32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 64.015 64 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 64.015 64(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: 64.
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;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 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.


64(10) =


100 0000(2)


3. Convert to binary (base 2) the fractional part: 0.015 64.

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) 0.015 64 × 2 = 0 + 0.031 28;
  • 2) 0.031 28 × 2 = 0 + 0.062 56;
  • 3) 0.062 56 × 2 = 0 + 0.125 12;
  • 4) 0.125 12 × 2 = 0 + 0.250 24;
  • 5) 0.250 24 × 2 = 0 + 0.500 48;
  • 6) 0.500 48 × 2 = 1 + 0.000 96;
  • 7) 0.000 96 × 2 = 0 + 0.001 92;
  • 8) 0.001 92 × 2 = 0 + 0.003 84;
  • 9) 0.003 84 × 2 = 0 + 0.007 68;
  • 10) 0.007 68 × 2 = 0 + 0.015 36;
  • 11) 0.015 36 × 2 = 0 + 0.030 72;
  • 12) 0.030 72 × 2 = 0 + 0.061 44;
  • 13) 0.061 44 × 2 = 0 + 0.122 88;
  • 14) 0.122 88 × 2 = 0 + 0.245 76;
  • 15) 0.245 76 × 2 = 0 + 0.491 52;
  • 16) 0.491 52 × 2 = 0 + 0.983 04;
  • 17) 0.983 04 × 2 = 1 + 0.966 08;
  • 18) 0.966 08 × 2 = 1 + 0.932 16;
  • 19) 0.932 16 × 2 = 1 + 0.864 32;
  • 20) 0.864 32 × 2 = 1 + 0.728 64;
  • 21) 0.728 64 × 2 = 1 + 0.457 28;
  • 22) 0.457 28 × 2 = 0 + 0.914 56;
  • 23) 0.914 56 × 2 = 1 + 0.829 12;
  • 24) 0.829 12 × 2 = 1 + 0.658 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:


0.015 64(10) =


0.0000 0100 0000 0000 1111 1011(2)


5. Positive number before normalization:

64.015 64(10) =


100 0000.0000 0100 0000 0000 1111 1011(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:


64.015 64(10) =


100 0000.0000 0100 0000 0000 1111 1011(2) =


100 0000.0000 0100 0000 0000 1111 1011(2) × 20 =


1.0000 0000 0001 0000 0000 0011 1110 11(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):
1.0000 0000 0001 0000 0000 0011 1110 11


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. 000 0000 0000 1000 0000 0001 111 1011 =


000 0000 0000 1000 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) =
1000 0101


Mantissa (23 bits) =
000 0000 0000 1000 0000 0001


The base ten decimal number 64.015 64 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1000 0101 - 000 0000 0000 1000 0000 0001

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