-0.000 000 000 903 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 903(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 903(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 903| = 0.000 000 000 903


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 903.

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) 0.000 000 000 903 × 2 = 0 + 0.000 000 001 806;
  • 2) 0.000 000 001 806 × 2 = 0 + 0.000 000 003 612;
  • 3) 0.000 000 003 612 × 2 = 0 + 0.000 000 007 224;
  • 4) 0.000 000 007 224 × 2 = 0 + 0.000 000 014 448;
  • 5) 0.000 000 014 448 × 2 = 0 + 0.000 000 028 896;
  • 6) 0.000 000 028 896 × 2 = 0 + 0.000 000 057 792;
  • 7) 0.000 000 057 792 × 2 = 0 + 0.000 000 115 584;
  • 8) 0.000 000 115 584 × 2 = 0 + 0.000 000 231 168;
  • 9) 0.000 000 231 168 × 2 = 0 + 0.000 000 462 336;
  • 10) 0.000 000 462 336 × 2 = 0 + 0.000 000 924 672;
  • 11) 0.000 000 924 672 × 2 = 0 + 0.000 001 849 344;
  • 12) 0.000 001 849 344 × 2 = 0 + 0.000 003 698 688;
  • 13) 0.000 003 698 688 × 2 = 0 + 0.000 007 397 376;
  • 14) 0.000 007 397 376 × 2 = 0 + 0.000 014 794 752;
  • 15) 0.000 014 794 752 × 2 = 0 + 0.000 029 589 504;
  • 16) 0.000 029 589 504 × 2 = 0 + 0.000 059 179 008;
  • 17) 0.000 059 179 008 × 2 = 0 + 0.000 118 358 016;
  • 18) 0.000 118 358 016 × 2 = 0 + 0.000 236 716 032;
  • 19) 0.000 236 716 032 × 2 = 0 + 0.000 473 432 064;
  • 20) 0.000 473 432 064 × 2 = 0 + 0.000 946 864 128;
  • 21) 0.000 946 864 128 × 2 = 0 + 0.001 893 728 256;
  • 22) 0.001 893 728 256 × 2 = 0 + 0.003 787 456 512;
  • 23) 0.003 787 456 512 × 2 = 0 + 0.007 574 913 024;
  • 24) 0.007 574 913 024 × 2 = 0 + 0.015 149 826 048;
  • 25) 0.015 149 826 048 × 2 = 0 + 0.030 299 652 096;
  • 26) 0.030 299 652 096 × 2 = 0 + 0.060 599 304 192;
  • 27) 0.060 599 304 192 × 2 = 0 + 0.121 198 608 384;
  • 28) 0.121 198 608 384 × 2 = 0 + 0.242 397 216 768;
  • 29) 0.242 397 216 768 × 2 = 0 + 0.484 794 433 536;
  • 30) 0.484 794 433 536 × 2 = 0 + 0.969 588 867 072;
  • 31) 0.969 588 867 072 × 2 = 1 + 0.939 177 734 144;
  • 32) 0.939 177 734 144 × 2 = 1 + 0.878 355 468 288;
  • 33) 0.878 355 468 288 × 2 = 1 + 0.756 710 936 576;
  • 34) 0.756 710 936 576 × 2 = 1 + 0.513 421 873 152;
  • 35) 0.513 421 873 152 × 2 = 1 + 0.026 843 746 304;
  • 36) 0.026 843 746 304 × 2 = 0 + 0.053 687 492 608;
  • 37) 0.053 687 492 608 × 2 = 0 + 0.107 374 985 216;
  • 38) 0.107 374 985 216 × 2 = 0 + 0.214 749 970 432;
  • 39) 0.214 749 970 432 × 2 = 0 + 0.429 499 940 864;
  • 40) 0.429 499 940 864 × 2 = 0 + 0.858 999 881 728;
  • 41) 0.858 999 881 728 × 2 = 1 + 0.717 999 763 456;
  • 42) 0.717 999 763 456 × 2 = 1 + 0.435 999 526 912;
  • 43) 0.435 999 526 912 × 2 = 0 + 0.871 999 053 824;
  • 44) 0.871 999 053 824 × 2 = 1 + 0.743 998 107 648;
  • 45) 0.743 998 107 648 × 2 = 1 + 0.487 996 215 296;
  • 46) 0.487 996 215 296 × 2 = 0 + 0.975 992 430 592;
  • 47) 0.975 992 430 592 × 2 = 1 + 0.951 984 861 184;
  • 48) 0.951 984 861 184 × 2 = 1 + 0.903 969 722 368;
  • 49) 0.903 969 722 368 × 2 = 1 + 0.807 939 444 736;
  • 50) 0.807 939 444 736 × 2 = 1 + 0.615 878 889 472;
  • 51) 0.615 878 889 472 × 2 = 1 + 0.231 757 778 944;
  • 52) 0.231 757 778 944 × 2 = 0 + 0.463 515 557 888;
  • 53) 0.463 515 557 888 × 2 = 0 + 0.927 031 115 776;
  • 54) 0.927 031 115 776 × 2 = 1 + 0.854 062 231 552;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


5. 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:


0.000 000 000 903(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1110 0000 1101 1011 1110 01(2)

6. Positive number before normalization:

0.000 000 000 903(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1110 0000 1101 1011 1110 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 903(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1110 0000 1101 1011 1110 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1110 0000 1101 1011 1110 01(2) × 20 =

1.1111 0000 0110 1101 1111 001(2) × 2-31


8. 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 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1111 0000 0110 1101 1111 001


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. 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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. 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. 111 1000 0011 0110 1111 1001 =

111 1000 0011 0110 1111 1001


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

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

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
111 1000 0011 0110 1111 1001


Decimal number -0.000 000 000 903 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 111 1000 0011 0110 1111 1001

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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