-0.000 000 000 742 147 675 999 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 999(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
-0.000 000 000 742 147 675 999(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:

|-0.000 000 000 742 147 675 999| = 0.000 000 000 742 147 675 999


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: 0.000 000 000 742 147 675 999.

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) 0.000 000 000 742 147 675 999 × 2 = 0 + 0.000 000 001 484 295 351 998;
  • 2) 0.000 000 001 484 295 351 998 × 2 = 0 + 0.000 000 002 968 590 703 996;
  • 3) 0.000 000 002 968 590 703 996 × 2 = 0 + 0.000 000 005 937 181 407 992;
  • 4) 0.000 000 005 937 181 407 992 × 2 = 0 + 0.000 000 011 874 362 815 984;
  • 5) 0.000 000 011 874 362 815 984 × 2 = 0 + 0.000 000 023 748 725 631 968;
  • 6) 0.000 000 023 748 725 631 968 × 2 = 0 + 0.000 000 047 497 451 263 936;
  • 7) 0.000 000 047 497 451 263 936 × 2 = 0 + 0.000 000 094 994 902 527 872;
  • 8) 0.000 000 094 994 902 527 872 × 2 = 0 + 0.000 000 189 989 805 055 744;
  • 9) 0.000 000 189 989 805 055 744 × 2 = 0 + 0.000 000 379 979 610 111 488;
  • 10) 0.000 000 379 979 610 111 488 × 2 = 0 + 0.000 000 759 959 220 222 976;
  • 11) 0.000 000 759 959 220 222 976 × 2 = 0 + 0.000 001 519 918 440 445 952;
  • 12) 0.000 001 519 918 440 445 952 × 2 = 0 + 0.000 003 039 836 880 891 904;
  • 13) 0.000 003 039 836 880 891 904 × 2 = 0 + 0.000 006 079 673 761 783 808;
  • 14) 0.000 006 079 673 761 783 808 × 2 = 0 + 0.000 012 159 347 523 567 616;
  • 15) 0.000 012 159 347 523 567 616 × 2 = 0 + 0.000 024 318 695 047 135 232;
  • 16) 0.000 024 318 695 047 135 232 × 2 = 0 + 0.000 048 637 390 094 270 464;
  • 17) 0.000 048 637 390 094 270 464 × 2 = 0 + 0.000 097 274 780 188 540 928;
  • 18) 0.000 097 274 780 188 540 928 × 2 = 0 + 0.000 194 549 560 377 081 856;
  • 19) 0.000 194 549 560 377 081 856 × 2 = 0 + 0.000 389 099 120 754 163 712;
  • 20) 0.000 389 099 120 754 163 712 × 2 = 0 + 0.000 778 198 241 508 327 424;
  • 21) 0.000 778 198 241 508 327 424 × 2 = 0 + 0.001 556 396 483 016 654 848;
  • 22) 0.001 556 396 483 016 654 848 × 2 = 0 + 0.003 112 792 966 033 309 696;
  • 23) 0.003 112 792 966 033 309 696 × 2 = 0 + 0.006 225 585 932 066 619 392;
  • 24) 0.006 225 585 932 066 619 392 × 2 = 0 + 0.012 451 171 864 133 238 784;
  • 25) 0.012 451 171 864 133 238 784 × 2 = 0 + 0.024 902 343 728 266 477 568;
  • 26) 0.024 902 343 728 266 477 568 × 2 = 0 + 0.049 804 687 456 532 955 136;
  • 27) 0.049 804 687 456 532 955 136 × 2 = 0 + 0.099 609 374 913 065 910 272;
  • 28) 0.099 609 374 913 065 910 272 × 2 = 0 + 0.199 218 749 826 131 820 544;
  • 29) 0.199 218 749 826 131 820 544 × 2 = 0 + 0.398 437 499 652 263 641 088;
  • 30) 0.398 437 499 652 263 641 088 × 2 = 0 + 0.796 874 999 304 527 282 176;
  • 31) 0.796 874 999 304 527 282 176 × 2 = 1 + 0.593 749 998 609 054 564 352;
  • 32) 0.593 749 998 609 054 564 352 × 2 = 1 + 0.187 499 997 218 109 128 704;
  • 33) 0.187 499 997 218 109 128 704 × 2 = 0 + 0.374 999 994 436 218 257 408;
  • 34) 0.374 999 994 436 218 257 408 × 2 = 0 + 0.749 999 988 872 436 514 816;
  • 35) 0.749 999 988 872 436 514 816 × 2 = 1 + 0.499 999 977 744 873 029 632;
  • 36) 0.499 999 977 744 873 029 632 × 2 = 0 + 0.999 999 955 489 746 059 264;
  • 37) 0.999 999 955 489 746 059 264 × 2 = 1 + 0.999 999 910 979 492 118 528;
  • 38) 0.999 999 910 979 492 118 528 × 2 = 1 + 0.999 999 821 958 984 237 056;
  • 39) 0.999 999 821 958 984 237 056 × 2 = 1 + 0.999 999 643 917 968 474 112;
  • 40) 0.999 999 643 917 968 474 112 × 2 = 1 + 0.999 999 287 835 936 948 224;
  • 41) 0.999 999 287 835 936 948 224 × 2 = 1 + 0.999 998 575 671 873 896 448;
  • 42) 0.999 998 575 671 873 896 448 × 2 = 1 + 0.999 997 151 343 747 792 896;
  • 43) 0.999 997 151 343 747 792 896 × 2 = 1 + 0.999 994 302 687 495 585 792;
  • 44) 0.999 994 302 687 495 585 792 × 2 = 1 + 0.999 988 605 374 991 171 584;
  • 45) 0.999 988 605 374 991 171 584 × 2 = 1 + 0.999 977 210 749 982 343 168;
  • 46) 0.999 977 210 749 982 343 168 × 2 = 1 + 0.999 954 421 499 964 686 336;
  • 47) 0.999 954 421 499 964 686 336 × 2 = 1 + 0.999 908 842 999 929 372 672;
  • 48) 0.999 908 842 999 929 372 672 × 2 = 1 + 0.999 817 685 999 858 745 344;
  • 49) 0.999 817 685 999 858 745 344 × 2 = 1 + 0.999 635 371 999 717 490 688;
  • 50) 0.999 635 371 999 717 490 688 × 2 = 1 + 0.999 270 743 999 434 981 376;
  • 51) 0.999 270 743 999 434 981 376 × 2 = 1 + 0.998 541 487 998 869 962 752;
  • 52) 0.998 541 487 998 869 962 752 × 2 = 1 + 0.997 082 975 997 739 925 504;
  • 53) 0.997 082 975 997 739 925 504 × 2 = 1 + 0.994 165 951 995 479 851 008;
  • 54) 0.994 165 951 995 479 851 008 × 2 = 1 + 0.988 331 903 990 959 702 016;

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:


0.000 000 000 742 147 675 999(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 675 999(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:


0.000 000 000 742 147 675 999(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 -0.000 000 000 742 147 675 999 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 025 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 minutes

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