-0.000 000 000 742 147 685 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 685(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
-0.000 000 000 742 147 685(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:

|-0.000 000 000 742 147 685| = 0.000 000 000 742 147 685


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: 0.000 000 000 742 147 685.

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) 0.000 000 000 742 147 685 × 2 = 0 + 0.000 000 001 484 295 37;
  • 2) 0.000 000 001 484 295 37 × 2 = 0 + 0.000 000 002 968 590 74;
  • 3) 0.000 000 002 968 590 74 × 2 = 0 + 0.000 000 005 937 181 48;
  • 4) 0.000 000 005 937 181 48 × 2 = 0 + 0.000 000 011 874 362 96;
  • 5) 0.000 000 011 874 362 96 × 2 = 0 + 0.000 000 023 748 725 92;
  • 6) 0.000 000 023 748 725 92 × 2 = 0 + 0.000 000 047 497 451 84;
  • 7) 0.000 000 047 497 451 84 × 2 = 0 + 0.000 000 094 994 903 68;
  • 8) 0.000 000 094 994 903 68 × 2 = 0 + 0.000 000 189 989 807 36;
  • 9) 0.000 000 189 989 807 36 × 2 = 0 + 0.000 000 379 979 614 72;
  • 10) 0.000 000 379 979 614 72 × 2 = 0 + 0.000 000 759 959 229 44;
  • 11) 0.000 000 759 959 229 44 × 2 = 0 + 0.000 001 519 918 458 88;
  • 12) 0.000 001 519 918 458 88 × 2 = 0 + 0.000 003 039 836 917 76;
  • 13) 0.000 003 039 836 917 76 × 2 = 0 + 0.000 006 079 673 835 52;
  • 14) 0.000 006 079 673 835 52 × 2 = 0 + 0.000 012 159 347 671 04;
  • 15) 0.000 012 159 347 671 04 × 2 = 0 + 0.000 024 318 695 342 08;
  • 16) 0.000 024 318 695 342 08 × 2 = 0 + 0.000 048 637 390 684 16;
  • 17) 0.000 048 637 390 684 16 × 2 = 0 + 0.000 097 274 781 368 32;
  • 18) 0.000 097 274 781 368 32 × 2 = 0 + 0.000 194 549 562 736 64;
  • 19) 0.000 194 549 562 736 64 × 2 = 0 + 0.000 389 099 125 473 28;
  • 20) 0.000 389 099 125 473 28 × 2 = 0 + 0.000 778 198 250 946 56;
  • 21) 0.000 778 198 250 946 56 × 2 = 0 + 0.001 556 396 501 893 12;
  • 22) 0.001 556 396 501 893 12 × 2 = 0 + 0.003 112 793 003 786 24;
  • 23) 0.003 112 793 003 786 24 × 2 = 0 + 0.006 225 586 007 572 48;
  • 24) 0.006 225 586 007 572 48 × 2 = 0 + 0.012 451 172 015 144 96;
  • 25) 0.012 451 172 015 144 96 × 2 = 0 + 0.024 902 344 030 289 92;
  • 26) 0.024 902 344 030 289 92 × 2 = 0 + 0.049 804 688 060 579 84;
  • 27) 0.049 804 688 060 579 84 × 2 = 0 + 0.099 609 376 121 159 68;
  • 28) 0.099 609 376 121 159 68 × 2 = 0 + 0.199 218 752 242 319 36;
  • 29) 0.199 218 752 242 319 36 × 2 = 0 + 0.398 437 504 484 638 72;
  • 30) 0.398 437 504 484 638 72 × 2 = 0 + 0.796 875 008 969 277 44;
  • 31) 0.796 875 008 969 277 44 × 2 = 1 + 0.593 750 017 938 554 88;
  • 32) 0.593 750 017 938 554 88 × 2 = 1 + 0.187 500 035 877 109 76;
  • 33) 0.187 500 035 877 109 76 × 2 = 0 + 0.375 000 071 754 219 52;
  • 34) 0.375 000 071 754 219 52 × 2 = 0 + 0.750 000 143 508 439 04;
  • 35) 0.750 000 143 508 439 04 × 2 = 1 + 0.500 000 287 016 878 08;
  • 36) 0.500 000 287 016 878 08 × 2 = 1 + 0.000 000 574 033 756 16;
  • 37) 0.000 000 574 033 756 16 × 2 = 0 + 0.000 001 148 067 512 32;
  • 38) 0.000 001 148 067 512 32 × 2 = 0 + 0.000 002 296 135 024 64;
  • 39) 0.000 002 296 135 024 64 × 2 = 0 + 0.000 004 592 270 049 28;
  • 40) 0.000 004 592 270 049 28 × 2 = 0 + 0.000 009 184 540 098 56;
  • 41) 0.000 009 184 540 098 56 × 2 = 0 + 0.000 018 369 080 197 12;
  • 42) 0.000 018 369 080 197 12 × 2 = 0 + 0.000 036 738 160 394 24;
  • 43) 0.000 036 738 160 394 24 × 2 = 0 + 0.000 073 476 320 788 48;
  • 44) 0.000 073 476 320 788 48 × 2 = 0 + 0.000 146 952 641 576 96;
  • 45) 0.000 146 952 641 576 96 × 2 = 0 + 0.000 293 905 283 153 92;
  • 46) 0.000 293 905 283 153 92 × 2 = 0 + 0.000 587 810 566 307 84;
  • 47) 0.000 587 810 566 307 84 × 2 = 0 + 0.001 175 621 132 615 68;
  • 48) 0.001 175 621 132 615 68 × 2 = 0 + 0.002 351 242 265 231 36;
  • 49) 0.002 351 242 265 231 36 × 2 = 0 + 0.004 702 484 530 462 72;
  • 50) 0.004 702 484 530 462 72 × 2 = 0 + 0.009 404 969 060 925 44;
  • 51) 0.009 404 969 060 925 44 × 2 = 0 + 0.018 809 938 121 850 88;
  • 52) 0.018 809 938 121 850 88 × 2 = 0 + 0.037 619 876 243 701 76;
  • 53) 0.037 619 876 243 701 76 × 2 = 0 + 0.075 239 752 487 403 52;
  • 54) 0.075 239 752 487 403 52 × 2 = 0 + 0.150 479 504 974 807 04;

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:


0.000 000 000 742 147 685(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 685(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 685(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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):
1.1001 1000 0000 0000 0000 000


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. 100 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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) =
100 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 685 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 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