-0.000 000 000 742 147 676 132 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 132(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
-0.000 000 000 742 147 676 132(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:

|-0.000 000 000 742 147 676 132| = 0.000 000 000 742 147 676 132


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: 0.000 000 000 742 147 676 132.

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) 0.000 000 000 742 147 676 132 × 2 = 0 + 0.000 000 001 484 295 352 264;
  • 2) 0.000 000 001 484 295 352 264 × 2 = 0 + 0.000 000 002 968 590 704 528;
  • 3) 0.000 000 002 968 590 704 528 × 2 = 0 + 0.000 000 005 937 181 409 056;
  • 4) 0.000 000 005 937 181 409 056 × 2 = 0 + 0.000 000 011 874 362 818 112;
  • 5) 0.000 000 011 874 362 818 112 × 2 = 0 + 0.000 000 023 748 725 636 224;
  • 6) 0.000 000 023 748 725 636 224 × 2 = 0 + 0.000 000 047 497 451 272 448;
  • 7) 0.000 000 047 497 451 272 448 × 2 = 0 + 0.000 000 094 994 902 544 896;
  • 8) 0.000 000 094 994 902 544 896 × 2 = 0 + 0.000 000 189 989 805 089 792;
  • 9) 0.000 000 189 989 805 089 792 × 2 = 0 + 0.000 000 379 979 610 179 584;
  • 10) 0.000 000 379 979 610 179 584 × 2 = 0 + 0.000 000 759 959 220 359 168;
  • 11) 0.000 000 759 959 220 359 168 × 2 = 0 + 0.000 001 519 918 440 718 336;
  • 12) 0.000 001 519 918 440 718 336 × 2 = 0 + 0.000 003 039 836 881 436 672;
  • 13) 0.000 003 039 836 881 436 672 × 2 = 0 + 0.000 006 079 673 762 873 344;
  • 14) 0.000 006 079 673 762 873 344 × 2 = 0 + 0.000 012 159 347 525 746 688;
  • 15) 0.000 012 159 347 525 746 688 × 2 = 0 + 0.000 024 318 695 051 493 376;
  • 16) 0.000 024 318 695 051 493 376 × 2 = 0 + 0.000 048 637 390 102 986 752;
  • 17) 0.000 048 637 390 102 986 752 × 2 = 0 + 0.000 097 274 780 205 973 504;
  • 18) 0.000 097 274 780 205 973 504 × 2 = 0 + 0.000 194 549 560 411 947 008;
  • 19) 0.000 194 549 560 411 947 008 × 2 = 0 + 0.000 389 099 120 823 894 016;
  • 20) 0.000 389 099 120 823 894 016 × 2 = 0 + 0.000 778 198 241 647 788 032;
  • 21) 0.000 778 198 241 647 788 032 × 2 = 0 + 0.001 556 396 483 295 576 064;
  • 22) 0.001 556 396 483 295 576 064 × 2 = 0 + 0.003 112 792 966 591 152 128;
  • 23) 0.003 112 792 966 591 152 128 × 2 = 0 + 0.006 225 585 933 182 304 256;
  • 24) 0.006 225 585 933 182 304 256 × 2 = 0 + 0.012 451 171 866 364 608 512;
  • 25) 0.012 451 171 866 364 608 512 × 2 = 0 + 0.024 902 343 732 729 217 024;
  • 26) 0.024 902 343 732 729 217 024 × 2 = 0 + 0.049 804 687 465 458 434 048;
  • 27) 0.049 804 687 465 458 434 048 × 2 = 0 + 0.099 609 374 930 916 868 096;
  • 28) 0.099 609 374 930 916 868 096 × 2 = 0 + 0.199 218 749 861 833 736 192;
  • 29) 0.199 218 749 861 833 736 192 × 2 = 0 + 0.398 437 499 723 667 472 384;
  • 30) 0.398 437 499 723 667 472 384 × 2 = 0 + 0.796 874 999 447 334 944 768;
  • 31) 0.796 874 999 447 334 944 768 × 2 = 1 + 0.593 749 998 894 669 889 536;
  • 32) 0.593 749 998 894 669 889 536 × 2 = 1 + 0.187 499 997 789 339 779 072;
  • 33) 0.187 499 997 789 339 779 072 × 2 = 0 + 0.374 999 995 578 679 558 144;
  • 34) 0.374 999 995 578 679 558 144 × 2 = 0 + 0.749 999 991 157 359 116 288;
  • 35) 0.749 999 991 157 359 116 288 × 2 = 1 + 0.499 999 982 314 718 232 576;
  • 36) 0.499 999 982 314 718 232 576 × 2 = 0 + 0.999 999 964 629 436 465 152;
  • 37) 0.999 999 964 629 436 465 152 × 2 = 1 + 0.999 999 929 258 872 930 304;
  • 38) 0.999 999 929 258 872 930 304 × 2 = 1 + 0.999 999 858 517 745 860 608;
  • 39) 0.999 999 858 517 745 860 608 × 2 = 1 + 0.999 999 717 035 491 721 216;
  • 40) 0.999 999 717 035 491 721 216 × 2 = 1 + 0.999 999 434 070 983 442 432;
  • 41) 0.999 999 434 070 983 442 432 × 2 = 1 + 0.999 998 868 141 966 884 864;
  • 42) 0.999 998 868 141 966 884 864 × 2 = 1 + 0.999 997 736 283 933 769 728;
  • 43) 0.999 997 736 283 933 769 728 × 2 = 1 + 0.999 995 472 567 867 539 456;
  • 44) 0.999 995 472 567 867 539 456 × 2 = 1 + 0.999 990 945 135 735 078 912;
  • 45) 0.999 990 945 135 735 078 912 × 2 = 1 + 0.999 981 890 271 470 157 824;
  • 46) 0.999 981 890 271 470 157 824 × 2 = 1 + 0.999 963 780 542 940 315 648;
  • 47) 0.999 963 780 542 940 315 648 × 2 = 1 + 0.999 927 561 085 880 631 296;
  • 48) 0.999 927 561 085 880 631 296 × 2 = 1 + 0.999 855 122 171 761 262 592;
  • 49) 0.999 855 122 171 761 262 592 × 2 = 1 + 0.999 710 244 343 522 525 184;
  • 50) 0.999 710 244 343 522 525 184 × 2 = 1 + 0.999 420 488 687 045 050 368;
  • 51) 0.999 420 488 687 045 050 368 × 2 = 1 + 0.998 840 977 374 090 100 736;
  • 52) 0.998 840 977 374 090 100 736 × 2 = 1 + 0.997 681 954 748 180 201 472;
  • 53) 0.997 681 954 748 180 201 472 × 2 = 1 + 0.995 363 909 496 360 402 944;
  • 54) 0.995 363 909 496 360 402 944 × 2 = 1 + 0.990 727 818 992 720 805 888;

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:


0.000 000 000 742 147 676 132(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 132(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:


0.000 000 000 742 147 676 132(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 -0.000 000 000 742 147 676 132 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