Multiplication of Binary Numbers for JSS 1

Multiplication of Numbers in Base 2

Performance Objectives

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

1. multiply two-or three-digits binary numbers.

Introduction to Binary Multiplication

Binary multiplication is the mathematical process of finding the product of two numbers written in base 2.

The binary system uses only two digits: $0$ and $1$. Because of this, multiplying binary numbers is simpler than multiplying base 10 (decimal) numbers. There are no large multiplication tables to memorize. The process relies on a few basic rules and the method of long multiplication.

The Rules of Binary Multiplication

The fundamental rules for multiplying single binary digits are:

1. $0_2 \times 0_2 = 0_2$
2. $0_2 \times 1_2 = 0_2$
3. $1_2 \times 0_2 = 0_2$
4. $1_2 \times 1_2 = 1_2$

Binary Addition Rules for Long Multiplication

When multiplying numbers with two or three digits, partial products are written down and then added together. Therefore, performing binary multiplication requires a clear understanding of binary addition.

The basic rules of binary addition are:

1. $0_2 + 0_2 = 0_2$
2. $0_2 + 1_2 = 1_2$
3. $1_2 + 0_2 = 1_2$
4. $1_2 + 1_2 = 10_2$ (write down $0$ and carry $1$ to the next column on the left)
5. $1_2 + 1_2 + 1_2 = 11_2$ (write down $1$ and carry $1$ to the next column on the left)

Using Base 2 Blocks for Multiplication

Base 2 blocks, also known as binary blocks, are concrete mathematical tools used to represent powers of 2. These blocks help to visualize how binary numbers are built and multiplied.

The types of base 2 blocks include:

1. Unit blocks: Small single cubes representing $2^0$ (which equals $1$).
2. Long blocks: Rectangular blocks made of $2$ unit blocks joined together, representing $2^1$ (which equals $2$).
3. Flat blocks: Square blocks made of $4$ unit blocks ($2 \times 2$), representing $2^2$ (which equals $4$).

To multiply binary numbers using blocks, an area model is constructed. The length represents one binary number, and the width represents the other. The total area covered by the blocks represents the product.

Multiplying Two-Digit Binary Numbers

Multiplying two-digit binary numbers involves multiplying the multiplicand by each digit of the multiplier, shifting the second partial product one place to the left, and then adding the results.

Worked Example 1: Multiply $10_2$ by $11_2$.

Solution:

Set up the numbers vertically:

$$
\begin{array}{cccc}
& & 1 & 0_2 \\
\times & & 1 & 1_2 \\
\hline
\end{array}
$$

First, multiply $10_2$ by the rightmost digit ($1$) of the multiplier:

$$
10_2 \times 1 = 10_2
$$

Next, multiply $10_2$ by the leftmost digit ($1$) of the multiplier, shifting the result one place to the left:

$$
10_2 \times 10_2 = 100_2
$$

Now, write these partial products down and add them:

$$
\begin{array}{cccc}
& & 1 & 0 \\
\times & & 1 & 1 \\
\hline
& & 1 & 0 \\
& 1 & 0 & 0 \\
\hline
& 1 & 1 & 0_2 \\
\hline
\end{array}
$$

Therefore, $10_2 \times 11_2 = 110_2$.

Worked Example 2: Multiply $11_2$ by $11_2$.

Solution:

Set up the vertical multiplication:

$$
\begin{array}{cccc}
& & 1 & 1_2 \\
\times & & 1 & 1_2 \\
\hline
\end{array}
$$

Multiply $11_2$ by the rightmost $1$ to get the first partial product:

$$
11_2 \times 1 = 11_2
$$

Multiply $11_2$ by the leftmost $1$ and shift one place to the left to get the second partial product:

$$
11_2 \times 10_2 = 110_2
$$

Add the two partial products:

$$
\begin{array}{cccc}
& & 1 & 1 \\
\times & & 1 & 1 \\
\hline
& & 1 & 1 \\
& 1 & 1 & 0 \\
\hline
1 & 0 & 0 & 1_2 \\
\hline
\end{array}
$$

The addition steps for the columns from right to left are:

1. First column (right): $1 + 0 = 1$.
2. Second column: $1 + 1 = 10_2$. Write down $0$ and carry $1$ to the next column.
3. Third column: $1$ (carried over) $+ 1 = 10_2$. Write down $0$ and carry $1$ to the next column.
4. Fourth column: Write down the carried $1$.

Therefore, $11_2 \times 11_2 = 1001_2$.

Multiplying Three-Digit Binary Numbers

Multiplying three-digit binary numbers follows the same process of long multiplication. The partial products are shifted progressively to the left before being added together.

Worked Example 3: Multiply $101_2$ by $11_2$.

Solution:

Set up the vertical multiplication:

$$
\begin{array}{ccccc}
& & 1 & 0 & 1_2 \\
\times & & & 1 & 1_2 \\
\hline
\end{array}
$$

Multiply $101_2$ by the rightmost $1$ of the multiplier:

$$
101_2 \times 1 = 101_2
$$

Multiply $101_2$ by the leftmost $1$ of the multiplier and shift one place to the left:

$$
101_2 \times 10_2 = 1010_2
$$

Add the partial products:

$$
\begin{array}{ccccc}
& & 1 & 0 & 1 \\
\times & & & 1 & 1 \\
\hline
& & 1 & 0 & 1 \\
& 1 & 0 & 1 & 0 \\
\hline
& 1 & 1 & 1 & 1_2 \\
\hline
\end{array}
$$

The addition steps for the columns from right to left are:

1. First column (right): $1 + 0 = 1$.
2. Second column: $0 + 1 = 1$.
3. Third column: $1 + 0 = 1$.
4. Fourth column: Write down $1$.

Therefore, $101_2 \times 11_2 = 1111_2$.

Worked Example 4: Multiply $110_2$ by $101_2$.

Solution:

Set up the vertical multiplication:

$$
\begin{array}{cccccc}
& & 1 & 1 & 0_2 \\
\times & & 1 & 0 & 1_2 \\
\hline
\end{array}
$$

Multiply $110_2$ by each digit of the multiplier $101_2$:

1. Multiply by the rightmost $1$: $110_2 \times 1 = 110_2$.
2. Multiply by the middle $0$: $110_2 \times 0 = 000_2$ (shifted one place left).
3. Multiply by the leftmost $1$: $110_2 \times 1 = 110_2$ (shifted two places left).

Write down the partial products and add them:

$$
\begin{array}{cccccc}
& & 1 & 1 & 0 \\
\times & & 1 & 0 & 1 \\
\hline
& & 1 & 1 & 0 \\
& 0 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 & 0 \\
\hline
1 & 1 & 1 & 1 & 0_2 \\
\hline
\end{array}
$$

The addition steps for the columns from right to left are:

1. First column (right): $0 + 0 + 0 = 0$.
2. Second column: $1 + 0 + 0 = 1$.
3. Third column: $1 + 0 + 0 = 1$.
4. Fourth column: $0 + 0 + 1 = 1$.
5. Fifth column: Write down $1$.

Therefore, $110_2 \times 101_2 = 11110_2$.

Worked Example 5: Multiply $111_2$ by $111_2$.

Solution:

Set up the vertical multiplication:

$$
\begin{array}{cccccc}
& & 1 & 1 & 1_2 \\
\times & & 1 & 1 & 1_2 \\
\hline
\end{array}
$$

Multiply $111_2$ by each digit of the multiplier $111_2$:

1. Multiply by the rightmost $1$: $111_2 \times 1 = 111_2$.
2. Multiply by the middle $1$: $111_2 \times 1 = 111_2$ (shifted one place left).
3. Multiply by the leftmost $1$: $111_2 \times 1 = 111_2$ (shifted two places left).

Write down the partial products and add them:

$$
\begin{array}{cccccc}
& & 1 & 1 & 1 \\
& & 1 & 1 & 1 \\
\times & & 1 & 1 & 1 \\
\hline
& & 1 & 1 & 1 \\
& 1 & 1 & 1 & 0 \\
1 & 1 & 1 & 0 & 0 \\
\hline
1 & 1 & 0 & 0 & 0 & 1_2 \\
\hline
\end{array}
$$

The addition steps for the columns from right to left are:

1. First column (right): $1 + 0 + 0 = 1$.
2. Second column: $1 + 1 = 10_2$. Write down $0$ and carry $1$.
3. Third column: $1$ (carried over) $+ 1 + 1 + 1 = 100_2$. Write down $0$ and carry $10_2$ (which is $2$ in base 10) to the next column. This means carrying $0$ to the fourth column and $1$ to the fifth column.
4. Fourth column: $0$ (carried) $+ 1 + 1 = 10_2$. Write down $0$ and carry $1$ to the fifth column.
5. Fifth column: $1$ (carried from third column) $+ 1$ (carried from fourth column) $+ 1 = 11_2$. Write down $1$ and carry $1$ to the sixth column.
6. Sixth column: Write down the carried $1$.

Therefore, $111_2 \times 111_2 = 110001_2$.

Checking Answers by Converting to Base 10

A reliable way to verify binary multiplication is to convert the binary numbers to base 10 (decimal), perform the multiplication in base 10, and then convert the product back to base 2.

To verify the result of Worked Example 1 ($10_2 \times 11_2$):

1. Convert $10_2$ to base 10: $(1 \times 2^1) + (0 \times 2^0) = 2 + 0 = 2_{10}$.
2. Convert $11_2$ to base 10: $(1 \times 2^1) + (1 \times 2^0) = 2 + 1 = 3_{10}$.
3. Multiply the base 10 numbers: $2 \times 3 = 6_{10}$.
4. Convert $6_{10}$ back to base 2:
   $$
   \begin{array}{l|l|l}
   2 & 6 & \text{remainder } 0 \\
   2 & 3 & \text{remainder } 1 \\
   2 & 1 & \text{remainder } 1 \\
   & 0 &
   \end{array}
   $$
   Reading the remainders from bottom to top gives $110_2$.

The result matches the binary multiplication product of $110_2$.

Summary of Key Steps

The steps for multiplying binary numbers are:

1. Write the numbers vertically, aligning them to the right.
2. Multiply the multiplicand by each digit of the multiplier, starting from the rightmost digit.
3. Shift each subsequent partial product one place to the left.
4. Add the partial products using the rules of binary addition.
5. Verify the final answer by converting to base 10 if necessary.

Practice Exercises

Solve the following binary multiplication problems:

1. $10_2 \times 10_2$
2. $10_2 \times 11_2$
3. $11_2 \times 11_2$
4. $100_2 \times 10_2$
5. $101_2 \times 11_2$
6. $110_2 \times 11_2$
7. $111_2 \times 10_2$
8. $101_2 \times 101_2$
9. $110_2 \times 101_2$
10. $111_2 \times 111_2$

The answers and brief steps are:

1. $10_2 \times 10_2 = 100_2$ (In base 10: $2 \times 2 = 4$)
2. $10_2 \times 11_2 = 110_2$ (In base 10: $2 \times 3 = 6$)
3. $11_2 \times 11_2 = 1001_2$ (In base 10: $3 \times 3 = 9$)
4. $100_2 \times 10_2 = 1000_2$ (In base 10: $4 \times 2 = 8$)
5. $101_2 \times 11_2 = 1111_2$ (In base 10: $5 \times 3 = 15$)
6. $110_2 \times 11_2 = 10010_2$ (In base 10: $6 \times 3 = 18$)
7. $111_2 \times 10_2 = 1110_2$ (In base 10: $7 \times 2 = 14$)
8. $101_2 \times 101_2 = 11001_2$ (In base 10: $5 \times 5 = 25$)
9. $110_2 \times 101_2 = 11110_2$ (In base 10: $6 \times 5 = 30$)
10. $111_2 \times 111_2 = 110001_2$ (In base 10: $7 \times 7 = 49$)