-0.000 000 000 742 147 676 872 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 872(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
-0.000 000 000 742 147 676 872(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:

|-0.000 000 000 742 147 676 872| = 0.000 000 000 742 147 676 872


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: 0.000 000 000 742 147 676 872.

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) 0.000 000 000 742 147 676 872 × 2 = 0 + 0.000 000 001 484 295 353 744;
  • 2) 0.000 000 001 484 295 353 744 × 2 = 0 + 0.000 000 002 968 590 707 488;
  • 3) 0.000 000 002 968 590 707 488 × 2 = 0 + 0.000 000 005 937 181 414 976;
  • 4) 0.000 000 005 937 181 414 976 × 2 = 0 + 0.000 000 011 874 362 829 952;
  • 5) 0.000 000 011 874 362 829 952 × 2 = 0 + 0.000 000 023 748 725 659 904;
  • 6) 0.000 000 023 748 725 659 904 × 2 = 0 + 0.000 000 047 497 451 319 808;
  • 7) 0.000 000 047 497 451 319 808 × 2 = 0 + 0.000 000 094 994 902 639 616;
  • 8) 0.000 000 094 994 902 639 616 × 2 = 0 + 0.000 000 189 989 805 279 232;
  • 9) 0.000 000 189 989 805 279 232 × 2 = 0 + 0.000 000 379 979 610 558 464;
  • 10) 0.000 000 379 979 610 558 464 × 2 = 0 + 0.000 000 759 959 221 116 928;
  • 11) 0.000 000 759 959 221 116 928 × 2 = 0 + 0.000 001 519 918 442 233 856;
  • 12) 0.000 001 519 918 442 233 856 × 2 = 0 + 0.000 003 039 836 884 467 712;
  • 13) 0.000 003 039 836 884 467 712 × 2 = 0 + 0.000 006 079 673 768 935 424;
  • 14) 0.000 006 079 673 768 935 424 × 2 = 0 + 0.000 012 159 347 537 870 848;
  • 15) 0.000 012 159 347 537 870 848 × 2 = 0 + 0.000 024 318 695 075 741 696;
  • 16) 0.000 024 318 695 075 741 696 × 2 = 0 + 0.000 048 637 390 151 483 392;
  • 17) 0.000 048 637 390 151 483 392 × 2 = 0 + 0.000 097 274 780 302 966 784;
  • 18) 0.000 097 274 780 302 966 784 × 2 = 0 + 0.000 194 549 560 605 933 568;
  • 19) 0.000 194 549 560 605 933 568 × 2 = 0 + 0.000 389 099 121 211 867 136;
  • 20) 0.000 389 099 121 211 867 136 × 2 = 0 + 0.000 778 198 242 423 734 272;
  • 21) 0.000 778 198 242 423 734 272 × 2 = 0 + 0.001 556 396 484 847 468 544;
  • 22) 0.001 556 396 484 847 468 544 × 2 = 0 + 0.003 112 792 969 694 937 088;
  • 23) 0.003 112 792 969 694 937 088 × 2 = 0 + 0.006 225 585 939 389 874 176;
  • 24) 0.006 225 585 939 389 874 176 × 2 = 0 + 0.012 451 171 878 779 748 352;
  • 25) 0.012 451 171 878 779 748 352 × 2 = 0 + 0.024 902 343 757 559 496 704;
  • 26) 0.024 902 343 757 559 496 704 × 2 = 0 + 0.049 804 687 515 118 993 408;
  • 27) 0.049 804 687 515 118 993 408 × 2 = 0 + 0.099 609 375 030 237 986 816;
  • 28) 0.099 609 375 030 237 986 816 × 2 = 0 + 0.199 218 750 060 475 973 632;
  • 29) 0.199 218 750 060 475 973 632 × 2 = 0 + 0.398 437 500 120 951 947 264;
  • 30) 0.398 437 500 120 951 947 264 × 2 = 0 + 0.796 875 000 241 903 894 528;
  • 31) 0.796 875 000 241 903 894 528 × 2 = 1 + 0.593 750 000 483 807 789 056;
  • 32) 0.593 750 000 483 807 789 056 × 2 = 1 + 0.187 500 000 967 615 578 112;
  • 33) 0.187 500 000 967 615 578 112 × 2 = 0 + 0.375 000 001 935 231 156 224;
  • 34) 0.375 000 001 935 231 156 224 × 2 = 0 + 0.750 000 003 870 462 312 448;
  • 35) 0.750 000 003 870 462 312 448 × 2 = 1 + 0.500 000 007 740 924 624 896;
  • 36) 0.500 000 007 740 924 624 896 × 2 = 1 + 0.000 000 015 481 849 249 792;
  • 37) 0.000 000 015 481 849 249 792 × 2 = 0 + 0.000 000 030 963 698 499 584;
  • 38) 0.000 000 030 963 698 499 584 × 2 = 0 + 0.000 000 061 927 396 999 168;
  • 39) 0.000 000 061 927 396 999 168 × 2 = 0 + 0.000 000 123 854 793 998 336;
  • 40) 0.000 000 123 854 793 998 336 × 2 = 0 + 0.000 000 247 709 587 996 672;
  • 41) 0.000 000 247 709 587 996 672 × 2 = 0 + 0.000 000 495 419 175 993 344;
  • 42) 0.000 000 495 419 175 993 344 × 2 = 0 + 0.000 000 990 838 351 986 688;
  • 43) 0.000 000 990 838 351 986 688 × 2 = 0 + 0.000 001 981 676 703 973 376;
  • 44) 0.000 001 981 676 703 973 376 × 2 = 0 + 0.000 003 963 353 407 946 752;
  • 45) 0.000 003 963 353 407 946 752 × 2 = 0 + 0.000 007 926 706 815 893 504;
  • 46) 0.000 007 926 706 815 893 504 × 2 = 0 + 0.000 015 853 413 631 787 008;
  • 47) 0.000 015 853 413 631 787 008 × 2 = 0 + 0.000 031 706 827 263 574 016;
  • 48) 0.000 031 706 827 263 574 016 × 2 = 0 + 0.000 063 413 654 527 148 032;
  • 49) 0.000 063 413 654 527 148 032 × 2 = 0 + 0.000 126 827 309 054 296 064;
  • 50) 0.000 126 827 309 054 296 064 × 2 = 0 + 0.000 253 654 618 108 592 128;
  • 51) 0.000 253 654 618 108 592 128 × 2 = 0 + 0.000 507 309 236 217 184 256;
  • 52) 0.000 507 309 236 217 184 256 × 2 = 0 + 0.001 014 618 472 434 368 512;
  • 53) 0.001 014 618 472 434 368 512 × 2 = 0 + 0.002 029 236 944 868 737 024;
  • 54) 0.002 029 236 944 868 737 024 × 2 = 0 + 0.004 058 473 889 737 474 048;

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:


0.000 000 000 742 147 676 872(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 872(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:


0.000 000 000 742 147 676 872(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 -0.000 000 000 742 147 676 872 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