32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.012 069 940 567 05 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.012 069 940 567 05(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: 0.012 069 940 567 05.

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) 0.012 069 940 567 05 × 2 = 0 + 0.024 139 881 134 1;
  • 2) 0.024 139 881 134 1 × 2 = 0 + 0.048 279 762 268 2;
  • 3) 0.048 279 762 268 2 × 2 = 0 + 0.096 559 524 536 4;
  • 4) 0.096 559 524 536 4 × 2 = 0 + 0.193 119 049 072 8;
  • 5) 0.193 119 049 072 8 × 2 = 0 + 0.386 238 098 145 6;
  • 6) 0.386 238 098 145 6 × 2 = 0 + 0.772 476 196 291 2;
  • 7) 0.772 476 196 291 2 × 2 = 1 + 0.544 952 392 582 4;
  • 8) 0.544 952 392 582 4 × 2 = 1 + 0.089 904 785 164 8;
  • 9) 0.089 904 785 164 8 × 2 = 0 + 0.179 809 570 329 6;
  • 10) 0.179 809 570 329 6 × 2 = 0 + 0.359 619 140 659 2;
  • 11) 0.359 619 140 659 2 × 2 = 0 + 0.719 238 281 318 4;
  • 12) 0.719 238 281 318 4 × 2 = 1 + 0.438 476 562 636 8;
  • 13) 0.438 476 562 636 8 × 2 = 0 + 0.876 953 125 273 6;
  • 14) 0.876 953 125 273 6 × 2 = 1 + 0.753 906 250 547 2;
  • 15) 0.753 906 250 547 2 × 2 = 1 + 0.507 812 501 094 4;
  • 16) 0.507 812 501 094 4 × 2 = 1 + 0.015 625 002 188 8;
  • 17) 0.015 625 002 188 8 × 2 = 0 + 0.031 250 004 377 6;
  • 18) 0.031 250 004 377 6 × 2 = 0 + 0.062 500 008 755 2;
  • 19) 0.062 500 008 755 2 × 2 = 0 + 0.125 000 017 510 4;
  • 20) 0.125 000 017 510 4 × 2 = 0 + 0.250 000 035 020 8;
  • 21) 0.250 000 035 020 8 × 2 = 0 + 0.500 000 070 041 6;
  • 22) 0.500 000 070 041 6 × 2 = 1 + 0.000 000 140 083 2;
  • 23) 0.000 000 140 083 2 × 2 = 0 + 0.000 000 280 166 4;
  • 24) 0.000 000 280 166 4 × 2 = 0 + 0.000 000 560 332 8;
  • 25) 0.000 000 560 332 8 × 2 = 0 + 0.000 001 120 665 6;
  • 26) 0.000 001 120 665 6 × 2 = 0 + 0.000 002 241 331 2;
  • 27) 0.000 002 241 331 2 × 2 = 0 + 0.000 004 482 662 4;
  • 28) 0.000 004 482 662 4 × 2 = 0 + 0.000 008 965 324 8;
  • 29) 0.000 008 965 324 8 × 2 = 0 + 0.000 017 930 649 6;
  • 30) 0.000 017 930 649 6 × 2 = 0 + 0.000 035 861 299 2;

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:


0.012 069 940 567 05(10) =

0.0000 0011 0001 0111 0000 0100 0000 00(2)


5. Positive number before normalization:

0.012 069 940 567 05(10) =

0.0000 0011 0001 0111 0000 0100 0000 00(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right, so that only one non zero digit remains to the left of it:


0.012 069 940 567 05(10) =

0.0000 0011 0001 0111 0000 0100 0000 00(2) =

0.0000 0011 0001 0111 0000 0100 0000 00(2) × 20 =

1.1000 1011 1000 0010 0000 000(2) × 2-7


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): -7

Mantissa (not normalized):
1.1000 1011 1000 0010 0000 000


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-7 + 2(8-1) - 1 =

(-7 + 127)(10) =

120(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;
  • 120 ÷ 2 = 60 + 0;
  • 60 ÷ 2 = 30 + 0;
  • 30 ÷ 2 = 15 + 0;
  • 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) =

120(10) =

0111 1000(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 0101 1100 0001 0000 0000 =

100 0101 1100 0001 0000 0000


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 1000

Mantissa (23 bits) =
100 0101 1100 0001 0000 0000


The base ten decimal number 0.012 069 940 567 05 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0111 1000 - 100 0101 1100 0001 0000 0000

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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