32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 86.132 870 320 9 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 86.132 870 320 9(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: 86.
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;
  • 86 ÷ 2 = 43 + 0;
  • 43 ÷ 2 = 21 + 1;
  • 21 ÷ 2 = 10 + 1;
  • 10 ÷ 2 = 5 + 0;
  • 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.


86(10) =


101 0110(2)


3. Convert to binary (base 2) the fractional part: 0.132 870 320 9.

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) 0.132 870 320 9 × 2 = 0 + 0.265 740 641 8;
  • 2) 0.265 740 641 8 × 2 = 0 + 0.531 481 283 6;
  • 3) 0.531 481 283 6 × 2 = 1 + 0.062 962 567 2;
  • 4) 0.062 962 567 2 × 2 = 0 + 0.125 925 134 4;
  • 5) 0.125 925 134 4 × 2 = 0 + 0.251 850 268 8;
  • 6) 0.251 850 268 8 × 2 = 0 + 0.503 700 537 6;
  • 7) 0.503 700 537 6 × 2 = 1 + 0.007 401 075 2;
  • 8) 0.007 401 075 2 × 2 = 0 + 0.014 802 150 4;
  • 9) 0.014 802 150 4 × 2 = 0 + 0.029 604 300 8;
  • 10) 0.029 604 300 8 × 2 = 0 + 0.059 208 601 6;
  • 11) 0.059 208 601 6 × 2 = 0 + 0.118 417 203 2;
  • 12) 0.118 417 203 2 × 2 = 0 + 0.236 834 406 4;
  • 13) 0.236 834 406 4 × 2 = 0 + 0.473 668 812 8;
  • 14) 0.473 668 812 8 × 2 = 0 + 0.947 337 625 6;
  • 15) 0.947 337 625 6 × 2 = 1 + 0.894 675 251 2;
  • 16) 0.894 675 251 2 × 2 = 1 + 0.789 350 502 4;
  • 17) 0.789 350 502 4 × 2 = 1 + 0.578 701 004 8;
  • 18) 0.578 701 004 8 × 2 = 1 + 0.157 402 009 6;
  • 19) 0.157 402 009 6 × 2 = 0 + 0.314 804 019 2;
  • 20) 0.314 804 019 2 × 2 = 0 + 0.629 608 038 4;
  • 21) 0.629 608 038 4 × 2 = 1 + 0.259 216 076 8;
  • 22) 0.259 216 076 8 × 2 = 0 + 0.518 432 153 6;
  • 23) 0.518 432 153 6 × 2 = 1 + 0.036 864 307 2;
  • 24) 0.036 864 307 2 × 2 = 0 + 0.073 728 614 4;

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:


0.132 870 320 9(10) =


0.0010 0010 0000 0011 1100 1010(2)


5. Positive number before normalization:

86.132 870 320 9(10) =


101 0110.0010 0010 0000 0011 1100 1010(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:


86.132 870 320 9(10) =


101 0110.0010 0010 0000 0011 1100 1010(2) =


101 0110.0010 0010 0000 0011 1100 1010(2) × 20 =


1.0101 1000 1000 1000 0000 1111 0010 10(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):
1.0101 1000 1000 1000 0000 1111 0010 10


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. 010 1100 0100 0100 0000 0111 100 1010 =


010 1100 0100 0100 0000 0111


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) =
010 1100 0100 0100 0000 0111


The base ten decimal number 86.132 870 320 9 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1000 0101 - 010 1100 0100 0100 0000 0111

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