) 3/50      (ii) 2/9

  • (i) 3/50:
    Dividing 3 by 50 using long division:
    3 ÷ 50 = 0.06 (with remainder 0).
    Since the remainder becomes 0, it is a terminating decimal: 0.06.
  • (ii) 2/9:
    Dividing 2 by 9 using long division:
    2 ÷ 9 = 0.2222... where remainder 2 repeats endlessly.
    Since the division loops infinitely, it is a non-terminating and repeating decimal: 0.2̅ (or 0.222...).
Q2

End-of-Chapter Exercises

2. Prove that √5 is an irrational number.

Proof by Contradiction:

  1. Assume to the contrary that √5 is a rational number.
  2. Therefore, it can be expressed in simplest form as √5 = p/q, where p and q are non-zero co-prime integers (sharing no common factor other than 1).
  3. Squaring both sides gives: 5 = p² / q² ⇒ p² = 5q².
  4. This means p² is divisible by 5, which implies that p must also be divisible by 5 (since 5 is a prime number). Let p = 5k for some integer k.
  5. Substitute p = 5k back into the equation:
    (5k)² = 5q² ⇒ 25k² = 5q² ⇒ q² = 5k².
  6. This indicates that q² is divisible by 5, which means q must also be divisible by 5.
  7. Hence, both p and q have at least 5 as a common factor. This directly contradicts our initial statement that p and q are co-prime.
  8. Therefore, our assumption is false, and √5 is an irrational number.
Q4

End-of-Chapter Exercises

4. Locate the following rational numbers on the number line: (i) 0.532, (ii) 1.15̅

  • (i) 0.532:
    Lies strictly between 0 and 1. Locate 0.5 (the midpoint), divide the tenth interval from 0.5 to 0.6 into hundredths to find 0.53, and mark 0.532 just right of 0.53.
  • (ii) 1.15̅ (1.1555...):
    In fraction form, 10x = 11.555..., 100x = 115.555... ⇒ 90x = 104 ⇒ x = 104/90 = 52/45 = 1 7/45 ≈ 1.156.
    Lies between 1 and 2, situated slightly past the midpoint of the interval between 1.1 and 1.2 (near 1.16).
Q5

End-of-Chapter Exercises

5. Find 6 rational numbers between 3 and 4.

To find 6 rational numbers, multiply the numerator and denominator by (6 + 1) = 7:

  • 3 = 3 × 7/7 = 21/7
  • 4 = 4 × 7/7 = 28/7

Six distinct rational numbers between 21/7 and 28/7 are:

22/7, 23/7, 24/7, 25/7, 26/7, 27/7

Q6

End-of-Chapter Exercises

6. Find 5 rational numbers between 2/5 and 3/5.

Multiply numerator and denominator of both fractions by (5 + 1) = 6:

  • 2/5 = (2 × 6) / (5 × 6) = 12/30
  • 3/5 = (3 × 6) / (5 × 6) = 18/30

Five distinct rational numbers between 12/30 and 18/30 are:

13/30, 14/30 (or 7/15), 15/30 (or 1/2), 16/30 (or 8/15), 17/30

Q7

End-of-Chapter Exercises

7. Find 5 rational numbers between 1/6 and 2/5.

First find the common denominator of 6 and 5, which is 30:

  • 1/6 = 5/30
  • 2/5 = 12/30

Since the integer gap between 5 and 12 is large enough (6, 7, 8, 9, 10, 11), five rational numbers are:

6/30 (or 1/5), 7/30, 8/30 (or 4/15), 9/30 (or 3/10), 10/30 (or 1/3)

Q8

End-of-Chapter Exercises

8. If x/3 + x/5 = 16/15, find the rational number x.

Step-by-step solution:

  1. Combine terms over the common denominator 15:
    (5x + 3x) / 15 = 16/15
  2. Simplify the numerator:
    8x / 15 = 16/15
  3. Multiply both sides by 15:
    8x = 16
  4. Divide by 8:
    x = 16/8 = 2
Q9

End-of-Chapter Exercises

9. Let a and b be two non-zero rational numbers such that a + 1/b = 0. Without assigning any numerical values, determine whether ab is positive or negative. Justify your answer.

Solution:

  • Given: a + 1/b = 0 ⇒ a = -1/b
  • Multiplying both sides by b: ab = -1
  • Since -1 is strictly less than 0, the product ab is negative.
Q10

End-of-Chapter Exercises

10. A rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place. Show that such a number can be written in the form p/10⁴ where p is an integer not divisible by 10. Is it necessary that the denominator of this rational number, when written in the lowest form, is divisible by 2⁴ or 5⁴? Give reasons.

  • Representation: A number with 4 decimal places can be written as p / 10⁴. If p were divisible by 10, the final 0 would cancel out with 10⁴, leaving fewer than 4 decimal places. Thus, p is not divisible by 10.
  • Divisibility by 2⁴ or 5⁴: Yes. In lowest terms, p / (2⁴ × 5⁴) cannot cancel both prime factors 2 and 5 simultaneously because 10 does not divide p. Therefore, at least one of the prime factor powers 2⁴ or 5⁴ must remain fully intact in the denominator.
Q11

End-of-Chapter Exercises

11. Without performing division, determine whether the decimal expansion of 18/125 is terminating or non-terminating. If it terminates, state the number of decimal places.

  • Fraction in lowest terms: 18/125 (18 and 125 share no common factors).
  • Prime factorization of denominator: 125 = 5³ = 2° × 5³.
  • Since the denominator contains only powers of 5, the decimal expansion is terminating.
  • The highest power between 2 and 5 is 3 (5³), so it terminates after 3 decimal places (18/125 = 144/1000 = 0.144).
Q12

End-of-Chapter Exercises

12. A rational number in its lowest form has denominator 2³ × 5. How many decimal places will its decimal expansion have? Explain your answer.

  • The denominator is 2³ × 5¹.
  • To express the denominator as a power of 10, multiply numerator and denominator by 5²:
    (2³ × 5¹) × 5² = 2³ × 5³ = 10³ = 1000.
  • Because the maximum exponent of 2 and 5 is 3, the decimal expansion will terminate after exactly 3 decimal places.
Q13

End-of-Chapter Exercises

13. Let a = 7/12 and b = 5/6. Express both a and b in the form k₁/m and k₂/m where k₁, k₂ and m are integers and k₂ - k₁ > 6. Using the same denominator m, write exactly five distinct rational numbers lying between a and b keeping an integer numerator. Explain why the condition k₂ - k₁ > n + 1 is necessary to find n such rational numbers between the two rational numbers a and b using this method.

  • Equalizing Denominators:
    Choose common denominator m = 24:
    a = 7/12 = 14/24 (k₁ = 14)
    b = 5/6 = 20/24 (k₂ = 20)
    Here, k₂ - k₁ = 20 - 14 = 6 (not strictly > 6). Let's take m = 36:
    a = 21/36 (k₁ = 21)
    b = 30/36 (k₂ = 30)
    Now, k₂ - k₁ = 30 - 21 = 9 > 6.
  • Five rational numbers between them:
    22/36, 23/36, 24/36, 25/36, 26/36
  • Explanation: Between two integers k₁ and k₂, the number of strictly intermediate integers is (k₂ - k₁ - 1). For this count to be at least n, we must have k₂ - k₁ - 1 ≥ n ⇒ k₂ - k₁ > n (or ≥ n + 1).
Q14

End-of-Chapter Exercises

14. Three rational numbers x, y, z satisfy x + y + z = 0 and xy + yz + zx = 0. Show that all the rational numbers x, y, z must be simultaneously zero.

Proof:

  1. Consider the algebraic identity:
    (x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
  2. Substitute the given values x + y + z = 0 and xy + yz + zx = 0:
    0² = x² + y² + z² + 2(0)
    x² + y² + z² = 0
  3. Since x, y, z are rational numbers, their squares x², y², z² must all be non-negative real values (≥ 0).
  4. The sum of non-negative terms can only equal 0 if each individual term is 0:
    x² = 0 ⇒ x = 0
    y² = 0 ⇒ y = 0
    z² = 0 ⇒ z = 0
  5. Thus, x, y, and z must be simultaneously zero.
Q15

End-of-Chapter Exercises

15. Show that the rational number (a + b)/2 lies between the rational numbers a and b.

Let a < b:

  1. Add a to both sides of the inequality:
    a + a < a + b ⇒ 2a < a + b ⇒ a < (a + b)/2
  2. Add b to both sides of the original inequality:
    a + b < b + b ⇒ a + b < 2b ⇒ (a + b)/2 < b
  3. Combining both results gives:
    a < (a + b)/2 < b

Hence, the average (a + b)/2 always lies strictly between a and b.