Arithmetic Operations in Non-Decimal Bases
Addition, multiplication, and subtraction of two numbers written in the same non-decimal base are performed similarly to operations in base 10. However, two critical principles must be taken into account:
- First, during addition and multiplication, whenever a positional value reaches or exceeds the base value, the multiples of the base value are carried over as a carry to the next higher positional column.
- Second, during subtraction, when borrowing 1 from the next higher column on the left, its contribution to the current column is equal to the value of the base.
Examples:
- $(2)_6 + (3)_6=(5)_6$
- $(31)_6 + (14)_6=(45)_6$
- $(3)_6+(3)_6=(10)_6$
Explanation: $3+3 = 6$. Since the digit 6 cannot be written in base 6, we calculate how many times the base fits into 6. It fits exactly once, meaning we have a carry of 1. Subtracting this 1 multiple of 6 from our sum leaves a remainder of 0. This 0 is written in the units place. The carry of 1 is moved to the next higher column (the 6s place). Since there are no further digits to add, the 6s place contains just 1, giving the final result $(10)_6$. Let’s look at a multi-digit addition example:
\[ \begin{array}{r}
(3214)_6 \\
+\quad (235)_6 \\
\hline
(3453)_6
\end{array} \]
In the units column, adding 4 and 5 yields 9. Since 9 is greater than the base 6, we divide 9 by 6: 6 fits into 9 once, resulting in a carry of 1. Subtracting one instance of the base from 9 ($9 – 6$) leaves a remainder of 3, which is written down in the units place. The carry of 1 is added to the next column (the 6s place). In that column, $1 + 3 = 4$, and adding our carry of 1 results in 5. Since 5 is less than the base 6, it is written down as is. This process is systematically applied to the remaining columns from right to left.
Multiplication
\[ \begin{array}{r}
(343)_5 \\
\times\quad (34)_5 \\
\hline
(3032)_5 \\
+\quad (2134)_5\phantom{0} \\
\hline
(24422)_5
\end{array} \]
Explanation: First, multiplying $4 \times 3 = 12 = (22)_5$. The unit digit 2 is written down, and the 2 in the 5s place is kept as a carry. Next, $4 \times 4 = 16$; adding our carry of 2 gives $18 = (33)_5$. The unit digit 3 is written down, and 3 is kept as a carry. Then, $4 \times 3 = 12$; adding our carry of 3 yields $15 = (30)_5$. The 0 is recorded, and the 3 is carried down to the next position, completing the first partial product: $(343)_5 \times (4)_5 = (3032)_5$.
Following the same method for the next digit:
\[ (343)_5 \times (3)_5 = (2134)_5 \]
Just as in base 10, the second partial product is shifted one column to the left. Adding these two partial products together yields the final multiplication result: $(24422)_5$.
Let’s look at two more examples:
\[ \begin{array}{c@{\qquad}c}
\begin{array}{r}
(101101)_2 \\
\times\quad (101)_2 \\
\hline
(101101)_2 \\
+\quad (101101)_2\phantom{00} \\
\hline
(11100001)_2
\end{array}
&
\begin{array}{r}
(3454)_6 \\
\times\quad (345)_6 \\
\hline
(31042)_6 \\
(23144)_6\phantom{0} \\
+\quad (15250)_6\phantom{00} \\
\hline
(2231522)_6
\end{array}
\end{array}
\]
Subtraction
\[ \begin{array}{r}
(30324)_5 \\
-\quad (12343)_5 \\
\hline
(12431)_5
\end{array} \]
Explanation: In the units column, $4 – 3 = 1$, which is written down. In the 5s column, we cannot subtract 4 from 2, so we borrow 1 from the 25s column to the left. Since its value equals the base (5), it is added to our current digit: $2 + 5 = 7$. Now, $7 – 4 = 3$, which is written down in the 5s place. Due to the borrow, the digit in the 25s column is reduced from 3 to 2. Since we cannot subtract 3 from 2, we must borrow from the left. The adjacent 125s column contains a 0, so we move to the next column (625s place) to borrow 1. This 1 adds 5 to the 125s column, turning the 0 into a 5. We then borrow 1 from this 5 (leaving 4 behind) and transfer it to our 25s column, where it contributes 5. Adding this to our remaining 2 yields $2 + 5 = 7$. Now, $7 – 3 = 4$, which is written down in the 25s column. Moving left, the remaining 4 in the 125s column minus 2 yields $4 – 2 = 2$. Finally, in the leftmost 625s column, the 3 was reduced to 2 by the original borrow chain; $2 – 1 = 1$ is recorded, completing the result: $(12431)_5$.
\[
\begin{array}{c@{\qquad}c@{\qquad}c}
\begin{array}{r}
(6543)_7 \\
-\quad (2343)_7 \\
\hline
(4200)_7
\end{array}
&
\begin{array}{r}
(1234)_5 \\
-\quad (321)_5 \\
\hline
(0413)_5
\end{array}
&
\begin{arraypre}
\begin{array}{r}
(53\text{C}47)_{15} \\
-\quad (4\text{A}35)_{15} \\
\hline
(4E212)_{15}
\end{array}
\end{arraypre}
\end{array}
\]
Division
Division within non-decimal bases can be evaluated directly from left to right using estimation based on the base-10 equivalent values of the divisor and dividend, where the subsequent multiplication and subtraction steps are carried out within the designated base system.
Alternatively, both numbers can be converted into base 10 to perform standard division. The resulting quotient and remainder are then converted back into the target base.
Example:
Let’s solve the division problem: $(41)_5 \div (12)_5$
\[
\begin{array}{l}
\textbf{Step 1: Convert Numbers to Base 10}\\
(41)_5 = 4 \times 5^1 + 1 \times 5^0 = 21_{10}\\
(12)_5 = 1 \times 5^1 + 2 \times 5^0 = 7_{10}\\
\\
\textbf{Step 2: Perform Division in Base 10}\\
21 \div 7 = 3_{10}\\
\\
\textbf{Step 3: Convert the Result back to Base 5}\\
\text{Since the integer value 3 is represented by the same digit in both base 10 and base 5, no conversion is needed.}\\
\text{Verification: } (12)_5 \times (3)_5 = (41)_5
\end{array}
\]
Example:
Let’s solve the division problem: $(114332)_5 \div (23)_5$
First, convert the numbers into base 10:
Divisor:
\[ (23)_5 = 2 \times 5^1 + 3 \times 5^0 = 10 + 3 = 13_{10} \]
Dividend:
\[ (114332)_5 = 1 \times 5^5 + 1 \times 5^4 + 4 \times 5^3 + 3 \times 5^2 + 3 \times 5^1 + 2 \times 5^0 \]
\[ = 3125 + 625 + 500 + 75 + 15 + 2 = 4342_{10} \]
\[
\begin{array}{l}
\textbf{Step 2: Perform Division in Base 10}\\
4342 \div 13 = 334_{10}\\
\\
\textbf{Step 3: Convert the Result back to Base 5}\\
334_{10} = 2 \times 5^3 + 3 \times 5^2 + 1 \times 5^1 + 4 \times 5^0 = (2314)_5
\end{array}
\]