To add the two binary numbers \(101011_2\) and \(10111_2\), we can line them up and add them similar to how we would with decimal numbers, remembering to carry over when the sum exceeds \(1\).
Let's align the numbers:
101011
+ 010111
Now, we add from right to left:
- \(1 + 1 = 10_2\) (write \(0\) and carry \(1\))
- \(1 + 1 + 1 = 11_2\) (write \(1\) and carry \(1\))
- \(0 + 1 + 0 = 1_2\) (write \(1\))
- \(1 + 0 = 1_2\) (write \(1\))
- \(0 + 1 = 1_2\) (write \(1\))
- \(1 + 0 = 1_2\) (write \(1\))
Now, putting it all together, we get:
101011
+ 010111
-----------
1001010
Thus, the result of \(101011_2 + 10111_2\) is \(1001010_2\).
Now, if needed, we can check our work by converting the binary numbers to decimal:
-
\(101011_2 = 1 \cdot 2^5 + 0 \cdot 2^4 + 1 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0 = 32 + 0 + 8 + 0 + 2 + 1 = 43_{10}\)
-
\(10111_2 = 1 \cdot 2^4 + 0 \cdot 2^3 + 1 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0 = 16 + 0 + 4 + 2 + 1 = 23_{10}\)
Now, if we add these decimal values:
\(43 + 23 = 66_{10}\)
Finally, let's convert \(1001010_2\) back to decimal to verify:
- \(1001010_2 = 1 \cdot 2^6 + 0 \cdot 2^5 + 0 \cdot 2^4 + 1 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2^1 + 0 \cdot 2^0 = 64 + 0 + 0 + 8 + 0 + 2 + 0 = 66_{10}\)
The decimal confirms our binary addition is correct.
The final result is:
\[ 1001010_2 \]