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Chapter-3 The World of Numbers

EXERCISE SET 3.1

Question 1

Problem: A merchant in the port city of Lothal is exchanging bags of spices for copper ingots. He receives 15 ingots for every 2 bags of spices. If he brings 12 bags of spices to the market, how many copper ingots will he leave with?

Solution:

  • Given: Number of copper ingots received for 2 bags of spices = 15 ingots. Total bags of spices brought = 12 bags.
  • Step 1: Find the rate of exchange per bag of spices.
    Rate = 15 ingots / 2 bags = 7.5 ingots per bag.
  • Step 2: Calculate total ingots for 12 bags.
    Total Ingots = 12 bags multiplied by (15 / 2) ingots/bag = 6 multiplied by 15 = 90 ingots.
  • Final Answer: The merchant will leave with 90 copper ingots.

Question 2

Problem: Look at the sequence of numbers on one column of the Ishango bone: 11, 13, 17, 19. What do these numbers have in common? List the next three numbers that fit this pattern.

Solution:

  • Observation: The given numbers are 11, 13, 17, 19. None of these numbers have any factors other than 1 and themselves.
  • Common Property: They are consecutive prime numbers starting from 11.
  • Next Three Numbers: The prime numbers immediately following 19 are 23, 29, and 31.
  • Final Answer: Common property: Prime numbers. Next three numbers: 23, 29, 31.

Question 3

Problem: We know that Natural Numbers are closed under addition. Are they closed under subtraction? Provide a couple of examples to justify your answer.

Solution:

  • Definition: A set is closed under an operation if performing that operation on any two elements in the set always produces an element that is also in that set.
  • Analysis: Natural numbers (1, 2, 3, and so on) are not closed under subtraction because subtracting a larger natural number from a smaller one results in a negative integer, which is not a natural number.
  • Example 1: Let a = 5 and b = 3. Then a minus b = 5 minus 3 = 2, which is a natural number.
  • Example 2: Let a = 3 and b = 5. Then a minus b = 3 minus 5 = -2. Since -2 is not a natural number, the property of closure fails.
  • Final Answer: No, natural numbers are not closed under subtraction.

Question 4

Problem: Ancient Indians used the joints of their fingers to count. Each finger has 3 joints, and the thumb is used to count them. How many can you count on one hand? How does this relate to the ancient base-12 counting systems?

Solution:

  • Calculation: One hand has 4 fingers (excluding the thumb). Each of the 4 fingers has 3 distinct joints.
    Total count = 4 fingers multiplied by 3 joints/finger = 12 joints.
  • Relation to Base-12: This method forms the physical basis for the base-12 counting system. Instead of counting 10 digits (base-10), ancient cultures used the thumb to count the 12 segments of the other four fingers, allowing them to easily count up to 12 on a single hand.
  • Final Answer: You can count up to 12 on one hand, which serves as the foundation for base-12 systems.

EXERCISE SET 3.2

Question 1

Problem: The temperature in Ladakh is recorded as 4 degrees Celsius at noon. By midnight, it drops by 15 degrees Celsius. What is the midnight temperature?

Solution:

  • Given: Initial temperature = 4 degrees Celsius, Temperature drop = 15 degrees Celsius.
  • Calculation: Midnight Temperature = Initial Temperature minus Drop = 4 minus 15 = -11 degrees Celsius.
  • Final Answer: The midnight temperature is -11 degrees Celsius.

Question 2

Problem: A spice trader takes a loan (debt) of Rupees 850. The next day, he makes a profit of Rupees 1,200. The following week, he incurs a loss of Rupees 450. Write this sequence as an equation using integers and calculate his final financial standing.

Solution:

  • Sign Convention: Debt or Loss is represented by negative integers; Profit is represented by positive integers.
    Loan = -850, Profit = +1200, Loss = -450.
  • Equation: Final Standing = (-850) + 1200 + (-450)
  • Calculation:
    = -850 – 450 + 1200
    = -1300 + 1200 = -100.
  • Final Answer: The equation is (-850) + 1200 + (-450) = -100. His final financial standing is a net debt of Rupees 100.

Question 3

Problem: Calculate the following using Brahmagupta’s laws:

Solution:

  • (i) (-12) multiplied by 5: A negative times a positive is negative. -60
  • (ii) (-8) multiplied by (-7): A negative times a negative is positive. 56
  • (iii) 0 minus (-14): Subtracting a negative is adding a positive. 0 + 14 = 14
  • (iv) (-20) divided by 4: A negative divided by a positive is negative. -5

Question 4

Problem: Explain, using a real-world example of debt, why subtracting a negative number is the same as adding a positive number (for example, 10 minus (-5) = 15).

Solution:

  • Concept: Think of a negative number as a debt. If someone has cash worth Rupees 10 but also a debt of Rupees 5, their net worth is evaluated. Now, if the debt of Rupees 5 is removed or forgiven (which mathematically means subtracting -5), their net worth increases directly to Rupees 15.
  • Analogy: Removing a debt has the exact same financial effect as giving that person Rupees 5 in cash (adding +5). Hence, 10 minus (-5) = 10 + 5 = 15.

EXERCISE SET 3.3

Question 1

Problem: Prove that the following rational numbers are equal:

Solution:

  • (i) 2/3 and 4/6: Simplifying 4/6 by dividing numerator and denominator by 2 gives (4 divided by 2) / (6 divided by 2) = 2/3. Hence, 2/3 = 4/6.
  • (ii) 9/3 and 10/8: 9/3 = 3 and 10/8 = 5/4. These are not equal.
  • (iii) -3/5 and -6/10: Multiplying the numerator and denominator of -3/5 by 2 yields (-3 multiplied by 2) / (5 multiplied by 2) = -6/10. Hence, they are equal.

Question 2

Problem: Find the sum:

Solution:

  • (i) 7/12 + 5/8: LCM of 12 and 8 is 24.
    (7 multiplied by 2)/24 + (5 multiplied by 3)/24 = (14 + 15)/24 = 29/24
  • (ii) 2/5 + 3/10: LCM of 5 and 10 is 10.
    4/10 + 3/10 = 7/10
  • (iii) -4/7 + 3/14: LCM of 7 and 14 is 14.
    -8/14 + 3/14 = -5/14

Question 3

Problem: Find the difference:

Solution:

  • (i) 5/6 – 1/4: LCM of 6 and 4 is 12.
    10/12 – 3/12 = 7/12
  • (ii) 11/8 – 3/4: LCM of 8 and 4 is 8.
    11/8 – 6/8 = 5/8
  • (iii) -7/9 – (-2/3): -7/9 + 6/9 = -1/9

Question 4

Problem: Find the product:

Solution:

  • (i) 2/3 multiplied by 3/10 = 6/30 = 1/5
  • (ii) 7/11 multiplied by 5/8 = 35/88
  • (iii) -4/7 multiplied by 5/14 = -20/98 = -10/49

Question 5

Problem: Find the quotient:

Solution:

  • (i) 2/3 divided by 3/10 = 2/3 multiplied by 10/3 = 20/9
  • (ii) 7/11 divided by 5/8 = 7/11 multiplied by 8/5 = 56/55
  • (iii) -4/7 divided by 5/14 = -4/7 multiplied by 14/5 = -4 multiplied by 2/5 = -8/5

Question 6

Problem: Show that: (1/2 + 3/4) multiplied by 8/3 = (1/2 multiplied by 8/3) + (3/4 multiplied by 8/3)

Solution:

  • LHS: (2/4 + 3/4) multiplied by 8/3 = 5/4 multiplied by 8/3 = 40/12 = 10/3
  • RHS: (1/2 multiplied by 8/3) + (3/4 multiplied by 8/3) = 4/3 + 2 = 4/3 + 6/3 = 10/3
  • Since LHS = RHS = 10/3, the statement is verified.

Question 7

Problem: Simplify using distributive property: 7/9 multiplied by (6/7 – 3/4)

Solution:

  • Apply the property:
    = (7/9 multiplied by 6/7) minus (7/9 multiplied by 3/4)
    = 6/9 – 21/36 = 2/3 – 7/12
  • LCM of 3 and 12 is 12:
    = 8/12 – 7/12 = 1/12

Question 8

Problem: Find the rational number x such that: 5/6 multiplied by (x + 3/5) = 5/6 x + 1/2

Solution:

  • Expand LHS: 5/6 x + (5/6 multiplied by 3/5) = 5/6 x + 3/6 = 5/6 x + 1/2
  • The equation becomes: 5/6 x + 1/2 = 5/6 x + 1/2
  • Since both sides are identically equal for any value of x, this equation is an identity. x can be any rational number.

EXERCISE SET 3.4

Question 1

Problem: Represent 2/3, -5/4, and 1 and 1/2 on a single number line.

Solution:

  • Convert numbers to comparable formats: 2/3 is about 0.67, -5/4 = -1.25, 1 and 1/2 = 3/2 = 1.5.
  • Draw a real line marking 0, 1, 2, -1, -2.
  • -5/4 lies between -1 and -2, exactly one-quarter of the way to the left from -1.
  • 2/3 lies between 0 and 1, two-thirds of the way to the right from 0.
  • 1.5 lies exactly halfway between 1 and 2.

Question 2

Problem: Find three distinct rational numbers that lie strictly between -1/4 and -1/2.

Solution:

  • Express with a common denominator of 16:
    -1/2 = -8/16 and -1/4 = -4/16
  • The integers between -8 and -4 are -7, -6, -5.
  • Therefore, three rational numbers are: -7/16, -6/16 (or -3/8), -5/16.

Question 3

Problem: Simplify: (-1/4) + (5/12)

Solution:

  • LCM of 4 and 12 is 12:
    -3/12 + 5/12 = 2/12 = 1/6

Question 4

Problem: A tailor has 15 and 3/4 metres of fine silk. If making one kurta requires 2 and 1/4 metres of silk, exactly how many kurtas can he make?

Solution:

  • Total Silk = 15 and 3/4 = 63/4 metres.
  • Silk per kurta = 2 and 1/4 = 9/4 metres.
  • Number of Kurtas = 63/4 divided by 9/4 = 63/4 multiplied by 4/9 = 63/9 = 7.
  • Final Answer: He can make exactly 7 kurtas.

Question 5

Problem: Find three rational numbers between 3.1415 and 3.1416.

Solution:

  • Add a decimal place to create space: 3.14150 and 3.14160.
  • Three numbers in between are: 3.14151, 3.14152, 3.14153.

Question 6

Problem: Can you think of other way(s) to find a rational number between any two rational numbers?

Solution:

  • Method 1 (Mean Method): Find the average of the two numbers: x = (a + b) / 2.
  • Method 2 (Scaling up Denominators): Convert both rational numbers to equivalent fractions with a much larger common denominator to find intermediate values directly.

EXERCISE SET 3.5

Question 1

Problem: Without long division, determine terminating/repeating types for 7/20, 4/15, 13/250. Then check via long division.

Solution:

  • 7/20: Denominator 20 = 2 square multiplied by 5 (only 2 and 5) yields a Terminating decimal. Division: 7 divided by 20 = 0.35.
  • 4/15: Denominator 15 = 3 multiplied by 5 (contains factor 3) yields a Repeating decimal. Division: 4 divided by 15 = 0.2666….
  • 13/250: Denominator 250 = 2 multiplied by 5 cubed (only 2 and 5) yields a Terminating decimal. Division: 13 divided by 250 = 0.052.

Question 2

Problem: Perform long division for 1/13 and investigate cyclic properties.

Solution:

  • 1/13 = 0.076923… (repeating block of 6 digits 076923).
  • 2/13 = 0.153846… (repeating block 153846), 3/13 = 0.230769… (repeating block 230769).
  • Observation: Yes, it displays cyclic behavior. The sequence of digits remains identical in its cyclical arrangement but starts from a different digit depending on the remainder.

Question 3

Problem: Classify as rational or irrational:

Solution:

  • (i) Square root of 81 = 9 = 9/1 which is Rational
  • (ii) Square root of 12 = 2 multiplied by square root of 3 which is Irrational
  • (iii) 0.33333… = 1/3 which is Rational
  • (iv) 0.1234512345… = 12345/99999 which is Rational
  • (v) 1.010010001… has a non-terminating, non-repeating pattern which is Irrational
  • (vi) 23.560185612239874790120 is a terminating decimal which is Rational

Question 4

Problem: Explain algebraically why 0.9999… = 1.

Solution:

  • Let x = 0.9999… (Equation 1)
  • Multiply both sides by 10: 10x = 9.9999… (Equation 2)
  • Subtract Equation 1 from Equation 2:
    10x minus x = 9.9999… minus 0.9999…
    9x = 9 which means x = 1.
  • Hence, 0.9999… is exactly equal to 1.

Question 5

Problem: Find more numbers whose reciprocals produce cyclic repeating blocks.

Solution:

  • Full-period prime numbers (primes n where the repeating decimal period of 1/n is n minus 1) exhibit this property. Examples include 7, 17, 19, 23, 29.

END-OF-CHAPTER EXERCISES

Question 1

Solution:

  • (i) 3/50 = 0.06 (Terminating)
  • (ii) 2/9 = 0.2222… (Non-terminating repeating)

Question 2

Problem: Prove that the square root of 5 is an irrational number.

Solution:

  • Assume to the contrary that the square root of 5 is rational. Therefore, square root of 5 = p/q, where p and q are co-prime integers (q is not 0).
  • Squaring both sides: 5 = p square / q square, which means p square = 5 multiplied by q square. This means p square is divisible by 5, so p must be divisible by 5. Let p = 5k.
  • Substitute p = 5k into the equation: (5k) square = 5 multiplied by q square, which simplifies to 25 multiplied by k square = 5 multiplied by q square, or q square = 5 multiplied by k square. This implies q square is divisible by 5, so q must also be divisible by 5.
  • This contradicts our assumption that p and q are co-prime (they share a common factor of 5). Thus, the square root of 5 must be irrational.

Question 3

Problem: Convert to p/q form:

DecimalFractional Form (p/q)
(i) 12.6126/10 = 63/5
(ii) 0.0120120/10000 = 3/250
(iii) 3.05252…1511/495
(iv) 1.23535…1223/990
(v) 0.2323…23/99
(vi) 2.0555…(205 – 20) / 90 = 37/18
(vii) 2.12555…(2125 – 212) / 900 = 1913/900
(viii) 3.12555…(3125 – 312) / 900 = 2813/900
(ix) 2.162521625/10000 = 173/80

Question 4 & 5

Question 5 Solution: Find 6 rational numbers between 3 and 4.

  • Write 3 and 4 with denominator 7: 3 = 21/7 and 4 = 28/7.
  • Six rational numbers are: 22/7, 23/7, 24/7, 25/7, 26/7, 27/7.

Question 6

Problem: Find 5 rational numbers between 2/5 and 3/5.

Solution:

  • Multiply numerator and denominator by 6: 12/30 and 18/30.
  • Five numbers in between are: 13/30, 14/30, 15/30, 16/30, 17/30.

Question 7

Problem: Find 5 rational numbers between 1/6 and 2/5.

Solution:

  • Make denominators identical (LCM = 30): 5/30 and 12/30.
  • Five numbers in between are: 6/30, 7/30, 8/30, 9/30, 10/30.

Question 8

Problem: If x/3 + x/5 = 16/15, find x.

Solution:

  • Combine fractions on LHS: (5x + 3x) / 15 = 16/15, which means 8x / 15 = 16/15
  • Multiply both sides by 15: 8x = 16, which means x = 2.

Question 9

Problem: Let a, b be non-zero rationals such that a + 1/b = 0. Determine if the product ab is positive or negative.

Solution:

  • a + 1/b = 0 means a = -1/b
  • Cross-multiplying by b gives: ab = -1.
  • Since -1 is less than 0, the product ab is always negative.

Question 10

Solution: If the last non-zero digit occurs at the 4th decimal place, x = p / 10 to the power of 4. In reduced form, the denominator will be 2 to the power of a multiplied by 5 to the power of b where the maximum of a and b is 4. It is not required to be divisible by both 2 to the power of 4 and 5 to the power of 4, but it must be divisible by at least one of them.

Question 11

Solution: 18/125 = 18 / (5 cubed). Since the denominator has only 5 as a prime factor, it is terminating and will have exactly 3 decimal places.

Question 12

Solution: Denominator is 2 to the power of 9 multiplied by 5 to the power of 1. The number of decimal places is determined by the highest exponent of 2 or 5. Thus, it will have 9 decimal places.

Question 13

Solution: a = 7/12, b = 5/6 = 10/12. Scale the denominator up to m = 24: a = 14/24, b = 20/24. Here the difference between numerators is 20 minus 14 = 6. To find n integers between them, the gap must satisfy the condition that the difference is greater than n + 1.

Question 14

Problem: If x + y + z = 0 and xy + yz + zx = 0, show x = y = z = 0.

Solution:

  • Use identity: (x + y + z) squared = x squared + y squared + z squared + 2 multiplied by (xy + yz + zx)
  • Substitute given values: 0 squared = x squared + y squared + z squared + 2 multiplied by 0, which means x squared + y squared + z squared = 0
  • Since squares of real rational numbers are non-negative, their sum can only equal zero if each term is zero: x = 0, y = 0, z = 0.

Question 15

Solution: Since a is less than b, it follows that a + a is less than a + b, meaning 2a is less than a + b, so a is less than (a + b) / 2. Similarly, a + b is less than b + b, meaning (a + b) / 2 is less than b. Hence, (a + b) / 2 lies strictly between a and b.

Question 16

Problem: Find the lengths of the hypotenuses of all right triangles in the square root spiral.

Solution:

  • Triangle 1: Hypotenuse = square root of (1 squared + 1 squared) = square root of 2
  • Triangle 2: Hypotenuse = square root of (square root of 2 squared + 1 squared) = square root of (2 + 1) = square root of 3
  • Triangle 3: Hypotenuse = square root of (square root of 3 squared + 1 squared) = square root of (3 + 1) = square root of 4 = 2
  • Following this iterative application of the Pythagorean theorem, successive hypotenuse lengths are: square root of 5, square root of 6, square root of 7, square root of 8, 3, square root of 10, square root of 11, square root of 12, square root of 13, square root of 14, square root of 15, 4, square root of 17.