-0.000 000 000 742 147 676 113 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 113(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
-0.000 000 000 742 147 676 113(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:

|-0.000 000 000 742 147 676 113| = 0.000 000 000 742 147 676 113


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: 0.000 000 000 742 147 676 113.

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) 0.000 000 000 742 147 676 113 × 2 = 0 + 0.000 000 001 484 295 352 226;
  • 2) 0.000 000 001 484 295 352 226 × 2 = 0 + 0.000 000 002 968 590 704 452;
  • 3) 0.000 000 002 968 590 704 452 × 2 = 0 + 0.000 000 005 937 181 408 904;
  • 4) 0.000 000 005 937 181 408 904 × 2 = 0 + 0.000 000 011 874 362 817 808;
  • 5) 0.000 000 011 874 362 817 808 × 2 = 0 + 0.000 000 023 748 725 635 616;
  • 6) 0.000 000 023 748 725 635 616 × 2 = 0 + 0.000 000 047 497 451 271 232;
  • 7) 0.000 000 047 497 451 271 232 × 2 = 0 + 0.000 000 094 994 902 542 464;
  • 8) 0.000 000 094 994 902 542 464 × 2 = 0 + 0.000 000 189 989 805 084 928;
  • 9) 0.000 000 189 989 805 084 928 × 2 = 0 + 0.000 000 379 979 610 169 856;
  • 10) 0.000 000 379 979 610 169 856 × 2 = 0 + 0.000 000 759 959 220 339 712;
  • 11) 0.000 000 759 959 220 339 712 × 2 = 0 + 0.000 001 519 918 440 679 424;
  • 12) 0.000 001 519 918 440 679 424 × 2 = 0 + 0.000 003 039 836 881 358 848;
  • 13) 0.000 003 039 836 881 358 848 × 2 = 0 + 0.000 006 079 673 762 717 696;
  • 14) 0.000 006 079 673 762 717 696 × 2 = 0 + 0.000 012 159 347 525 435 392;
  • 15) 0.000 012 159 347 525 435 392 × 2 = 0 + 0.000 024 318 695 050 870 784;
  • 16) 0.000 024 318 695 050 870 784 × 2 = 0 + 0.000 048 637 390 101 741 568;
  • 17) 0.000 048 637 390 101 741 568 × 2 = 0 + 0.000 097 274 780 203 483 136;
  • 18) 0.000 097 274 780 203 483 136 × 2 = 0 + 0.000 194 549 560 406 966 272;
  • 19) 0.000 194 549 560 406 966 272 × 2 = 0 + 0.000 389 099 120 813 932 544;
  • 20) 0.000 389 099 120 813 932 544 × 2 = 0 + 0.000 778 198 241 627 865 088;
  • 21) 0.000 778 198 241 627 865 088 × 2 = 0 + 0.001 556 396 483 255 730 176;
  • 22) 0.001 556 396 483 255 730 176 × 2 = 0 + 0.003 112 792 966 511 460 352;
  • 23) 0.003 112 792 966 511 460 352 × 2 = 0 + 0.006 225 585 933 022 920 704;
  • 24) 0.006 225 585 933 022 920 704 × 2 = 0 + 0.012 451 171 866 045 841 408;
  • 25) 0.012 451 171 866 045 841 408 × 2 = 0 + 0.024 902 343 732 091 682 816;
  • 26) 0.024 902 343 732 091 682 816 × 2 = 0 + 0.049 804 687 464 183 365 632;
  • 27) 0.049 804 687 464 183 365 632 × 2 = 0 + 0.099 609 374 928 366 731 264;
  • 28) 0.099 609 374 928 366 731 264 × 2 = 0 + 0.199 218 749 856 733 462 528;
  • 29) 0.199 218 749 856 733 462 528 × 2 = 0 + 0.398 437 499 713 466 925 056;
  • 30) 0.398 437 499 713 466 925 056 × 2 = 0 + 0.796 874 999 426 933 850 112;
  • 31) 0.796 874 999 426 933 850 112 × 2 = 1 + 0.593 749 998 853 867 700 224;
  • 32) 0.593 749 998 853 867 700 224 × 2 = 1 + 0.187 499 997 707 735 400 448;
  • 33) 0.187 499 997 707 735 400 448 × 2 = 0 + 0.374 999 995 415 470 800 896;
  • 34) 0.374 999 995 415 470 800 896 × 2 = 0 + 0.749 999 990 830 941 601 792;
  • 35) 0.749 999 990 830 941 601 792 × 2 = 1 + 0.499 999 981 661 883 203 584;
  • 36) 0.499 999 981 661 883 203 584 × 2 = 0 + 0.999 999 963 323 766 407 168;
  • 37) 0.999 999 963 323 766 407 168 × 2 = 1 + 0.999 999 926 647 532 814 336;
  • 38) 0.999 999 926 647 532 814 336 × 2 = 1 + 0.999 999 853 295 065 628 672;
  • 39) 0.999 999 853 295 065 628 672 × 2 = 1 + 0.999 999 706 590 131 257 344;
  • 40) 0.999 999 706 590 131 257 344 × 2 = 1 + 0.999 999 413 180 262 514 688;
  • 41) 0.999 999 413 180 262 514 688 × 2 = 1 + 0.999 998 826 360 525 029 376;
  • 42) 0.999 998 826 360 525 029 376 × 2 = 1 + 0.999 997 652 721 050 058 752;
  • 43) 0.999 997 652 721 050 058 752 × 2 = 1 + 0.999 995 305 442 100 117 504;
  • 44) 0.999 995 305 442 100 117 504 × 2 = 1 + 0.999 990 610 884 200 235 008;
  • 45) 0.999 990 610 884 200 235 008 × 2 = 1 + 0.999 981 221 768 400 470 016;
  • 46) 0.999 981 221 768 400 470 016 × 2 = 1 + 0.999 962 443 536 800 940 032;
  • 47) 0.999 962 443 536 800 940 032 × 2 = 1 + 0.999 924 887 073 601 880 064;
  • 48) 0.999 924 887 073 601 880 064 × 2 = 1 + 0.999 849 774 147 203 760 128;
  • 49) 0.999 849 774 147 203 760 128 × 2 = 1 + 0.999 699 548 294 407 520 256;
  • 50) 0.999 699 548 294 407 520 256 × 2 = 1 + 0.999 399 096 588 815 040 512;
  • 51) 0.999 399 096 588 815 040 512 × 2 = 1 + 0.998 798 193 177 630 081 024;
  • 52) 0.998 798 193 177 630 081 024 × 2 = 1 + 0.997 596 386 355 260 162 048;
  • 53) 0.997 596 386 355 260 162 048 × 2 = 1 + 0.995 192 772 710 520 324 096;
  • 54) 0.995 192 772 710 520 324 096 × 2 = 1 + 0.990 385 545 421 040 648 192;

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:


0.000 000 000 742 147 676 113(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 113(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:


0.000 000 000 742 147 676 113(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 -0.000 000 000 742 147 676 113 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