All Exams Test series for 1 year @ ₹349 only
Question

Multiplication of two binary numbers 11.01 and 10.11 is:

The correct answer is 1000.1111

Binary Multiplication Fundamentals

Binary multiplication is a fundamental operation in digital electronics and computer arithmetic. It is similar to decimal multiplication but uses only two digits: 0 and 1. When performing binary multiplication of numbers with fractional parts (like 11.01 and 10.11), we treat them initially as whole numbers and then place the decimal point (or radix point) in the final product based on the total number of fractional bits in the original operands.

Let's break down the process of multiplying the two binary numbers: 11.01 and 10.11.

Step-by-Step Binary Multiplication

To multiply the binary numbers \(11.01\) and \(10.11\), we follow these steps:

  1. Remove the Decimal Points: First, remove the decimal points from both binary numbers and treat them as integers.
    • \(11.01 \rightarrow 1101\)
    • \(10.11 \rightarrow 1011\)
  2. Count Decimal Places: Count the total number of digits after the decimal point in both original numbers.
    • For \(11.01\), there are 2 digits after the decimal point (\(01\)).
    • For \(10.11\), there are 2 digits after the decimal point (\(11\)).
    • Total decimal places \( = 2 + 2 = 4\). This means our final product will have 4 digits after the decimal point.
  3. Perform Integer Multiplication: Multiply the integer equivalents of the binary numbers (\(1101\) and \(1011\)) using standard binary multiplication rules.
    • When multiplying by '1', the result is the multiplicand itself.
    • When multiplying by '0', the result is all zeros.
    • Shift each subsequent partial product to the left, just like in decimal multiplication.
    \[ \begin{array}{rcccccccc} & & & & 1 & 1 & 0 & 1 & \text{ (Multiplicand: 11.01)} \\ \times & & & & 1 & 0 & 1 & 1 & \text{ (Multiplier: 10.11)} \\ \cline{2-9} & & & & 1 & 1 & 0 & 1 & \text{ (1101 \(\times\) 1)} \\ & & & 1 & 1 & 0 & 1 & 0 & \text{ (1101 \(\times\) 1, shifted left by 1)} \\ & & 0 & 0 & 0 & 0 & 0 & 0 & \text{ (1101 \(\times\) 0, shifted left by 2)} \\ + & 1 & 1 & 0 & 1 & 0 & 0 & 0 & \text{ (1101 \(\times\) 1, shifted left by 3)} \\ \cline{2-9} & 1 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ \end{array} \]

    Now, let's sum the partial products:

    \[ \begin{array}{r} 1101 \\ 11010 \\ 000000 \\ +\ 1101000 \\ \cline{1-1} 10001111 \\ \end{array} \]

    The sum of the partial products is \(10001111\).

  4. Place the Decimal Point: Insert the decimal point 4 places from the right (as determined in step 2) into the final binary product.
    • The integer product is \(10001111\).
    • Counting 4 places from the right gives us \(1000.1111\).

Final Binary Product

Therefore, the multiplication of \(11.01\) and \(10.11\) in binary results in \(1000.1111\). This process demonstrates the systematic approach to binary arithmetic, which is a core concept for understanding how computers perform calculations. Mastering binary number multiplication is crucial for students studying digital systems.

The result aligns with the understanding of binary numbers and their operations.

Was this answer helpful?

Important Questions from Types of Number System

  1. The logic XOR operation of (4AC0) 16 and (B53F) 16 results________

  2. What is the Hexadecimal equivalent of 160?

  3. Find out the decimal equivalent of (1011100) 2.

  4. (1101 0001) 2 binary number is same as (   ) 8 octal number.
  5. The MSB of a binary number has a weight of 512. The number consists of ______.
Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App