32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 91.402 701 95 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 91.402 701 95(10) converted and written in 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 91.
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;
  • 91 ÷ 2 = 45 + 1;
  • 45 ÷ 2 = 22 + 1;
  • 22 ÷ 2 = 11 + 0;
  • 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.


91(10) =


101 1011(2)


3. Convert to binary (base 2) the fractional part: 0.402 701 95.

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) 0.402 701 95 × 2 = 0 + 0.805 403 9;
  • 2) 0.805 403 9 × 2 = 1 + 0.610 807 8;
  • 3) 0.610 807 8 × 2 = 1 + 0.221 615 6;
  • 4) 0.221 615 6 × 2 = 0 + 0.443 231 2;
  • 5) 0.443 231 2 × 2 = 0 + 0.886 462 4;
  • 6) 0.886 462 4 × 2 = 1 + 0.772 924 8;
  • 7) 0.772 924 8 × 2 = 1 + 0.545 849 6;
  • 8) 0.545 849 6 × 2 = 1 + 0.091 699 2;
  • 9) 0.091 699 2 × 2 = 0 + 0.183 398 4;
  • 10) 0.183 398 4 × 2 = 0 + 0.366 796 8;
  • 11) 0.366 796 8 × 2 = 0 + 0.733 593 6;
  • 12) 0.733 593 6 × 2 = 1 + 0.467 187 2;
  • 13) 0.467 187 2 × 2 = 0 + 0.934 374 4;
  • 14) 0.934 374 4 × 2 = 1 + 0.868 748 8;
  • 15) 0.868 748 8 × 2 = 1 + 0.737 497 6;
  • 16) 0.737 497 6 × 2 = 1 + 0.474 995 2;
  • 17) 0.474 995 2 × 2 = 0 + 0.949 990 4;
  • 18) 0.949 990 4 × 2 = 1 + 0.899 980 8;
  • 19) 0.899 980 8 × 2 = 1 + 0.799 961 6;
  • 20) 0.799 961 6 × 2 = 1 + 0.599 923 2;
  • 21) 0.599 923 2 × 2 = 1 + 0.199 846 4;
  • 22) 0.199 846 4 × 2 = 0 + 0.399 692 8;
  • 23) 0.399 692 8 × 2 = 0 + 0.799 385 6;
  • 24) 0.799 385 6 × 2 = 1 + 0.598 771 2;

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:


0.402 701 95(10) =


0.0110 0111 0001 0111 0111 1001(2)


5. Positive number before normalization:

91.402 701 95(10) =


101 1011.0110 0111 0001 0111 0111 1001(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the left, so that only one non zero digit remains to the left of it:


91.402 701 95(10) =


101 1011.0110 0111 0001 0111 0111 1001(2) =


101 1011.0110 0111 0001 0111 0111 1001(2) × 20 =


1.0110 1101 1001 1100 0101 1101 1110 01(2) × 26


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): 6


Mantissa (not normalized):
1.0110 1101 1001 1100 0101 1101 1110 01


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


6 + 2(8-1) - 1 =


(6 + 127)(10) =


133(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;
  • 133 ÷ 2 = 66 + 1;
  • 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) =


133(10) =


1000 0101(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 011 0110 1100 1110 0010 1110 111 1001 =


011 0110 1100 1110 0010 1110


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 0101


Mantissa (23 bits) =
011 0110 1100 1110 0010 1110


The base ten decimal number 91.402 701 95 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1000 0101 - 011 0110 1100 1110 0010 1110

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