In the binary equation (1p101) 2+ (10q1) 2= (100r00) 2
0, 1, 0
Solving the Binary Equation to Find p, q, and r
The problem asks us to find the binary digits p, q, and r that satisfy the given binary addition equation: \((1p101)_2 + (10q1)_2 = (100r00)_2\). Here, p, q, and r can only be 0 or 1, as they are binary digits.
To solve this, we need to perform binary addition column by column, starting from the rightmost digit (the least significant bit), and consider the carries.
Step-by-Step Binary Addition Analysis
Let's align the two binary numbers vertically for addition. Note that the second number has fewer digits, so we align it to the right.
| 1 | p | 1 | 0 | 1 | (1p101)\(_2\) | ||
| + | 1 | 0 | q | 1 | (10q1)\(_2\) | ||
| — | — | — | — | — | — | — | — |
| 1 | 0 | 0 | r | 0 | 0 | (100r00)\(_2\) |
Now, let's perform the binary addition from right to left, column by column:
- Rightmost Column (Column 1): We add the least significant bits.
\(\qquad 1 + 1\)
In binary, \(1 + 1 = 10_2\). This means the sum digit for this column is 0, and there is a carry of 1 to the next column on the left.
The sum given in the equation is \((100r00)_2\). The rightmost digit of the sum is indeed 0. This step is consistent.
Carry to Column 2 = 1.
- Second Column from Right (Column 2): We add the digits in this column plus the carry from Column 1.
\(\qquad 0 + q + \text{carry}(1)\)
The sum digit for this column in \((100r00)_2\) is 0. So, \(0 + q + 1\) must result in a binary number ending in 0.
If \(q=0\): \(0 + 0 + 1 = 1_2\). The sum digit would be 1, which doesn't match 0.
If \(q=1\): \(0 + 1 + 1 = 10_2\). The sum digit is 0, and there is a carry of 1 to the next column. This matches the required sum digit 0.
Therefore, q must be 1.
Carry to Column 3 = 1.
- Third Column from Right (Column 3): We add the digits in this column plus the carry from Column 2.
\(\qquad 1 + 0 + \text{carry}(1)\)
The sum digit for this column in \((100r00)_2\) is r. So, \(1 + 0 + 1 = r\).
\(1 + 0 + 1 = 10_2\). This means the sum digit is 0, and there is a carry of 1 to the next column.
Therefore, r must be 0.
Carry to Column 4 = 1.
- Fourth Column from Right (Column 4): We add the digits in this column plus the carry from Column 3.
\(\qquad p + 1 + \text{carry}(1)\)
The sum digit for this column in \((100r00)_2\) is 0. So, \(p + 1 + 1\) must result in a binary number ending in 0.
This is equivalent to \(p + 10_2\) ending in 0.
If \(p=0\): \(0 + 1 + 1 = 10_2\). The sum digit is 0, and there is a carry of 1. This matches the required sum digit 0.
If \(p=1\): \(1 + 1 + 1 = 11_2\). The sum digit is 1, which doesn't match 0.
Therefore, p must be 0.
Carry to Column 5 = 1.
- Fifth Column from Right (Column 5): We add the digits in this column plus the carry from Column 4.
\(\qquad 1 + \text{carry}(1)\)
The sum digit for this column in \((100r00)_2\) is 0. So, \(1 + 1 = 0\).
\(1 + 1 = 10_2\). This means the sum digit is 0, and there is a carry of 1 to the next column. This matches the required sum digit 0.
Carry to Column 6 = 1.
- Sixth Column from Right (Column 6): We add the implicit 0 from the shorter number plus the carry from Column 5.
\(\qquad 0 + \text{carry}(1)\)
The sum digit for this column in \((100r00)_2\) is 1. So, \(0 + 1 = 1\).
\(0 + 1 = 1_2\). This means the sum digit is 1. This matches the required sum digit 1.
Carry to subsequent columns = 0.
Based on the step-by-step binary addition, we have determined the values:
- p = 0
- q = 1
- r = 0
Confirming the Solution with Found Values
Let's substitute p=0, q=1, and r=0 back into the original equation and perform the addition:
The equation becomes: \((10101)_2 + (1011)_2 = (100000)_2\).
Performing the addition \((10101)_2 + (1011)_2\):
| Carry: | 1 | 1 | 1 | 1 | |||
| 1 | 0 | 1 | 0 | 1 | (10101)\(_2\) | ||
| + | 1 | 0 | 1 | 1 | (1011)\(_2\) | ||
| — | — | — | — | — | — | — | — |
| 1 | 0 | 0 | 0 | 0 | 0 | (100000)\(_2\) |
The result of the addition is \((100000)_2\).
The required sum was \((100r00)_2\). Substituting r=0, we get \((100000)_2\).
Since \((100000)_2 = (100000)_2\), the values p=0, q=1, and r=0 satisfy the equation.
The possible values of p, q, and r respectively are 0, 1, and 0.
Revision Table: Understanding Binary Addition
| Concept | Explanation | Example |
|---|---|---|
| Binary Digits | The only digits used are 0 and 1. | 0, 1 |
| Place Value | Each position represents a power of 2, increasing from right to left (\(2^0, 2^1, 2^2, \dots\)). | In \((101)_2\), the place values are \(2^2\) (4), \(2^1\) (2), \(2^0\) (1). The value is \(1 \times 4 + 0 \times 2 + 1 \times 1 = 5_{10}\). |
| Binary Addition Rules | Rules for adding two binary digits: |
|
| Carry | When the sum in a column is 2 (\(10_2\)) or 3 (\(11_2\)), a carry of 1 is generated and added to the next column to the left. | Adding \(1+1\) gives sum 0 and carry 1. |
Additional Information: Binary Number Systems
Binary numbers are fundamental in digital electronics and computing because they can be easily represented by the on/off states of switches or the high/low voltage levels in circuits. The base of the binary system is 2.
Understanding binary addition, subtraction, multiplication, and division is essential in computer science and related fields. Converting between binary and decimal (base 10) or hexadecimal (base 16) is also a common operation.
For example, converting a binary number like \((1011)_2\) to decimal:
\((1011)_2 = 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0\)
\((1011)_2 = 1 \times 8 + 0 \times 4 + 1 \times 2 + 1 \times 1\)
\((1011)_2 = 8 + 0 + 2 + 1 = 11_{10}\).
Converting a decimal number like \(13_{10}\) to binary can be done by repeatedly dividing by 2 and recording the remainders:
- \(13 \div 2 = 6\) remainder 1
- \(6 \div 2 = 3\) remainder 0
- \(3 \div 2 = 1\) remainder 1
- \(1 \div 2 = 0\) remainder 1
Reading the remainders from bottom up gives \((1101)_2\).
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