Question 6

We will first understand the Gumulgal number system and then define how to perform arithmetic operations using its unique terms.

Step 1 — Understanding the Gumulgal System

The Gumulgal system uses specific words for numbers. 'urapon' (ur) represents the number one. 'ukasar' (uk) represents the number two. The system extends by counting in twos. This means two 'urapon's make one 'ukasar'. So, ur + ur = uk. Numbers are formed by combining 'ukasar's and adding an 'urapon' if needed. For example, three is 'ukasar-urapon' (uk-ur). Four is 'ukasar-ukasar' (uk-uk). Five is 'ukasar-ukasar-urapon' (uk-uk-ur). Six is 'ukasar-ukasar-ukasar' (uk-uk-uk). We will continue this pattern for larger numbers.

Step 2 — Performing Addition

To add numbers, we combine all 'ukasar' and 'urapon' terms. Then, we replace every two 'urapon's with one 'ukasar'.

Let us evaluate (ukasar-ukasar-ukasar-ukasar-urapon) + (ukasar-ukasar-ukasar-urapon). The first number is uk-uk-uk-uk-ur. The second number is uk-uk-uk-ur.

We combine all terms: $(\text{uk-uk-uk-uk-ur}) + (\text{uk-uk-uk-ur})$

$= \text{uk-uk-uk-uk-uk-uk-uk-ur-ur}$

Now, we use the rule that ur + ur = uk. We replace the ur-ur with uk.

$= \text{uk-uk-uk-uk-uk-uk-uk-uk}$

This is eight 'ukasar' terms.

$\boxed{\text{ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar}}$

Step 3 — Performing Subtraction

To subtract numbers, we remove matching 'ukasar' and 'urapon' terms. If we need to subtract an 'urapon' but only have 'ukasar's, we can change one 'ukasar' into two 'urapon's.

Let us evaluate (ukasar-ukasar-ukasar-ukasar-urapon) − (ukasar-ukasar-ukasar-ukasar). The first number is uk-uk-uk-uk-ur. The second number is uk-uk-uk-uk.

We remove the matching 'ukasar' terms from both numbers. We have four 'ukasar's in the first number. We also have four 'ukasar's in the second number.

$(\text{uk-uk-uk-uk-ur}) - (\text{uk-uk-uk-uk})$

We remove the four 'ukasar's from both parts. This leaves only ur from the first number.

$= \text{ur}$

The problem asks for the answer in a specific form. We know that ur represents one. We also know that uk represents two. So, uk - ur means $2 - 1$, which is also one. Thus, ur is the same as uk - ur.

$\boxed{\text{ukasar-urapon}}$

Step 4 — Performing Multiplication

To multiply, we can think of it as repeated addition. Multiplying by 'ukasar' (two) means doubling the number. Multiplying by 'ukasar-ukasar' (four) means doubling the number twice.

Let us evaluate (ukasar-ukasar-ukasar-ukasar-urapon) × (ukasar-ukasar). The first number is uk-uk-uk-uk-ur. The second number is uk-uk. We can write uk-uk as uk + uk.

So, we need to calculate (uk-uk-uk-uk-ur) × (uk + uk). This is the same as (uk-uk-uk-uk-ur) × uk + (uk-uk-uk-uk-ur) × uk.

First, let us multiply (uk-uk-uk-uk-ur) by uk. When we multiply by uk (two), each uk becomes uk-uk. Also, ur becomes uk.

$(\text{uk-uk-uk-uk-ur}) \times \text{uk}$

$= (\text{uk-uk}) - (\text{uk-uk}) - (\text{uk-uk}) - (\text{uk-uk}) - \text{uk}$

$= \text{uk-uk-uk-uk-uk-uk-uk-uk-uk}$

This is nine 'ukasar' terms. Now, we add this result to itself.

$(\text{uk-uk-uk-uk-uk-uk-uk-uk-uk}) + (\text{uk-uk-uk-uk-uk-uk-uk-uk-uk})$

We combine all the 'ukasar' terms. There are nine 'ukasar's in each part. So, we have $9 + 9 = 18$ 'ukasar' terms in total.

$= \text{uk-uk-uk-uk-uk-uk-uk-uk-uk-uk-uk-uk-uk-uk-uk-uk-uk-uk}$

This is eighteen 'ukasar' terms.

$\boxed{\text{ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar}}$

Step 5 — Performing Division

To divide, we find out how many times the divisor fits into the dividend. We can group the dividend into blocks that match the divisor.

Let us evaluate (ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar) ÷ (ukasar-ukasar). The first number is uk-uk-uk-uk-uk-uk-uk-uk. The second number is uk-uk.

We want to see how many groups of uk-uk are in uk-uk-uk-uk-uk-uk-uk-uk. We can rewrite the first number by grouping its 'ukasar's into pairs.

$(\text{uk-uk-uk-uk-uk-uk-uk-uk}) \div (\text{uk-uk})$

$= [(\text{uk-uk}) - (\text{uk-uk}) - (\text{uk-uk}) - (\text{uk-uk})] \div (\text{uk-uk})$

We see that the first number contains four groups of uk-uk. Each group (uk-uk) divided by (uk-uk) gives ur (one). So, we have ur + ur + ur + ur.

$= \text{ur + ur + ur + ur}$

Now, we use the rule ur + ur = uk. So, ur + ur becomes uk. We have two such pairs.

$= \text{uk + uk}$

$= \text{uk-uk}$

This is two 'ukasar' terms.

$\boxed{\text{ukasar-ukasar}}$

Answer

(i) ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar (ii) ukasar-urapon (iii) ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar (iv) ukasar-ukasar