Question

Please show all work:

Represent the number (+46.5) as a 32 bit floating-point number using the IEEE standard 754 format. N.B. The attached ‘Appendix’ section may prove useful in the conversion process.

           

1. 0 1000 0111 0100 0100 0000 0000 0000 000
2. 0 1011 0100 0111 0100 0000 0000 0000 000
3. 0 1100 0110 0101 0100 0000 0000 0000 000
4. 0 1000 0100 0111 0100 0000 0000 0000 000
5. 0 1010 0100 0111 0100 0000 0000 0000 000

Answer 1

1. First, convert integer part to binary: 46.
Divide the number repeatedly by 2.
Keep track of each of the remainder.
We stop when we get a quotient that is equal to zero.
division = quotient + remainder;
46 ÷ 2 = 23 + 0;
23 ÷ 2 = 11 + 1;
11 ÷ 2 = 5 + 1;
5 ÷ 2 = 2 + 1;
2 ÷ 2 = 1 + 0;
1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the integer part of the number.
Take all the remainders starting from the bottom of the list constructed above.
46(10) =
10 1110(2)

3. Convert to the binary (base 2) the fractional part: 0.5.
Multiply it repeatedly by 2.
Keep track of each integer part of the results.
Stop when we get a fractional part that is equal to zero.

#) multiplying = integer + fractional part;
1) 0.5 × 2 = 1 + 0;

4. Construct the base 2 representation of the fractional part of the number.
Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:
0.5(10) =
0.1(2)

5. Positive number before normalization:
46.5(10) =
10 1110.1(2)

6. Normalize the binary representation of the number.
Shift the decimal mark 5 positions to the left so that only one non zero digit remains to the left of it:
46.5(10) =
10 1110.1(2) =
10 1110.1(2) × 20 =
1.0111 01(2) × 25

7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:
Sign: 0 (a positive number)
Exponent (unadjusted): 5
Mantissa (not normalized):
1.0111 01

8. Adjust the exponent.
Use the 8 bit excess/bias notation:
Exponent (adjusted) =
Exponent (unadjusted) + 2(8-1) - 1 =
5 + 2(8-1) - 1 =
(5 + 127)(10) =
132(10)

9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.
Use the same technique of repeatedly dividing by 2:
division = quotient + remainder;
132 ÷ 2 = 66 + 0;
66 ÷ 2 = 33 + 0;
33 ÷ 2 = 16 + 1;
16 ÷ 2 = 8 + 0;
8 ÷ 2 = 4 + 0;
4 ÷ 2 = 2 + 0;
2 ÷ 2 = 1 + 0;
1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.
Take all the remainders starting from the bottom of the list constructed above:
Exponent (adjusted) =
132(10) =
1000 0100(2)

11. Normalize the mantissa.
a) Remove the leftmost bit, since it's allways 1, and the decimal point, if the case.
b) Adjust its length to 23 bits, by adding the necessary number of zeros to the right.
Mantissa (normalized) =
1. 01 1101 0 0000 0000 0000 0000 =
011 1010 0000 0000 0000 0000

12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:
Sign (1 bit) =
0 (a positive number)
Exponent (8 bits) =
1000 0100
Mantissa (23 bits) =
011 1010 0000 0000 0000 0000

Final answer:
Number 46.5 converted from decimal system (base 10)
to
32 bit single precision IEEE 754 binary floating point:
0 - 1000 0100 - 011 1010 0000 0000 0000 0000