Question

Convert 3/7 to IEEE 754.

Convert 171.375 to IEEE754

Convert the float number 0x4024 0080 0000 0000 to decimal. What is the true exponent? Circle the true exponent.

0x3FAF 9000 0000 is an IEEE float number. Convert it to base 10. While doing the conversion answer two questions: What is the true exponent? What is the mantissa? Make these two values obvious by circling them.

Answer 1

Convert 3/7 to IEEE 754

3/7=0.42857142857

1. First, convert to the binary (base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.

• division = quotient + remainder;
• 0 ÷ 2 = 0 + 0;

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.

0₍₁₀₎ =

0₍₂₎

3. Convert to the binary (base 2) the fractional part: 0.428 571 428 57.

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) 0.428 571 428 57 × 2 = 0 + 0.857 142 857 14;
• 2) 0.857 142 857 14 × 2 = 1 + 0.714 285 714 28;
• 3) 0.714 285 714 28 × 2 = 1 + 0.428 571 428 56;
• 4) 0.428 571 428 56 × 2 = 0 + 0.857 142 857 12;
• 5) 0.857 142 857 12 × 2 = 1 + 0.714 285 714 24;
• 6) 0.714 285 714 24 × 2 = 1 + 0.428 571 428 48;
• 7) 0.428 571 428 48 × 2 = 0 + 0.857 142 856 96;
• 8) 0.857 142 856 96 × 2 = 1 + 0.714 285 713 92;
• 9) 0.714 285 713 92 × 2 = 1 + 0.428 571 427 84;
• 10) 0.428 571 427 84 × 2 = 0 + 0.857 142 855 68;
• 11) 0.857 142 855 68 × 2 = 1 + 0.714 285 711 36;
• 12) 0.714 285 711 36 × 2 = 1 + 0.428 571 422 72;
• 13) 0.428 571 422 72 × 2 = 0 + 0.857 142 845 44;
• 14) 0.857 142 845 44 × 2 = 1 + 0.714 285 690 88;
• 15) 0.714 285 690 88 × 2 = 1 + 0.428 571 381 76;
• 16) 0.428 571 381 76 × 2 = 0 + 0.857 142 763 52;
• 17) 0.857 142 763 52 × 2 = 1 + 0.714 285 527 04;
• 18) 0.714 285 527 04 × 2 = 1 + 0.428 571 054 08;
• 19) 0.428 571 054 08 × 2 = 0 + 0.857 142 108 16;
• 20) 0.857 142 108 16 × 2 = 1 + 0.714 284 216 32;
• 21) 0.714 284 216 32 × 2 = 1 + 0.428 568 432 64;
• 22) 0.428 568 432 64 × 2 = 0 + 0.857 136 865 28;
• 23) 0.857 136 865 28 × 2 = 1 + 0.714 273 730 56;
• 24) 0.714 273 730 56 × 2 = 1 + 0.428 547 461 12;
• 25) 0.428 547 461 12 × 2 = 0 + 0.857 094 922 24;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)

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:

0.428 571 428 57₍₁₀₎ =

0.0110 1101 1011 0110 1101 1011 0₍₂₎

5. Positive number before normalization:

0.428 571 428 57₍₁₀₎ =

0.0110 1101 1011 0110 1101 1011 0₍₂₎

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right so that only one non zero digit remains to the left of it:

0.428 571 428 57₍₁₀₎ =

0.0110 1101 1011 0110 1101 1011 0₍₂₎ =

0.0110 1101 1011 0110 1101 1011 0₍₂₎ × 2⁰ =

1.1011 0110 1101 1011 0110 110₍₂₎ × 2^-2

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): -2

Mantissa (not normalized):
1.1011 0110 1101 1011 0110 110

8. Adjust the exponent.

Use the 8 bit excess/bias notation:

Exponent (adjusted) =

Exponent (unadjusted) + 2^(8-1) - 1 =

-2 + 2^(8-1) - 1 =

(-2 + 127)₍₁₀₎ =

125₍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;
• 125 ÷ 2 = 62 + 1;
• 62 ÷ 2 = 31 + 0;
• 31 ÷ 2 = 15 + 1;
• 15 ÷ 2 = 7 + 1;
• 7 ÷ 2 = 3 + 1;
• 3 ÷ 2 = 1 + 1;
• 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) =

125₍₁₀₎ =

0111 1101₍₂₎

11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).

Mantissa (normalized) =

1. 101 1011 0110 1101 1011 0110 =

101 1011 0110 1101 1011 0110

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) =
0111 1101

Mantissa (23 bits) =
101 1011 0110 1101 1011 0110