32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.000 006 12 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.000 006 12(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: 0.000 006 12.

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) 0.000 006 12 × 2 = 0 + 0.000 012 24;
  • 2) 0.000 012 24 × 2 = 0 + 0.000 024 48;
  • 3) 0.000 024 48 × 2 = 0 + 0.000 048 96;
  • 4) 0.000 048 96 × 2 = 0 + 0.000 097 92;
  • 5) 0.000 097 92 × 2 = 0 + 0.000 195 84;
  • 6) 0.000 195 84 × 2 = 0 + 0.000 391 68;
  • 7) 0.000 391 68 × 2 = 0 + 0.000 783 36;
  • 8) 0.000 783 36 × 2 = 0 + 0.001 566 72;
  • 9) 0.001 566 72 × 2 = 0 + 0.003 133 44;
  • 10) 0.003 133 44 × 2 = 0 + 0.006 266 88;
  • 11) 0.006 266 88 × 2 = 0 + 0.012 533 76;
  • 12) 0.012 533 76 × 2 = 0 + 0.025 067 52;
  • 13) 0.025 067 52 × 2 = 0 + 0.050 135 04;
  • 14) 0.050 135 04 × 2 = 0 + 0.100 270 08;
  • 15) 0.100 270 08 × 2 = 0 + 0.200 540 16;
  • 16) 0.200 540 16 × 2 = 0 + 0.401 080 32;
  • 17) 0.401 080 32 × 2 = 0 + 0.802 160 64;
  • 18) 0.802 160 64 × 2 = 1 + 0.604 321 28;
  • 19) 0.604 321 28 × 2 = 1 + 0.208 642 56;
  • 20) 0.208 642 56 × 2 = 0 + 0.417 285 12;
  • 21) 0.417 285 12 × 2 = 0 + 0.834 570 24;
  • 22) 0.834 570 24 × 2 = 1 + 0.669 140 48;
  • 23) 0.669 140 48 × 2 = 1 + 0.338 280 96;
  • 24) 0.338 280 96 × 2 = 0 + 0.676 561 92;
  • 25) 0.676 561 92 × 2 = 1 + 0.353 123 84;
  • 26) 0.353 123 84 × 2 = 0 + 0.706 247 68;
  • 27) 0.706 247 68 × 2 = 1 + 0.412 495 36;
  • 28) 0.412 495 36 × 2 = 0 + 0.824 990 72;
  • 29) 0.824 990 72 × 2 = 1 + 0.649 981 44;
  • 30) 0.649 981 44 × 2 = 1 + 0.299 962 88;
  • 31) 0.299 962 88 × 2 = 0 + 0.599 925 76;
  • 32) 0.599 925 76 × 2 = 1 + 0.199 851 52;
  • 33) 0.199 851 52 × 2 = 0 + 0.399 703 04;
  • 34) 0.399 703 04 × 2 = 0 + 0.799 406 08;
  • 35) 0.799 406 08 × 2 = 1 + 0.598 812 16;
  • 36) 0.598 812 16 × 2 = 1 + 0.197 624 32;
  • 37) 0.197 624 32 × 2 = 0 + 0.395 248 64;
  • 38) 0.395 248 64 × 2 = 0 + 0.790 497 28;
  • 39) 0.790 497 28 × 2 = 1 + 0.580 994 56;
  • 40) 0.580 994 56 × 2 = 1 + 0.161 989 12;
  • 41) 0.161 989 12 × 2 = 0 + 0.323 978 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:


0.000 006 12(10) =

0.0000 0000 0000 0000 0110 0110 1010 1101 0011 0011 0(2)


5. Positive number before normalization:

0.000 006 12(10) =

0.0000 0000 0000 0000 0110 0110 1010 1101 0011 0011 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 18 positions to the right, so that only one non zero digit remains to the left of it:


0.000 006 12(10) =

0.0000 0000 0000 0000 0110 0110 1010 1101 0011 0011 0(2) =

0.0000 0000 0000 0000 0110 0110 1010 1101 0011 0011 0(2) × 20 =

1.1001 1010 1011 0100 1100 110(2) × 2-18


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): -18

Mantissa (not normalized):
1.1001 1010 1011 0100 1100 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-18 + 2(8-1) - 1 =

(-18 + 127)(10) =

109(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;
  • 109 ÷ 2 = 54 + 1;
  • 54 ÷ 2 = 27 + 0;
  • 27 ÷ 2 = 13 + 1;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

109(10) =

0110 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. 100 1101 0101 1010 0110 0110 =

100 1101 0101 1010 0110 0110


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) =
0110 1101

Mantissa (23 bits) =
100 1101 0101 1010 0110 0110


The base ten decimal number 0.000 006 12 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0110 1101 - 100 1101 0101 1010 0110 0110

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111