Question 1: Perform the following conversions: 231710 to hexadecimal 11100101112
ID: 3745346 • Letter: Q
Question
Question 1: Perform the following conversions:
231710 to hexadecimal
11100101112 to hexadecimal
10110100012 to decimal
59116 to decimal
31610 to binary
DFA016 to binary
Question 2: Show the representation of the following values in 8-bit two's complement notation:
125
-71
56
-26
Question 3: Perform the following additions of 8-bit two's complement numbers:
0001 1011 + 0100 1010
1010 0011 + 0001 0101
1000 0110 + 1001 0111
1101 1100 + 0011 0110
Question 4: Perform the following subtraction of 8-bit two's complement numbers:
0001 1011 - 0100 1010
1010 0011 - 0001 0101
1000 0110 - 1001 0111
1101 1100 - 0011 0110
Question 5: Convert the following decimal numbers to 32-bit IEEE floating point:
173.28125
-43.3125
Question 6: Convert the following 32-bit IEEE floating point numbers to decimal:
0100 0100 1000 1010 0010 0000 0000 0000
1100 0000 0010 1101 0111 0000 1010 0100
Explanation / Answer
Following procedure can be used to convert decimal number to any number (binary, octal and hexadecimal)
Question 1:
If 231710 is binary number, then
Decimal Number Operation Quotient Remainder Hexadecimal Result
231710 / 16 14481 14 = E E
14481 / 16 905 1 1E
905 / 16 56 9 91E
56 / 16 3 8 891E
3 / 16 0 3 3891E
(231710)10 = (3891E)16
If 231710 is octal number, so first convert octal to binary by converting each octal number to its equivalent three digit binary number.
(231710)8 = (010 011 001 111 001 000)2
And then convert binary number to hexadecimal by arranging binary digit into set of four bit.
(010 011 001 111 001 000)2 = (0001 0011 0011 1100 1000)2 = (133C8)16
(231710)8 = (133C8)16
Similarly, solving another problems.
Here, radix is not mentioned in given problems, So I have converted all possible radix to the asked radix.
(11100101112)10 = (2959E19F8)16
(11100101112)8 = (4900824A)16
(11100101112)8 = (1224770122)10
(11100101112)16 = (1172527124754)2
(59116)16 = (364822)2
(31610)10 = (111101101111010)2
(31610)8 = (011 001 110 001 000)2
(31610)16 = (0011 0001 0110 0001 0000)2
(DFA016)16 = (110111111010000000010110)2
Question 2: Converting to 8-Bit 2’s complement.
(125)10 = (01111101)2
Invert all digits: 0 1 1 1 1 1 0 1
1 0 0 0 0 0 1 0
Now add 1 to it.
1 0 0 0 0 0 1 0
+ 1
= 1 0 0 0 0 0 1 1
So, 125 in 2’s Complement is 10000011
Similarly,
-71 in 2’s complement is 10111001
56 in 2’s complement is 00111000
-26 in 2’s complement is 11100110
Question 3: Addition of 8-Bit 2’s complement number
Carry 1 1 1
0 0 0 1 1 0 1 1
+ 0 1 0 0 1 0 1 0
= 0 1 1 0 0 1 0 1
So, 0001 1011 + 0100 1010 = 0110 0101
Similarly, solving other problems as
1010 0011 + 0001 0101 = 01011 1000
1000 0110 + 1001 0111 = 010001 1101
1101 1100 + 0011 0110 = 010001 0010
Question 4: Subtraction of 8-Bit 2’s complement number
Borrow 1
0 0 0 1 1 0 1 1
- 0 1 0 0 1 0 1 0
1 1 0 1 0 0 0 1
Since, there is no digit to borrow, so the result is in negative number.
Convert it into 2’s complement,
1 1 0 1 0 0 0 1
Inverting digits,
0 0 1 0 1 1 1 0
+ 1
0 0 1 0 1 1 1 1
So,
0001 1011 - 0100 1010 = -00101111
Similarly solving other problems,
1010 0011 - 0001 0101 = 1000 1110
1000 0110 - 1001 0111 = -0001 0001
1101 1100 - 0011 0110 = 1010 0110
Question 5: Convert decimal to 32-bit IEEE floating point
173.28125
The 32-bit representation consists of three parts.
First bit indicate whether number is positive or negative. 0 for positive and 1 for negative.
Here number is positive, so first digit is 0.
Convert 173.28125 to binary number.
(173.28125)10 = (10101101.01001)2
Now shift decimal to 7 place left so that there is only one non-zero decimal place in number.
1.010110101001
Now, take the number after decimal place and add Zero’s to make it 23 bit
01011010100100000000000
This 23 bit represents the fraction.
For exponent bit,
We have shifted the decimal to 7 place.
Add 7 to a127 = 134.
Convert 134 into binary.
(134)10 = (10000110)2
So, final 32-bit IEEE floating point for 173.28125 is 0 10000110 01011010100100000000000
Question 6: Convert 32-bit IEEE floating point numbers to decimal
0 10001001 00010100010000000000000
As, first digit is 0, so its positive number.
For exponent,
10001001 = 137
So, number of decimal place to shift right is 137-127 = 10
Now fraction part, shift decimal 10 place right in 00010100010000000000000.
After shifting we get,
10001010001.0000000000000
Now, convert above binary number to decimal number.
(10001010001)2 = (1105)10
So, 0 10001001 00010100010000000000000 in decimal number is +1105.
Similarly, solving another problem;
1100 0000 0010 1101 0111 0000 1010 0100 in decimal is -2.71000003814697265625.
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