Number Base Conversion and Operations for SS 1

Number Base System (1)

Performance Objectives

By the end of the lesson, students should be able to:

1. convert from any base to base 10;
2. convert fractions in any base to base 10;
3. convert from one base to another;
4. solve number bases problems involving addition, subtraction, multiplication and division.

Introduction to Number Bases

A number base system is a mathematical way of representing numbers using a specific set of unique digits.

The most common system used in daily life is the decimal system (base 10), which uses ten digits from $0$ to $9$. However, other bases exist and are widely applied in fields such as computer science and digital electronics. For instance, computers use the binary system (base 2), which consists of only two digits: $0$ and $1$.

In any base $n$, the digits allowed range from $0$ to $n-1$. The base itself is never used as a single digit in that base system.

The allowed digits for some common bases are:

1. Base 2 (Binary): $0, 1$
2. Base 3 (Ternary): $0, 1, 2$
3. Base 5 (Quinary): $0, 1, 2, 3, 4$
4. Base 8 (Octal): $0, 1, 2, 3, 4, 5, 6, 7$
5. Base 10 (Decimal): $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$

Conversion from Any Base to Base 10

To convert a number from any base $n$ to base 10, the expansion method is used. This method relies on the place value of each digit in the number. Each digit is multiplied by the base raised to the power of its position index, starting from index $0$ for the rightmost digit (units place) and increasing by $1$ as one moves to the left.

Multiplication is used because each position in a non-decimal base represents a group of a specific power of that base. Multiplying the digit by its positional weight determines its equivalent value in the decimal (base 10) system.

Worked Example 1

Convert $134_5$ to base 10.

Solution:

Assign positional weights to the digits of $134_5$ from right to left, starting from $0$:

The digit $4$ is in position $0$.

The digit $3$ is in position $1$.

The digit $1$ is in position $2$.

Expand the number by multiplying each digit by $5$ raised to its positional power:

$$134_5 = (1 \times 5^2) + (3 \times 5^1) + (4 \times 5^0)$$

Simplify the powers of the base:

$$134_5 = (1 \times 25) + (3 \times 5) + (4 \times 1)$$

Calculate the products and sum them up:

$$134_5 = 25 + 15 + 4$$
$$134_5 = 44_{10}$$

Worked Example 2

Convert $101101_2$ to base 10.

Solution:

Assign positional weights to the digits of $101101_2$ from right to left, starting from $0$:

$$101101_2 = (1 \times 2^5) + (0 \times 2^4) + (1 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (1 \times 2^0)$$

Simplify the powers of the base:

$$101101_2 = (1 \times 32) + (0 \times 16) + (1 \times 8) + (1 \times 4) + (0 \times 2) + (1 \times 1)$$

Calculate the products and sum them up:

$$101101_2 = 32 + 0 + 8 + 4 + 0 + 1$$
$$101101_2 = 45_{10}$$

Conversion of Fractions in Any Base to Base 10

When a number contains a fractional part (digits to the right of the radix point), the place values continue to the right of the point with negative powers of the base. The first digit to the right of the point has a positional index of $-1$, the second has $-2$, the third has $-3$, and so on.

Worked Example 1

Convert $11.011_2$ to base 10.

Solution:

Assign positional weights to both the integer and fractional parts:

$$11.011_2 = (1 \times 2^1) + (1 \times 2^0) + (0 \times 2^{-1}) + (1 \times 2^{-2}) + (1 \times 2^{-3})$$

Convert the negative powers into fractions:

$$11.011_2 = (1 \times 2) + (1 \times 1) + \left(0 \times \frac{1}{2}\right) + \left(1 \times \frac{1}{4}\right) + \left(1 \times \frac{1}{8}\right)$$

Convert the fractions to decimals:

$$11.011_2 = 2 + 1 + 0 + 0.25 + 0.125$$

Sum the values:

$$11.011_2 = 3.375_{10}$$

Worked Example 2

Convert $23.12_4$ to base 10.

Solution:

Assign positional weights to both the integer and fractional parts:

$$23.12_4 = (2 \times 4^1) + (3 \times 4^0) + (1 \times 4^{-1}) + (2 \times 4^{-2})$$

Convert the negative powers into fractions:

$$23.12_4 = (2 \times 4) + (3 \times 1) + \left(1 \times \frac{1}{4}\right) + \left(2 \times \frac{1}{16}\right)$$

Convert the fractions to decimals:

$$23.12_4 = 8 + 3 + 0.25 + 0.125$$

Sum the values:

$$23.12_4 = 11.375_{10}$$

Conversion of Numbers from One Base to Another Base

To convert a number from a non-decimal base (Base A) to another non-decimal base (Base B), a two-step process is followed:

1. Convert the number from Base A to Base 10 using the expansion method.
2. Convert the resulting Base 10 number to Base B using successive division.

Successive division is used because it groups the decimal value into powers of the new base. The remainders generated at each step of the division represent the digits of the new base, read from the bottom (most significant digit) to the top (least significant digit).

Worked Example 1

Convert $324_5$ to base 8.

Solution:

Step 1: Convert $324_5$ to base 10 using expansion.

$$324_5 = (3 \times 5^2) + (2 \times 5^1) + (4 \times 5^0)$$
$$324_5 = (3 \times 25) + (2 \times 5) + (4 \times 1)$$
$$324_5 = 75 + 10 + 4$$
$$324_5 = 89_{10}$$

Step 2: Convert $89_{10}$ to base 8 using successive division.

Divide $89$ by $8$ repeatedly and record the remainders:

$$89 \div 8 = 11 \text{ remainder } 1$$
$$11 \div 8 = 1 \text{ remainder } 3$$
$$1 \div 8 = 0 \text{ remainder } 1$$

Read the remainders from bottom to top to get the final base 8 value:

$$131_8$$

Therefore, $324_5 = 131_8$.

Worked Example 2

Convert $10111_2$ to base 5.

Solution:

Step 1: Convert $10111_2$ to base 10 using expansion.

$$10111_2 = (1 \times 2^4) + (0 \times 2^3) + (1 \times 2^2) + (1 \times 2^1) + (1 \times 2^0)$$
$$10111_2 = (1 \times 16) + (0 \times 8) + (1 \times 4) + (1 \times 2) + (1 \times 1)$$
$$10111_2 = 16 + 0 + 4 + 2 + 1$$
$$10111_2 = 23_{10}$$

Step 2: Convert $23_{10}$ to base 5 using successive division.

Divide $23$ by $5$ repeatedly and record the remainders:

$$23 \div 5 = 4 \text{ remainder } 3$$
$$4 \div 5 = 0 \text{ remainder } 4$$

Read the remainders from bottom to top to get the final base 5 value:

$$43_5$$

Therefore, $10111_2 = 43_5$.

Arithmetic Operations on Number Bases

Addition of Number Bases

Addition in other bases follows the same vertical alignment and carrying principles as base 10. However, when the sum of the digits in a column is equal to or greater than the base $n$, the sum must be divided by $n$. The quotient is carried over to the next column to the left, and the remainder is written down in the current column.

Worked Example 1

Calculate $325_6 + 423_6$.

Solution:

Set up the addition vertically:

$$\begin{array}{r@{\quad}l}
325_6 \\
+\quad 423_6 \\
\hline
\end{array}$$

Perform the addition column by column from right to left:

Column 1 (Units): $5 + 3 = 8$. Since $8$ is greater than the base $6$, divide $8$ by $6$. This gives $1$ remainder $2$. Write down $2$ and carry $1$ to the next column.

Column 2: $2 + 2 + 1 \text{ (carry)} = 5$. Since $5$ is less than the base $6$, write down $5$ and carry $0$.

Column 3: $3 + 4 + 0 \text{ (carry)} = 7$. Since $7$ is greater than the base $6$, divide $7$ by $6$. This gives $1$ remainder $1$. Write down $1$ and carry $1$ to the next column.

Column 4: Write down the carried $1$.

The final sum is:

$$1152_6$$

Worked Example 2

Calculate $11011_2 + 1011_2$.

Solution:

Set up the addition vertically:

$$\begin{array}{r@{\quad}l}
11011_2 \\
+\quad 1011_2 \\
\hline
\end{array}$$

Perform the addition column by column from right to left:

Column 1: $1 + 1 = 2$. Divide $2$ by the base $2$, which gives $1$ remainder $0$. Write down $0$ and carry $1$.

Column 2: $1 + 1 + 1 \text{ (carry)} = 3$. Divide $3$ by the base $2$, which gives $1$ remainder $1$. Write down $1$ and carry $1$.

Column 3: $0 + 0 + 1 \text{ (carry)} = 1$. Write down $1$ and carry $0$.

Column 4: $1 + 1 + 0 \text{ (carry)} = 2$. Divide $2$ by the base $2$, which gives $1$ remainder $0$. Write down $0$ and carry $1$.

Column 5: $1 + 1 \text{ (carry)} = 2$. Divide $2$ by the base $2$, which gives $1$ remainder $0$. Write down $0$ and carry $1$.

Column 6: Write down the carried $1$.

The final sum is:

$$100110_2$$

Subtraction of Number Bases

Subtraction is performed vertically. When a digit in the minuend (top number) is smaller than the corresponding digit in the subtrahend (bottom number), borrowing is required from the next column to the left. In base $n$, borrowing $1$ from the left column transfers a value of $n$ (not 10) to the current column.

Worked Example 1

Calculate $465_7 – 352_7$.

Solution:

Set up the subtraction vertically:

$$\begin{array}{r@{\quad}l}
465_7 \\
-\quad 352_7 \\
\hline
\end{array}$$

Perform the subtraction column by column from right to left:

Column 1: $5 – 2 = 3$. Write down $3$.

Column 2: $6 – 5 = 1$. Write down $1$.

Column 3: $4 – 3 = 1$. Write down $1$.

The final difference is:

$$113_7$$

Worked Example 2

Calculate $312_4 – 123_4$.

Solution:

Set up the subtraction vertically:

$$\begin{array}{r@{\quad}l}
312_4 \\
-\quad 123_4 \\
\hline
\end{array}$$

Perform the subtraction column by column from right to left:

Column 1: $2 – 3$ is impossible. Borrow $1$ from the next column to the left. In base 4, this borrow adds $4$ to the current column, making it $2 + 4 = 6$. Now, calculate $6 – 3 = 3$. Write down $3$.

Column 2: The digit $1$ has been reduced to $0$ due to borrowing. Since $0 – 2$ is impossible, borrow $1$ from the next column to the left. This adds $4$ to the current column, making it $0 + 4 = 4$. Now, calculate $4 – 2 = 2$. Write down $2$.

Column 3: The digit $3$ has been reduced to $2$ due to borrowing. Calculate $2 – 1 = 1$. Write down $1$.

The final difference is:

$$123_4$$

Multiplication of Number Bases

Multiplication can be performed directly in the given base or by converting the numbers to base 10, multiplying, and converting the product back to the original base. Direct multiplication is performed by multiplying digits and converting any product equal to or greater than the base into the base system (carrying the quotient and writing the remainder).

Worked Example 1

Calculate $110_3 \times 11_3$.

Solution:

Set up the multiplication vertically:

$$\begin{array}{r@{\quad}l}
110_3 \\
\times\quad 11_3 \\
\hline
\end{array}$$

Multiply $110_3$ by the units digit $1_3$ of the multiplier:

$$1 \times 110_3 = 110_3$$

Multiply $110_3$ by the next digit $1_3$ of the multiplier, shifting the result one place to the left:

$$1 \times 110_3 = 1100_3 \text{ (with placeholder zero)}$$

Add the partial products vertically in base 3:

$$\begin{array}{r@{\quad}l}
110_3 \\
+\quad 1100_3 \\
\hline
\end{array}$$

Column 1: $0 + 0 = 0$.

Column 2: $1 + 0 = 1$.

Column 3: $1 + 1 = 2$. Since $2$ is a valid digit in base 3, write down $2$.

Column 4: Write down $1$.

The final product is:

$$1210_3$$

Worked Example 2

Calculate $23_5 \times 14_5$.

Solution:

Set up the multiplication vertically:

$$\begin{array}{r@{\quad}l}
23_5 \\
\times\quad 14_5 \\
\hline
\end{array}$$

Multiply $23_5$ by the units digit $4_5$ of the multiplier:

$4 \times 3 = 12$. In base 5, $12 \div 5 = 2$ remainder $2$. Write down $2$ and carry $2$.

$4 \times 2 = 8$. Add the carry: $8 + 2 = 10$. In base 5, $10 \div 5 = 2$ remainder $0$. Write down $0$ and carry $2$.

Write down the carried $2$. The first partial product is $202_5$.

Multiply $23_5$ by the next digit $1_5$ of the multiplier, shifting the result one place to the left:

$$1 \times 23_5 = 230_5 \text{ (with placeholder zero)}$$

Add the partial products vertically in base 5:

$$\begin{array}{r@{\quad}l}
202_5 \\
+\quad 230_5 \\
\hline
\end{array}$$

Column 1: $2 + 0 = 2$.

Column 2: $0 + 3 = 3$.

Column 3: $2 + 2 = 4$. Since $4$ is a valid digit in base 5, write down $4$.

The final product is:

$$432_5$$

Division of Number Bases

Division in non-decimal bases can be performed using long division directly in that base, or by converting the dividend and divisor to base 10, performing the division in base 10, and then converting the quotient back to the original base. The conversion method is highly reliable and less prone to errors.

Worked Example 1

Divide $20232_6$ by $4_6$.

Solution:

Step 1: Convert the dividend $20232_6$ to base 10.

$$20232_6 = (2 \times 6^4) + (0 \times 6^3) + (2 \times 6^2) + (3 \times 6^1) + (2 \times 6^0)$$
$$20232_6 = (2 \times 1296) + 0 + (2 \times 36) + (3 \times 6) + (2 \times 1)$$
$$20232_6 = 2592 + 0 + 72 + 18 + 2$$
$$20232_6 = 2684_{10}$$

Step 2: Convert the divisor $4_6$ to base 10.

$$4_6 = 4_{10}$$

Step 3: Divide the values in base 10.

$$2684_{10} \div 4_{10} = 671_{10}$$

Step 4: Convert the quotient $671_{10}$ back to base 6 using successive division.

Divide $671$ by $6$ repeatedly and record the remainders:

$$671 \div 6 = 111 \text{ remainder } 5$$
$$111 \div 6 = 18 \text{ remainder } 3$$
$$18 \div 6 = 3 \text{ remainder } 0$$
$$3 \div 6 = 0 \text{ remainder } 3$$

Read the remainders from bottom to top to get the final base 6 value:

$$3035_6$$

Therefore, $20232_6 \div 4_6 = 3035_6$.

Worked Example 2

Divide $110110_2$ by $11_2$.

Solution:

Step 1: Convert the dividend $110110_2$ to base 10.

$$110110_2 = (1 \times 2^5) + (1 \times 2^4) + (0 \times 2^3) + (1 \times 2^2) + (1 \times 2^1) + (0 \times 2^0)$$
$$110110_2 = 32 + 16 + 0 + 4 + 2 + 0$$
$$110110_2 = 54_{10}$$

Step 2: Convert the divisor $11_2$ to base 10.

$$11_2 = (1 \times 2^1) + (1 \times 2^0) = 2 + 1 = 3_{10}$$

Step 3: Divide the values in base 10.

$$54_{10} \div 3_{10} = 18_{10}$$

Step 4: Convert the quotient $18_{10}$ back to base 2 using successive division.

Divide $18$ by $2$ repeatedly and record the remainders:

$$18 \div 2 = 9 \text{ remainder } 0$$
$$9 \div 2 = 4 \text{ remainder } 1$$
$$4 \div 2 = 2 \text{ remainder } 0$$
$$2 \div 2 = 1 \text{ remainder } 0$$
$$1 \div 2 = 0 \text{ remainder } 1$$

Read the remainders from bottom to top to get the final base 2 value:

$$10010_2$$

Therefore, $110110_2 \div 11_2 = 10010_2$.