-0.000 000 000 742 147 676 203 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 203(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
-0.000 000 000 742 147 676 203(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:

|-0.000 000 000 742 147 676 203| = 0.000 000 000 742 147 676 203


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: 0.000 000 000 742 147 676 203.

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) 0.000 000 000 742 147 676 203 × 2 = 0 + 0.000 000 001 484 295 352 406;
  • 2) 0.000 000 001 484 295 352 406 × 2 = 0 + 0.000 000 002 968 590 704 812;
  • 3) 0.000 000 002 968 590 704 812 × 2 = 0 + 0.000 000 005 937 181 409 624;
  • 4) 0.000 000 005 937 181 409 624 × 2 = 0 + 0.000 000 011 874 362 819 248;
  • 5) 0.000 000 011 874 362 819 248 × 2 = 0 + 0.000 000 023 748 725 638 496;
  • 6) 0.000 000 023 748 725 638 496 × 2 = 0 + 0.000 000 047 497 451 276 992;
  • 7) 0.000 000 047 497 451 276 992 × 2 = 0 + 0.000 000 094 994 902 553 984;
  • 8) 0.000 000 094 994 902 553 984 × 2 = 0 + 0.000 000 189 989 805 107 968;
  • 9) 0.000 000 189 989 805 107 968 × 2 = 0 + 0.000 000 379 979 610 215 936;
  • 10) 0.000 000 379 979 610 215 936 × 2 = 0 + 0.000 000 759 959 220 431 872;
  • 11) 0.000 000 759 959 220 431 872 × 2 = 0 + 0.000 001 519 918 440 863 744;
  • 12) 0.000 001 519 918 440 863 744 × 2 = 0 + 0.000 003 039 836 881 727 488;
  • 13) 0.000 003 039 836 881 727 488 × 2 = 0 + 0.000 006 079 673 763 454 976;
  • 14) 0.000 006 079 673 763 454 976 × 2 = 0 + 0.000 012 159 347 526 909 952;
  • 15) 0.000 012 159 347 526 909 952 × 2 = 0 + 0.000 024 318 695 053 819 904;
  • 16) 0.000 024 318 695 053 819 904 × 2 = 0 + 0.000 048 637 390 107 639 808;
  • 17) 0.000 048 637 390 107 639 808 × 2 = 0 + 0.000 097 274 780 215 279 616;
  • 18) 0.000 097 274 780 215 279 616 × 2 = 0 + 0.000 194 549 560 430 559 232;
  • 19) 0.000 194 549 560 430 559 232 × 2 = 0 + 0.000 389 099 120 861 118 464;
  • 20) 0.000 389 099 120 861 118 464 × 2 = 0 + 0.000 778 198 241 722 236 928;
  • 21) 0.000 778 198 241 722 236 928 × 2 = 0 + 0.001 556 396 483 444 473 856;
  • 22) 0.001 556 396 483 444 473 856 × 2 = 0 + 0.003 112 792 966 888 947 712;
  • 23) 0.003 112 792 966 888 947 712 × 2 = 0 + 0.006 225 585 933 777 895 424;
  • 24) 0.006 225 585 933 777 895 424 × 2 = 0 + 0.012 451 171 867 555 790 848;
  • 25) 0.012 451 171 867 555 790 848 × 2 = 0 + 0.024 902 343 735 111 581 696;
  • 26) 0.024 902 343 735 111 581 696 × 2 = 0 + 0.049 804 687 470 223 163 392;
  • 27) 0.049 804 687 470 223 163 392 × 2 = 0 + 0.099 609 374 940 446 326 784;
  • 28) 0.099 609 374 940 446 326 784 × 2 = 0 + 0.199 218 749 880 892 653 568;
  • 29) 0.199 218 749 880 892 653 568 × 2 = 0 + 0.398 437 499 761 785 307 136;
  • 30) 0.398 437 499 761 785 307 136 × 2 = 0 + 0.796 874 999 523 570 614 272;
  • 31) 0.796 874 999 523 570 614 272 × 2 = 1 + 0.593 749 999 047 141 228 544;
  • 32) 0.593 749 999 047 141 228 544 × 2 = 1 + 0.187 499 998 094 282 457 088;
  • 33) 0.187 499 998 094 282 457 088 × 2 = 0 + 0.374 999 996 188 564 914 176;
  • 34) 0.374 999 996 188 564 914 176 × 2 = 0 + 0.749 999 992 377 129 828 352;
  • 35) 0.749 999 992 377 129 828 352 × 2 = 1 + 0.499 999 984 754 259 656 704;
  • 36) 0.499 999 984 754 259 656 704 × 2 = 0 + 0.999 999 969 508 519 313 408;
  • 37) 0.999 999 969 508 519 313 408 × 2 = 1 + 0.999 999 939 017 038 626 816;
  • 38) 0.999 999 939 017 038 626 816 × 2 = 1 + 0.999 999 878 034 077 253 632;
  • 39) 0.999 999 878 034 077 253 632 × 2 = 1 + 0.999 999 756 068 154 507 264;
  • 40) 0.999 999 756 068 154 507 264 × 2 = 1 + 0.999 999 512 136 309 014 528;
  • 41) 0.999 999 512 136 309 014 528 × 2 = 1 + 0.999 999 024 272 618 029 056;
  • 42) 0.999 999 024 272 618 029 056 × 2 = 1 + 0.999 998 048 545 236 058 112;
  • 43) 0.999 998 048 545 236 058 112 × 2 = 1 + 0.999 996 097 090 472 116 224;
  • 44) 0.999 996 097 090 472 116 224 × 2 = 1 + 0.999 992 194 180 944 232 448;
  • 45) 0.999 992 194 180 944 232 448 × 2 = 1 + 0.999 984 388 361 888 464 896;
  • 46) 0.999 984 388 361 888 464 896 × 2 = 1 + 0.999 968 776 723 776 929 792;
  • 47) 0.999 968 776 723 776 929 792 × 2 = 1 + 0.999 937 553 447 553 859 584;
  • 48) 0.999 937 553 447 553 859 584 × 2 = 1 + 0.999 875 106 895 107 719 168;
  • 49) 0.999 875 106 895 107 719 168 × 2 = 1 + 0.999 750 213 790 215 438 336;
  • 50) 0.999 750 213 790 215 438 336 × 2 = 1 + 0.999 500 427 580 430 876 672;
  • 51) 0.999 500 427 580 430 876 672 × 2 = 1 + 0.999 000 855 160 861 753 344;
  • 52) 0.999 000 855 160 861 753 344 × 2 = 1 + 0.998 001 710 321 723 506 688;
  • 53) 0.998 001 710 321 723 506 688 × 2 = 1 + 0.996 003 420 643 447 013 376;
  • 54) 0.996 003 420 643 447 013 376 × 2 = 1 + 0.992 006 841 286 894 026 752;

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:


0.000 000 000 742 147 676 203(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 203(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:


0.000 000 000 742 147 676 203(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 -0.000 000 000 742 147 676 203 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