-0.000 000 000 742 147 675 863 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 863(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
-0.000 000 000 742 147 675 863(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:

|-0.000 000 000 742 147 675 863| = 0.000 000 000 742 147 675 863


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: 0.000 000 000 742 147 675 863.

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) 0.000 000 000 742 147 675 863 × 2 = 0 + 0.000 000 001 484 295 351 726;
  • 2) 0.000 000 001 484 295 351 726 × 2 = 0 + 0.000 000 002 968 590 703 452;
  • 3) 0.000 000 002 968 590 703 452 × 2 = 0 + 0.000 000 005 937 181 406 904;
  • 4) 0.000 000 005 937 181 406 904 × 2 = 0 + 0.000 000 011 874 362 813 808;
  • 5) 0.000 000 011 874 362 813 808 × 2 = 0 + 0.000 000 023 748 725 627 616;
  • 6) 0.000 000 023 748 725 627 616 × 2 = 0 + 0.000 000 047 497 451 255 232;
  • 7) 0.000 000 047 497 451 255 232 × 2 = 0 + 0.000 000 094 994 902 510 464;
  • 8) 0.000 000 094 994 902 510 464 × 2 = 0 + 0.000 000 189 989 805 020 928;
  • 9) 0.000 000 189 989 805 020 928 × 2 = 0 + 0.000 000 379 979 610 041 856;
  • 10) 0.000 000 379 979 610 041 856 × 2 = 0 + 0.000 000 759 959 220 083 712;
  • 11) 0.000 000 759 959 220 083 712 × 2 = 0 + 0.000 001 519 918 440 167 424;
  • 12) 0.000 001 519 918 440 167 424 × 2 = 0 + 0.000 003 039 836 880 334 848;
  • 13) 0.000 003 039 836 880 334 848 × 2 = 0 + 0.000 006 079 673 760 669 696;
  • 14) 0.000 006 079 673 760 669 696 × 2 = 0 + 0.000 012 159 347 521 339 392;
  • 15) 0.000 012 159 347 521 339 392 × 2 = 0 + 0.000 024 318 695 042 678 784;
  • 16) 0.000 024 318 695 042 678 784 × 2 = 0 + 0.000 048 637 390 085 357 568;
  • 17) 0.000 048 637 390 085 357 568 × 2 = 0 + 0.000 097 274 780 170 715 136;
  • 18) 0.000 097 274 780 170 715 136 × 2 = 0 + 0.000 194 549 560 341 430 272;
  • 19) 0.000 194 549 560 341 430 272 × 2 = 0 + 0.000 389 099 120 682 860 544;
  • 20) 0.000 389 099 120 682 860 544 × 2 = 0 + 0.000 778 198 241 365 721 088;
  • 21) 0.000 778 198 241 365 721 088 × 2 = 0 + 0.001 556 396 482 731 442 176;
  • 22) 0.001 556 396 482 731 442 176 × 2 = 0 + 0.003 112 792 965 462 884 352;
  • 23) 0.003 112 792 965 462 884 352 × 2 = 0 + 0.006 225 585 930 925 768 704;
  • 24) 0.006 225 585 930 925 768 704 × 2 = 0 + 0.012 451 171 861 851 537 408;
  • 25) 0.012 451 171 861 851 537 408 × 2 = 0 + 0.024 902 343 723 703 074 816;
  • 26) 0.024 902 343 723 703 074 816 × 2 = 0 + 0.049 804 687 447 406 149 632;
  • 27) 0.049 804 687 447 406 149 632 × 2 = 0 + 0.099 609 374 894 812 299 264;
  • 28) 0.099 609 374 894 812 299 264 × 2 = 0 + 0.199 218 749 789 624 598 528;
  • 29) 0.199 218 749 789 624 598 528 × 2 = 0 + 0.398 437 499 579 249 197 056;
  • 30) 0.398 437 499 579 249 197 056 × 2 = 0 + 0.796 874 999 158 498 394 112;
  • 31) 0.796 874 999 158 498 394 112 × 2 = 1 + 0.593 749 998 316 996 788 224;
  • 32) 0.593 749 998 316 996 788 224 × 2 = 1 + 0.187 499 996 633 993 576 448;
  • 33) 0.187 499 996 633 993 576 448 × 2 = 0 + 0.374 999 993 267 987 152 896;
  • 34) 0.374 999 993 267 987 152 896 × 2 = 0 + 0.749 999 986 535 974 305 792;
  • 35) 0.749 999 986 535 974 305 792 × 2 = 1 + 0.499 999 973 071 948 611 584;
  • 36) 0.499 999 973 071 948 611 584 × 2 = 0 + 0.999 999 946 143 897 223 168;
  • 37) 0.999 999 946 143 897 223 168 × 2 = 1 + 0.999 999 892 287 794 446 336;
  • 38) 0.999 999 892 287 794 446 336 × 2 = 1 + 0.999 999 784 575 588 892 672;
  • 39) 0.999 999 784 575 588 892 672 × 2 = 1 + 0.999 999 569 151 177 785 344;
  • 40) 0.999 999 569 151 177 785 344 × 2 = 1 + 0.999 999 138 302 355 570 688;
  • 41) 0.999 999 138 302 355 570 688 × 2 = 1 + 0.999 998 276 604 711 141 376;
  • 42) 0.999 998 276 604 711 141 376 × 2 = 1 + 0.999 996 553 209 422 282 752;
  • 43) 0.999 996 553 209 422 282 752 × 2 = 1 + 0.999 993 106 418 844 565 504;
  • 44) 0.999 993 106 418 844 565 504 × 2 = 1 + 0.999 986 212 837 689 131 008;
  • 45) 0.999 986 212 837 689 131 008 × 2 = 1 + 0.999 972 425 675 378 262 016;
  • 46) 0.999 972 425 675 378 262 016 × 2 = 1 + 0.999 944 851 350 756 524 032;
  • 47) 0.999 944 851 350 756 524 032 × 2 = 1 + 0.999 889 702 701 513 048 064;
  • 48) 0.999 889 702 701 513 048 064 × 2 = 1 + 0.999 779 405 403 026 096 128;
  • 49) 0.999 779 405 403 026 096 128 × 2 = 1 + 0.999 558 810 806 052 192 256;
  • 50) 0.999 558 810 806 052 192 256 × 2 = 1 + 0.999 117 621 612 104 384 512;
  • 51) 0.999 117 621 612 104 384 512 × 2 = 1 + 0.998 235 243 224 208 769 024;
  • 52) 0.998 235 243 224 208 769 024 × 2 = 1 + 0.996 470 486 448 417 538 048;
  • 53) 0.996 470 486 448 417 538 048 × 2 = 1 + 0.992 940 972 896 835 076 096;
  • 54) 0.992 940 972 896 835 076 096 × 2 = 1 + 0.985 881 945 793 670 152 192;

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:


0.000 000 000 742 147 675 863(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 675 863(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:


0.000 000 000 742 147 675 863(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 -0.000 000 000 742 147 675 863 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