Formula: Number of Boolean functions of
Condition: We are told the total functions must be ≤ 4:
Minimum
✅ Final Answer:
The floating-point binary number is
We need to convert it into an 8-bit fraction and a 6-bit exponent format.
We start by normalizing the binary number into scientific notation of the form:
Converting
The exponent is
The exponent is
The fractional part of the normalized binary number is
The floating-point binary number
Exponent:
Format: 10-bit signed integer using 2's complement representation.
0000100011
1111011100
1111011101
✅ Final Answer:
1111011101
–35 in 10-bit 2's complement: 1111011101
1. Two Zeros Exist
• Positive Zero: 0000
• Negative Zero: 1111
? This creates ambiguity, because logically both are zero but they have different bit patterns.
2. End-Around Carry Needed
When adding numbers, if a carry comes out of the MSB, it must be added back to the LSB.
(+5) = 0101
(-5) = 1010 (1’s complement of 0101)
----------------
Add: 1111 → End-around carry = 1
Final: 0000 (after adding carry)
? Result is zero, but note that two different zeros are possible.
3. Hardware Complexity
Extra logic is required so that +0
and -0
are treated the same, making design slower and costlier.
That’s why modern systems use 2’s complement. It has only one zero and simplifies arithmetic operations.
Total Size: 32 bits
The exponent is stored in 8-bit biased form with a bias of 127.
Stored exponent = Actual exponent + 127
Example:
Actual exponent = 3
Stored exponent = 3 + 127 = 130 = 10000010
In IEEE single precision, the exponent is represented in 8-bit biased (excess-127) notation.
Given the 2's complement binary number: (011010)
Find its equivalent hexadecimal representation.
Since the number has 6 bits, group the bits into two groups of 4 bits (add leading zeros if needed):
Now, split into two nibbles (4 bits each):
Convert each nibble to hexadecimal:
-
-
Therefore, the hexadecimal representation is: 1A
Question: Find the base r such that
(312)r ÷ (20)r = (13.1)r
(3r² + r + 2) / (2r) = r + 3 + 1/r
3r² + r + 2 = 2r² + 6r + 2
r² - 5r = 0
⇒ r(r - 5) = 0
Since base must be > 1, the valid solution is:
Base (radix) = 5
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and More.