-0.000 000 000 742 147 675 854 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 854(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
-0.000 000 000 742 147 675 854(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:

|-0.000 000 000 742 147 675 854| = 0.000 000 000 742 147 675 854


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: 0.000 000 000 742 147 675 854.

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) 0.000 000 000 742 147 675 854 × 2 = 0 + 0.000 000 001 484 295 351 708;
  • 2) 0.000 000 001 484 295 351 708 × 2 = 0 + 0.000 000 002 968 590 703 416;
  • 3) 0.000 000 002 968 590 703 416 × 2 = 0 + 0.000 000 005 937 181 406 832;
  • 4) 0.000 000 005 937 181 406 832 × 2 = 0 + 0.000 000 011 874 362 813 664;
  • 5) 0.000 000 011 874 362 813 664 × 2 = 0 + 0.000 000 023 748 725 627 328;
  • 6) 0.000 000 023 748 725 627 328 × 2 = 0 + 0.000 000 047 497 451 254 656;
  • 7) 0.000 000 047 497 451 254 656 × 2 = 0 + 0.000 000 094 994 902 509 312;
  • 8) 0.000 000 094 994 902 509 312 × 2 = 0 + 0.000 000 189 989 805 018 624;
  • 9) 0.000 000 189 989 805 018 624 × 2 = 0 + 0.000 000 379 979 610 037 248;
  • 10) 0.000 000 379 979 610 037 248 × 2 = 0 + 0.000 000 759 959 220 074 496;
  • 11) 0.000 000 759 959 220 074 496 × 2 = 0 + 0.000 001 519 918 440 148 992;
  • 12) 0.000 001 519 918 440 148 992 × 2 = 0 + 0.000 003 039 836 880 297 984;
  • 13) 0.000 003 039 836 880 297 984 × 2 = 0 + 0.000 006 079 673 760 595 968;
  • 14) 0.000 006 079 673 760 595 968 × 2 = 0 + 0.000 012 159 347 521 191 936;
  • 15) 0.000 012 159 347 521 191 936 × 2 = 0 + 0.000 024 318 695 042 383 872;
  • 16) 0.000 024 318 695 042 383 872 × 2 = 0 + 0.000 048 637 390 084 767 744;
  • 17) 0.000 048 637 390 084 767 744 × 2 = 0 + 0.000 097 274 780 169 535 488;
  • 18) 0.000 097 274 780 169 535 488 × 2 = 0 + 0.000 194 549 560 339 070 976;
  • 19) 0.000 194 549 560 339 070 976 × 2 = 0 + 0.000 389 099 120 678 141 952;
  • 20) 0.000 389 099 120 678 141 952 × 2 = 0 + 0.000 778 198 241 356 283 904;
  • 21) 0.000 778 198 241 356 283 904 × 2 = 0 + 0.001 556 396 482 712 567 808;
  • 22) 0.001 556 396 482 712 567 808 × 2 = 0 + 0.003 112 792 965 425 135 616;
  • 23) 0.003 112 792 965 425 135 616 × 2 = 0 + 0.006 225 585 930 850 271 232;
  • 24) 0.006 225 585 930 850 271 232 × 2 = 0 + 0.012 451 171 861 700 542 464;
  • 25) 0.012 451 171 861 700 542 464 × 2 = 0 + 0.024 902 343 723 401 084 928;
  • 26) 0.024 902 343 723 401 084 928 × 2 = 0 + 0.049 804 687 446 802 169 856;
  • 27) 0.049 804 687 446 802 169 856 × 2 = 0 + 0.099 609 374 893 604 339 712;
  • 28) 0.099 609 374 893 604 339 712 × 2 = 0 + 0.199 218 749 787 208 679 424;
  • 29) 0.199 218 749 787 208 679 424 × 2 = 0 + 0.398 437 499 574 417 358 848;
  • 30) 0.398 437 499 574 417 358 848 × 2 = 0 + 0.796 874 999 148 834 717 696;
  • 31) 0.796 874 999 148 834 717 696 × 2 = 1 + 0.593 749 998 297 669 435 392;
  • 32) 0.593 749 998 297 669 435 392 × 2 = 1 + 0.187 499 996 595 338 870 784;
  • 33) 0.187 499 996 595 338 870 784 × 2 = 0 + 0.374 999 993 190 677 741 568;
  • 34) 0.374 999 993 190 677 741 568 × 2 = 0 + 0.749 999 986 381 355 483 136;
  • 35) 0.749 999 986 381 355 483 136 × 2 = 1 + 0.499 999 972 762 710 966 272;
  • 36) 0.499 999 972 762 710 966 272 × 2 = 0 + 0.999 999 945 525 421 932 544;
  • 37) 0.999 999 945 525 421 932 544 × 2 = 1 + 0.999 999 891 050 843 865 088;
  • 38) 0.999 999 891 050 843 865 088 × 2 = 1 + 0.999 999 782 101 687 730 176;
  • 39) 0.999 999 782 101 687 730 176 × 2 = 1 + 0.999 999 564 203 375 460 352;
  • 40) 0.999 999 564 203 375 460 352 × 2 = 1 + 0.999 999 128 406 750 920 704;
  • 41) 0.999 999 128 406 750 920 704 × 2 = 1 + 0.999 998 256 813 501 841 408;
  • 42) 0.999 998 256 813 501 841 408 × 2 = 1 + 0.999 996 513 627 003 682 816;
  • 43) 0.999 996 513 627 003 682 816 × 2 = 1 + 0.999 993 027 254 007 365 632;
  • 44) 0.999 993 027 254 007 365 632 × 2 = 1 + 0.999 986 054 508 014 731 264;
  • 45) 0.999 986 054 508 014 731 264 × 2 = 1 + 0.999 972 109 016 029 462 528;
  • 46) 0.999 972 109 016 029 462 528 × 2 = 1 + 0.999 944 218 032 058 925 056;
  • 47) 0.999 944 218 032 058 925 056 × 2 = 1 + 0.999 888 436 064 117 850 112;
  • 48) 0.999 888 436 064 117 850 112 × 2 = 1 + 0.999 776 872 128 235 700 224;
  • 49) 0.999 776 872 128 235 700 224 × 2 = 1 + 0.999 553 744 256 471 400 448;
  • 50) 0.999 553 744 256 471 400 448 × 2 = 1 + 0.999 107 488 512 942 800 896;
  • 51) 0.999 107 488 512 942 800 896 × 2 = 1 + 0.998 214 977 025 885 601 792;
  • 52) 0.998 214 977 025 885 601 792 × 2 = 1 + 0.996 429 954 051 771 203 584;
  • 53) 0.996 429 954 051 771 203 584 × 2 = 1 + 0.992 859 908 103 542 407 168;
  • 54) 0.992 859 908 103 542 407 168 × 2 = 1 + 0.985 719 816 207 084 814 336;

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:


0.000 000 000 742 147 675 854(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 675 854(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(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:


0.000 000 000 742 147 675 854(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 0111 1111 1111 1111 111


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 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


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 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 675 854 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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