32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.292 968 79 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.292 968 79(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: 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 binary (base 2) the fractional part: 0.292 968 79.

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) 0.292 968 79 × 2 = 0 + 0.585 937 58;
  • 2) 0.585 937 58 × 2 = 1 + 0.171 875 16;
  • 3) 0.171 875 16 × 2 = 0 + 0.343 750 32;
  • 4) 0.343 750 32 × 2 = 0 + 0.687 500 64;
  • 5) 0.687 500 64 × 2 = 1 + 0.375 001 28;
  • 6) 0.375 001 28 × 2 = 0 + 0.750 002 56;
  • 7) 0.750 002 56 × 2 = 1 + 0.500 005 12;
  • 8) 0.500 005 12 × 2 = 1 + 0.000 010 24;
  • 9) 0.000 010 24 × 2 = 0 + 0.000 020 48;
  • 10) 0.000 020 48 × 2 = 0 + 0.000 040 96;
  • 11) 0.000 040 96 × 2 = 0 + 0.000 081 92;
  • 12) 0.000 081 92 × 2 = 0 + 0.000 163 84;
  • 13) 0.000 163 84 × 2 = 0 + 0.000 327 68;
  • 14) 0.000 327 68 × 2 = 0 + 0.000 655 36;
  • 15) 0.000 655 36 × 2 = 0 + 0.001 310 72;
  • 16) 0.001 310 72 × 2 = 0 + 0.002 621 44;
  • 17) 0.002 621 44 × 2 = 0 + 0.005 242 88;
  • 18) 0.005 242 88 × 2 = 0 + 0.010 485 76;
  • 19) 0.010 485 76 × 2 = 0 + 0.020 971 52;
  • 20) 0.020 971 52 × 2 = 0 + 0.041 943 04;
  • 21) 0.041 943 04 × 2 = 0 + 0.083 886 08;
  • 22) 0.083 886 08 × 2 = 0 + 0.167 772 16;
  • 23) 0.167 772 16 × 2 = 0 + 0.335 544 32;
  • 24) 0.335 544 32 × 2 = 0 + 0.671 088 64;
  • 25) 0.671 088 64 × 2 = 1 + 0.342 177 28;

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:


0.292 968 79(10) =


0.0100 1011 0000 0000 0000 0000 1(2)


5. Positive number before normalization:

0.292 968 79(10) =


0.0100 1011 0000 0000 0000 0000 1(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:


0.292 968 79(10) =


0.0100 1011 0000 0000 0000 0000 1(2) =


0.0100 1011 0000 0000 0000 0000 1(2) × 20 =


1.0010 1100 0000 0000 0000 001(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):
1.0010 1100 0000 0000 0000 001


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. 001 0110 0000 0000 0000 0001 =


001 0110 0000 0000 0000 0001


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) =
001 0110 0000 0000 0000 0001


The base ten decimal number 0.292 968 79 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0111 1101 - 001 0110 0000 0000 0000 0001

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