Based on Maths Class 8 "A Story of Numbers," Chapter-3: Solutions to the "Figure it Out" questions from pages 60 and 61.

Figure it Out 

1. A group of indigenous people in a Pacific island use different sequences of number names to count different objects. 

Why do you think they do this? 

As discussed in the text, early counting was tied to specific, practical needs like tracking livestock, food, or goods for trade. A plausible reason for using different sequences of number names for different objects is for classification. This practice embeds information about the type of object being counted directly into the number name itself. For example, there might be one set of number words for counting long, thin objects like canoes, another for round objects like coconuts, and a third for living things like fish or people. This is similar to the concept of noun classifiers found in many languages, where a word is used to specify the class or category of the noun being counted. This method adds a layer of context and specificity that a single, abstract number system would lack.

2. Consider the extension of the Gumulgal number system beyond 6 in the same way of counting by 2s. Come up with ways of performing the different arithmetic operations (+, –, ×, ÷) for numbers occurring in this system, without using Hindu numerals. 

Use this to evaluate the following:

(i) (ukasar-ukasar-ukasar-ukasar-urapon) + (ukasar-ukasarukasar-urapon) 

(ii) (ukasar-ukasar-ukasar-ukasar-urapon) – (ukasar-ukasarukasar) 

(iii) (ukasar-ukasar-ukasar-ukasar-urapon) × (ukasar-ukasar) 

(iv) (ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar) ÷ (ukasar-ukasar) 

The Gumulgal system described on page 56 is an additive system where urapon = 1 and ukasar = 2. Numbers are formed by adding ukasar as many times as possible, and then adding urapon if the number is odd.

Defining the Arithmetic Operations:

• Addition (+): Combine all the ukasar and urapon terms from both numbers. Then, simplify by replacing every pair of urapons (urapon-urapon) with one ukasar.
• Subtraction (–): Cancel out the terms of the second number from the first. If you need to subtract an urapon but there isn't one, you can convert one ukasar from the first number into an urapon-urapon pair and then proceed.
• Multiplication (×): This is repeated addition. To multiply A × B, add number A to itself B times.
• Division (÷): This is repeated subtraction. To divide A ÷ B, count how many times you can subtract number B from number A until you can't subtract anymore. That count is the answer.

Evaluations:

First, let's decode the numbers in the problems:

• ukasar-ukasar-ukasar-ukasar-urapon = (4 × 2) + 1 = 9
• ukasar-ukasar-ukasar-urapon = (3 × 2) + 1 = 7
• ukasar-ukasar-ukasar = (3 × 2) = 6
• ukasar-ukasar = (2 × 2) = 4
• ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar = (8 × 2) = 16

(i) (ukasar-ukasar-ukasar-ukasar-urapon) + (ukasar-ukasar-ukasar-urapon)
This is 9 + 7.

• Operation: Combine (4 ukasar, 1 urapon) and (3 ukasar, 1 urapon). This gives a total of 7 ukasar and 2 urapon.
• Simplify: The 2 urapons combine to make 1 ukasar.
• Result: We now have 7 ukasar + 1 ukasar = 8 ukasar. This is equal to 16.
• Answer: ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar

(ii) (ukasar-ukasar-ukasar-ukasar-urapon) – (ukasar-ukasar-ukasar)
This is 9 – 6.

• Operation: From (4 ukasar, 1 urapon), subtract 3 ukasar.
• Result: We are left with 1 ukasar and 1 urapon. This is equal to 3.
• Answer: ukasar-urapon

(iii) (ukasar-ukasar-ukasar-ukasar-urapon) × (ukasar-ukasar)
This is 9 × 4.

• Operation: We must add the representation of 9 to itself 4 times.
  • 9 + 9 = (4 ukasar, 1 urapon) + (4 ukasar, 1 urapon) = 8 ukasar, 2 urapon = 9 ukasar (which is 18).
  • 18 + 9 = (9 ukasar) + (4 ukasar, 1 urapon) = 13 ukasar, 1 urapon (which is 27).
  • 27 + 9 = (13 ukasar, 1 urapon) + (4 ukasar, 1 urapon) = 17 ukasar, 2 urapon = 18 ukasar (which is 36).
• Result: The final result is 18 ukasars.
• Answer: ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar

(iv) (ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar) ÷ (ukasar-ukasar)
This is 16 ÷ 4.

• Operation: We count how many times we can subtract 4 (which is ukasar-ukasar) from 16 (which is ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar).
  1. 16 – 4 = 12 (ukasar-ukasar-ukasar-ukasar-ukasar-ukasar)
  2. 12 – 4 = 8 (ukasar-ukasar-ukasar-ukasar)
  3. 8 – 4 = 4 (ukasar-ukasar)
  4. 4 – 4 = 0
• We were able to subtract it 4 times. The number 4 is represented as ukasar-ukasar.
• Answer: ukasar-ukasar

3. Identify the features of the Hindu number system that make it efficient when compared to the Roman number system.

Based on the text, the Hindu number system is more efficient than the Roman system due to the following key features:

1. Place Value System: The value of a digit is determined by its position within the number. In the number 375, the '3' represents 3 hundreds, the '7' represents 7 tens, and the '5' represents 5 ones. The Roman system is primarily additive (e.g., XXX is 10+10+10), and lacks this powerful positional structure.
2. The Digit Zero (0): The Hindu system includes a symbol for zero, which serves two critical functions. It is a placeholder to unambiguously distinguish between numbers like 31, 301, and 310. It is also a number in its own right, with arithmetic properties that are essential for computation. The Roman system has no concept of zero, making representation and calculation difficult.
3. A Small, Fixed Set of Numerals: The Hindu system uses only ten digits (0-9) to represent any number, no matter how large. The Roman system is not as scalable; it requires new symbols for larger landmark numbers (I, V, X, L, C, D, M) and would need to invent more symbols to represent ever-larger quantities.
4. A Consistent Base: The landmark numbers in the Hindu system are all powers of a single base (base-10). This makes arithmetic, especially multiplication and division, systematic and algorithmic. As shown in the text, multiplying two landmark numbers (e.g., 102×103) results in another predictable landmark number (105). In the Roman system, the landmark numbers (1, 5, 10, 50, 100...) do not have a consistent power relationship, which complicates arithmetic.

4. Using the ideas discussed in this section, try refining the number system you might have made earlier. 

This question refers to the activity on page 54, "Try making your own number system." While no system has been made, we can outline how one would refine a simple, self-created number system using the evolutionary steps described in the chapter.

Let's assume one initially created a simple tally system, like Method 1 (using sticks or marks). The refinement process would follow these steps:

1. Introduce Grouping and Landmark Numbers: Instead of an endless series of tally marks, introduce a new symbol for a group of a certain size, similar to how 'V' was used for 5 in the Roman system. For example, a new symbol could represent a group of 5 tallies. This makes larger numbers easier to read.
2. Establish a Consistent Base: Instead of having landmark numbers at irregular intervals (like the Roman 1, 5, 10, 50), choose a single number as a base (e.g., base-5 or base-10). Create new symbols only for powers of that base. For a base-5 system, you would have symbols for 1, 5, 25, 125, and so on. This is the key idea from the Egyptian system.
3. Implement a Place Value System: To avoid needing an infinite number of symbols for the powers of your base, adopt the idea of place value from the Mesopotamian and Hindu systems. Now, the position of a numeral determines its value. In our base-5 system, the rightmost position would be for units (1s), the position to its left for 5s, the next for 25s, and so on. You would only need numerals for 1 through 4.
4. Invent a Symbol for Zero: A place value system is ambiguous without a way to show an "empty" position. The final and most crucial refinement is to invent a symbol for zero to act as a placeholder, as perfected in the Hindu system. In our base-5 system, this symbol would allow us to distinguish between 5 (10_5), 25 (100_5), and 6 (11_5). This makes the system capable of representing any number unambiguously with a finite set of symbols and enables efficient, modern arithmetic.