Understanding Binary Subtraction

This question asks us to find the difference between two binary numbers, 10010000 and 1111001. Subtracting binary numbers is similar to subtracting decimal numbers, but instead of borrowing 10, we borrow 2.

Performing Binary Subtraction Step-by-Step

To subtract 1111001 from 10010000, we align the numbers vertically, padding the shorter number with leading zeros to match the length of the longer number (8 bits in this case).

Original numbers:

10010000
 1111001

Padded for subtraction:

10010000
01111001

Now, we subtract column by column from right to left, starting from the least significant bit.

Let's detail the borrowing process more clearly:

  0 1 1 1 0 1 1 (10)  <-- Effective digits after borrowing
  1 0 0 1 0 0 0 0
- 0 1 1 1 1 0 0 1
------------------
  0 0 0 1 0 1 1 1

Explanation of borrows:

• Starting from the rightmost column (position 0), we need to calculate \(0 - 1\). We must borrow. We look to the left. The first '1' is at position 4.
• We borrow 1 from position 4. The digit at position 4 becomes 0. This borrowed 1 becomes 2 (binary 10) at position 3.
• We borrow 1 from position 3 (which is now 2). It becomes 1. This borrowed 1 becomes 2 at position 2.
• We borrow 1 from position 2 (which is now 2). It becomes 1. This borrowed 1 becomes 2 at position 1.
• We borrow 1 from position 1 (which is now 2). It becomes 1. This borrowed 1 becomes 2 at position 0.
• Now, at position 0, we have \(2 - 1 = 1\). The result bit is 1.
• At position 1, we have \(1 - 0 = 1\). The result bit is 1.
• At position 2, we have \(1 - 0 = 1\). The result bit is 1.
• At position 3, we have \(1 - 1 = 0\). The result bit is 0.
• At position 4, the digit was 1, but became 0 due to the initial borrow. We need to calculate \(0 - 1\). We must borrow again.
• We look left. Positions 5 and 6 are 0. Position 7 is 1.
• We borrow 1 from position 7. It becomes 0. This borrowed 1 becomes 2 at position 6.
• We borrow 1 from position 6 (now 2). It becomes 1. This borrowed 1 becomes 2 at position 5.
• We borrow 1 from position 5 (now 2). It becomes 1. This borrowed 1 becomes 2 at position 4.
• Now, at position 4, we have \(2 - 1 = 1\). The result bit is 1.
• At position 5, we have \(1 - 1 = 0\). The result bit is 0.
• At position 6, we have \(1 - 1 = 0\). The result bit is 0.
• At position 7, the digit was 1, but became 0 due to borrowing. We have \(0 - 0 = 0\). The result bit is 0.

Combining the result bits from left to right gives us 00010111. Leading zeros can be dropped.

The difference is 10111.

Verification using Decimal Conversion

We can convert the binary numbers to decimal to verify our subtraction.

• Binary \(10010000_2\):
• \(1 \times 2^7 + 0 \times 2^6 + 0 \times 2^5 + 1 \times 2^4 + 0 \times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 0 \times 2^0\)
• \(= 1 \times 128 + 0 \times 64 + 0 \times 32 + 1 \times 16 + 0 \times 8 + 0 \times 4 + 0 \times 2 + 0 \times 1\)
• \(= 128 + 0 + 0 + 16 + 0 + 0 + 0 + 0 = 144_{10}\)
• Binary \(1111001_2\):
• \(1 \times 2^6 + 1 \times 2^5 + 1 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 1 \times 2^0\)
• \(= 1 \times 64 + 1 \times 32 + 1 \times 16 + 1 \times 8 + 0 \times 4 + 0 \times 2 + 1 \times 1\)
• \(= 64 + 32 + 16 + 8 + 0 + 0 + 1 = 121_{10}\)

Decimal difference: \(144 - 121 = 23_{10}\).

Now, let's convert our binary result 10111 back to decimal:

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

The decimal result matches, confirming our binary subtraction is correct.

The difference between the two binary numbers 10010000 and 1111001 is 10111.

Revision Table: Binary Subtraction Basics

Additional Information: Representing Negative Binary Numbers

When the subtrahend is larger than the minuend, the result of binary subtraction is negative. Negative binary numbers can be represented using methods like Sign-Magnitude, One's Complement, or Two's Complement. Two's Complement is the most common method used in computers for representing negative numbers and performing arithmetic operations, including subtraction.

In Two's Complement subtraction, the subtraction \(A - B\) is performed as \(A + (\text{Two's Complement of } B)\). This converts subtraction into an addition operation, simplifying hardware design in digital systems.