32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 1 027 788.421 023 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 1 027 788.421 023(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: 1 027 788.
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;
  • 1 027 788 ÷ 2 = 513 894 + 0;
  • 513 894 ÷ 2 = 256 947 + 0;
  • 256 947 ÷ 2 = 128 473 + 1;
  • 128 473 ÷ 2 = 64 236 + 1;
  • 64 236 ÷ 2 = 32 118 + 0;
  • 32 118 ÷ 2 = 16 059 + 0;
  • 16 059 ÷ 2 = 8 029 + 1;
  • 8 029 ÷ 2 = 4 014 + 1;
  • 4 014 ÷ 2 = 2 007 + 0;
  • 2 007 ÷ 2 = 1 003 + 1;
  • 1 003 ÷ 2 = 501 + 1;
  • 501 ÷ 2 = 250 + 1;
  • 250 ÷ 2 = 125 + 0;
  • 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;

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.


1 027 788(10) =

1111 1010 1110 1100 1100(2)


3. Convert to binary (base 2) the fractional part: 0.421 023.

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) 0.421 023 × 2 = 0 + 0.842 046;
  • 2) 0.842 046 × 2 = 1 + 0.684 092;
  • 3) 0.684 092 × 2 = 1 + 0.368 184;
  • 4) 0.368 184 × 2 = 0 + 0.736 368;
  • 5) 0.736 368 × 2 = 1 + 0.472 736;
  • 6) 0.472 736 × 2 = 0 + 0.945 472;
  • 7) 0.945 472 × 2 = 1 + 0.890 944;
  • 8) 0.890 944 × 2 = 1 + 0.781 888;
  • 9) 0.781 888 × 2 = 1 + 0.563 776;
  • 10) 0.563 776 × 2 = 1 + 0.127 552;
  • 11) 0.127 552 × 2 = 0 + 0.255 104;
  • 12) 0.255 104 × 2 = 0 + 0.510 208;
  • 13) 0.510 208 × 2 = 1 + 0.020 416;
  • 14) 0.020 416 × 2 = 0 + 0.040 832;
  • 15) 0.040 832 × 2 = 0 + 0.081 664;
  • 16) 0.081 664 × 2 = 0 + 0.163 328;
  • 17) 0.163 328 × 2 = 0 + 0.326 656;
  • 18) 0.326 656 × 2 = 0 + 0.653 312;
  • 19) 0.653 312 × 2 = 1 + 0.306 624;
  • 20) 0.306 624 × 2 = 0 + 0.613 248;
  • 21) 0.613 248 × 2 = 1 + 0.226 496;
  • 22) 0.226 496 × 2 = 0 + 0.452 992;
  • 23) 0.452 992 × 2 = 0 + 0.905 984;
  • 24) 0.905 984 × 2 = 1 + 0.811 968;

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:


0.421 023(10) =

0.0110 1011 1100 1000 0010 1001(2)


5. Positive number before normalization:

1 027 788.421 023(10) =

1111 1010 1110 1100 1100.0110 1011 1100 1000 0010 1001(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 19 positions to the left, so that only one non zero digit remains to the left of it:


1 027 788.421 023(10) =

1111 1010 1110 1100 1100.0110 1011 1100 1000 0010 1001(2) =

1111 1010 1110 1100 1100.0110 1011 1100 1000 0010 1001(2) × 20 =

1.1111 0101 1101 1001 1000 1101 0111 1001 0000 0101 001(2) × 219


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): 19

Mantissa (not normalized):
1.1111 0101 1101 1001 1000 1101 0111 1001 0000 0101 001


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

19 + 2(8-1) - 1 =

(19 + 127)(10) =

146(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;
  • 146 ÷ 2 = 73 + 0;
  • 73 ÷ 2 = 36 + 1;
  • 36 ÷ 2 = 18 + 0;
  • 18 ÷ 2 = 9 + 0;
  • 9 ÷ 2 = 4 + 1;
  • 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) =

146(10) =

1001 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. 111 1010 1110 1100 1100 0110 1011 1100 1000 0010 1001 =

111 1010 1110 1100 1100 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) =
1001 0010

Mantissa (23 bits) =
111 1010 1110 1100 1100 0110


The base ten decimal number 1 027 788.421 023 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1001 0010 - 111 1010 1110 1100 1100 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