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Sagot :
Certainly! Let's solve the problem step-by-step. Below is the detailed solution:
First, let's clearly identify the binary numbers we are working with.
[tex]\[ \text{Binary number 1: } 1011001_2 \][/tex]
[tex]\[ \text{Binary number 2: } 100_2 \][/tex]
However, it appears there's a typo with binary1 representation, it should rather be [tex]\( 110011_2 \)[/tex].
### Step 1: Convert Binary to Decimal
We need to convert these binary numbers to their decimal (base-10) equivalents.
For [tex]\( 1011001_2 \)[/tex]:
[tex]\[ \begin{align*} 1 \cdot 2^6 + 0 \cdot 2^5 + 1 \cdot 2^4 + 1 \cdot 2^3 + 0 \cdot 2^2 + 0 \cdot 2^1 + 1 \cdot 2^0 & = 1 \cdot 64 + 0 \cdot 32 + 1 \cdot 16 + 1 \cdot 8 + 0 \cdot 4 + 0 \cdot 2 + 1 \cdot 1 \\ & = 64 + 0 + 16 + 8 + 0 + 0 + 1 \\ & = 89 \end{align*} \][/tex]
For [tex]\( 100_2 \)[/tex]:
[tex]\[ \begin{align*} 1 \cdot 2^2 + 0 \cdot 2^1 + 0 \cdot 2^0 & = 1 \cdot 4 + 0 \cdot 2 + 0 \cdot 1 \\ & = 4 \end{align*} \][/tex]
So, the decimal equivalents are:
[tex]\[ 1011001_2 = 89_{10} \][/tex]
[tex]\[ 100_2 = 4_{10} \][/tex]
### Step 2: Sum the Decimal Numbers
Next, we add the decimal numbers together.
[tex]\[ 89 + 4 = 93 \][/tex]
### Step 3: Convert the Sum Back to Binary
Now we need to convert the sum (93 in decimal) back to its binary form.
To convert 93 into binary, we repeatedly divide by 2 and record the remainders:
[tex]\[ \begin{align*} 93 \div 2 & = 46 \quad \text{remainder } 1 \\ 46 \div 2 & = 23 \quad \text{remainder } 0 \\ 23 \div 2 & = 11 \quad \text{remainder } 1 \\ 11 \div 2 & = 5 \quad \text{remainder } 1 \\ 5 \div 2 & = 2 \quad \text{remainder } 1 \\ 2 \div 2 & = 1 \quad \text{remainder } 0 \\ 1 \div 2 & = 0 \quad \text{remainder } 1 \\ \end{align*} \][/tex]
Reading the remainders from bottom to top, we get [tex]\( 1011101_2 \)[/tex].
### Final Answer
[tex]\[ (1011001_2) + (100_2) = 1011101_2 \][/tex]
Thus:
- [tex]\( 1011001_2 = 89_{10} \)[/tex]
- [tex]\( 100_2 = 4_{10} \)[/tex]
- [tex]\( 89 + 4 = 93 \rightarrow 1011101_2 \)[/tex]
### Summary:
[tex]\( (1011001_2 + 100_2 = 1011101_2) \)[/tex]
First, let's clearly identify the binary numbers we are working with.
[tex]\[ \text{Binary number 1: } 1011001_2 \][/tex]
[tex]\[ \text{Binary number 2: } 100_2 \][/tex]
However, it appears there's a typo with binary1 representation, it should rather be [tex]\( 110011_2 \)[/tex].
### Step 1: Convert Binary to Decimal
We need to convert these binary numbers to their decimal (base-10) equivalents.
For [tex]\( 1011001_2 \)[/tex]:
[tex]\[ \begin{align*} 1 \cdot 2^6 + 0 \cdot 2^5 + 1 \cdot 2^4 + 1 \cdot 2^3 + 0 \cdot 2^2 + 0 \cdot 2^1 + 1 \cdot 2^0 & = 1 \cdot 64 + 0 \cdot 32 + 1 \cdot 16 + 1 \cdot 8 + 0 \cdot 4 + 0 \cdot 2 + 1 \cdot 1 \\ & = 64 + 0 + 16 + 8 + 0 + 0 + 1 \\ & = 89 \end{align*} \][/tex]
For [tex]\( 100_2 \)[/tex]:
[tex]\[ \begin{align*} 1 \cdot 2^2 + 0 \cdot 2^1 + 0 \cdot 2^0 & = 1 \cdot 4 + 0 \cdot 2 + 0 \cdot 1 \\ & = 4 \end{align*} \][/tex]
So, the decimal equivalents are:
[tex]\[ 1011001_2 = 89_{10} \][/tex]
[tex]\[ 100_2 = 4_{10} \][/tex]
### Step 2: Sum the Decimal Numbers
Next, we add the decimal numbers together.
[tex]\[ 89 + 4 = 93 \][/tex]
### Step 3: Convert the Sum Back to Binary
Now we need to convert the sum (93 in decimal) back to its binary form.
To convert 93 into binary, we repeatedly divide by 2 and record the remainders:
[tex]\[ \begin{align*} 93 \div 2 & = 46 \quad \text{remainder } 1 \\ 46 \div 2 & = 23 \quad \text{remainder } 0 \\ 23 \div 2 & = 11 \quad \text{remainder } 1 \\ 11 \div 2 & = 5 \quad \text{remainder } 1 \\ 5 \div 2 & = 2 \quad \text{remainder } 1 \\ 2 \div 2 & = 1 \quad \text{remainder } 0 \\ 1 \div 2 & = 0 \quad \text{remainder } 1 \\ \end{align*} \][/tex]
Reading the remainders from bottom to top, we get [tex]\( 1011101_2 \)[/tex].
### Final Answer
[tex]\[ (1011001_2) + (100_2) = 1011101_2 \][/tex]
Thus:
- [tex]\( 1011001_2 = 89_{10} \)[/tex]
- [tex]\( 100_2 = 4_{10} \)[/tex]
- [tex]\( 89 + 4 = 93 \rightarrow 1011101_2 \)[/tex]
### Summary:
[tex]\( (1011001_2 + 100_2 = 1011101_2) \)[/tex]
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