32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 10.024 920 89 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 10.024 920 89(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: 10.
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;
  • 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.


10(10) =


1010(2)


3. Convert to binary (base 2) the fractional part: 0.024 920 89.

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) 0.024 920 89 × 2 = 0 + 0.049 841 78;
  • 2) 0.049 841 78 × 2 = 0 + 0.099 683 56;
  • 3) 0.099 683 56 × 2 = 0 + 0.199 367 12;
  • 4) 0.199 367 12 × 2 = 0 + 0.398 734 24;
  • 5) 0.398 734 24 × 2 = 0 + 0.797 468 48;
  • 6) 0.797 468 48 × 2 = 1 + 0.594 936 96;
  • 7) 0.594 936 96 × 2 = 1 + 0.189 873 92;
  • 8) 0.189 873 92 × 2 = 0 + 0.379 747 84;
  • 9) 0.379 747 84 × 2 = 0 + 0.759 495 68;
  • 10) 0.759 495 68 × 2 = 1 + 0.518 991 36;
  • 11) 0.518 991 36 × 2 = 1 + 0.037 982 72;
  • 12) 0.037 982 72 × 2 = 0 + 0.075 965 44;
  • 13) 0.075 965 44 × 2 = 0 + 0.151 930 88;
  • 14) 0.151 930 88 × 2 = 0 + 0.303 861 76;
  • 15) 0.303 861 76 × 2 = 0 + 0.607 723 52;
  • 16) 0.607 723 52 × 2 = 1 + 0.215 447 04;
  • 17) 0.215 447 04 × 2 = 0 + 0.430 894 08;
  • 18) 0.430 894 08 × 2 = 0 + 0.861 788 16;
  • 19) 0.861 788 16 × 2 = 1 + 0.723 576 32;
  • 20) 0.723 576 32 × 2 = 1 + 0.447 152 64;
  • 21) 0.447 152 64 × 2 = 0 + 0.894 305 28;
  • 22) 0.894 305 28 × 2 = 1 + 0.788 610 56;
  • 23) 0.788 610 56 × 2 = 1 + 0.577 221 12;
  • 24) 0.577 221 12 × 2 = 1 + 0.154 442 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:


0.024 920 89(10) =


0.0000 0110 0110 0001 0011 0111(2)


5. Positive number before normalization:

10.024 920 89(10) =


1010.0000 0110 0110 0001 0011 0111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the left, so that only one non zero digit remains to the left of it:


10.024 920 89(10) =


1010.0000 0110 0110 0001 0011 0111(2) =


1010.0000 0110 0110 0001 0011 0111(2) × 20 =


1.0100 0000 1100 1100 0010 0110 111(2) × 23


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): 3


Mantissa (not normalized):
1.0100 0000 1100 1100 0010 0110 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


3 + 2(8-1) - 1 =


(3 + 127)(10) =


130(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;
  • 130 ÷ 2 = 65 + 0;
  • 65 ÷ 2 = 32 + 1;
  • 32 ÷ 2 = 16 + 0;
  • 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) =


130(10) =


1000 0010(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 0000 0110 0110 0001 0011 0111 =


010 0000 0110 0110 0001 0011


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 0010


Mantissa (23 bits) =
010 0000 0110 0110 0001 0011


The base ten decimal number 10.024 920 89 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1000 0010 - 010 0000 0110 0110 0001 0011

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