-0.000 000 000 742 147 676 647 88 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 647 88(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
-0.000 000 000 742 147 676 647 88(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:

|-0.000 000 000 742 147 676 647 88| = 0.000 000 000 742 147 676 647 88


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: 0.000 000 000 742 147 676 647 88.

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) 0.000 000 000 742 147 676 647 88 × 2 = 0 + 0.000 000 001 484 295 353 295 76;
  • 2) 0.000 000 001 484 295 353 295 76 × 2 = 0 + 0.000 000 002 968 590 706 591 52;
  • 3) 0.000 000 002 968 590 706 591 52 × 2 = 0 + 0.000 000 005 937 181 413 183 04;
  • 4) 0.000 000 005 937 181 413 183 04 × 2 = 0 + 0.000 000 011 874 362 826 366 08;
  • 5) 0.000 000 011 874 362 826 366 08 × 2 = 0 + 0.000 000 023 748 725 652 732 16;
  • 6) 0.000 000 023 748 725 652 732 16 × 2 = 0 + 0.000 000 047 497 451 305 464 32;
  • 7) 0.000 000 047 497 451 305 464 32 × 2 = 0 + 0.000 000 094 994 902 610 928 64;
  • 8) 0.000 000 094 994 902 610 928 64 × 2 = 0 + 0.000 000 189 989 805 221 857 28;
  • 9) 0.000 000 189 989 805 221 857 28 × 2 = 0 + 0.000 000 379 979 610 443 714 56;
  • 10) 0.000 000 379 979 610 443 714 56 × 2 = 0 + 0.000 000 759 959 220 887 429 12;
  • 11) 0.000 000 759 959 220 887 429 12 × 2 = 0 + 0.000 001 519 918 441 774 858 24;
  • 12) 0.000 001 519 918 441 774 858 24 × 2 = 0 + 0.000 003 039 836 883 549 716 48;
  • 13) 0.000 003 039 836 883 549 716 48 × 2 = 0 + 0.000 006 079 673 767 099 432 96;
  • 14) 0.000 006 079 673 767 099 432 96 × 2 = 0 + 0.000 012 159 347 534 198 865 92;
  • 15) 0.000 012 159 347 534 198 865 92 × 2 = 0 + 0.000 024 318 695 068 397 731 84;
  • 16) 0.000 024 318 695 068 397 731 84 × 2 = 0 + 0.000 048 637 390 136 795 463 68;
  • 17) 0.000 048 637 390 136 795 463 68 × 2 = 0 + 0.000 097 274 780 273 590 927 36;
  • 18) 0.000 097 274 780 273 590 927 36 × 2 = 0 + 0.000 194 549 560 547 181 854 72;
  • 19) 0.000 194 549 560 547 181 854 72 × 2 = 0 + 0.000 389 099 121 094 363 709 44;
  • 20) 0.000 389 099 121 094 363 709 44 × 2 = 0 + 0.000 778 198 242 188 727 418 88;
  • 21) 0.000 778 198 242 188 727 418 88 × 2 = 0 + 0.001 556 396 484 377 454 837 76;
  • 22) 0.001 556 396 484 377 454 837 76 × 2 = 0 + 0.003 112 792 968 754 909 675 52;
  • 23) 0.003 112 792 968 754 909 675 52 × 2 = 0 + 0.006 225 585 937 509 819 351 04;
  • 24) 0.006 225 585 937 509 819 351 04 × 2 = 0 + 0.012 451 171 875 019 638 702 08;
  • 25) 0.012 451 171 875 019 638 702 08 × 2 = 0 + 0.024 902 343 750 039 277 404 16;
  • 26) 0.024 902 343 750 039 277 404 16 × 2 = 0 + 0.049 804 687 500 078 554 808 32;
  • 27) 0.049 804 687 500 078 554 808 32 × 2 = 0 + 0.099 609 375 000 157 109 616 64;
  • 28) 0.099 609 375 000 157 109 616 64 × 2 = 0 + 0.199 218 750 000 314 219 233 28;
  • 29) 0.199 218 750 000 314 219 233 28 × 2 = 0 + 0.398 437 500 000 628 438 466 56;
  • 30) 0.398 437 500 000 628 438 466 56 × 2 = 0 + 0.796 875 000 001 256 876 933 12;
  • 31) 0.796 875 000 001 256 876 933 12 × 2 = 1 + 0.593 750 000 002 513 753 866 24;
  • 32) 0.593 750 000 002 513 753 866 24 × 2 = 1 + 0.187 500 000 005 027 507 732 48;
  • 33) 0.187 500 000 005 027 507 732 48 × 2 = 0 + 0.375 000 000 010 055 015 464 96;
  • 34) 0.375 000 000 010 055 015 464 96 × 2 = 0 + 0.750 000 000 020 110 030 929 92;
  • 35) 0.750 000 000 020 110 030 929 92 × 2 = 1 + 0.500 000 000 040 220 061 859 84;
  • 36) 0.500 000 000 040 220 061 859 84 × 2 = 1 + 0.000 000 000 080 440 123 719 68;
  • 37) 0.000 000 000 080 440 123 719 68 × 2 = 0 + 0.000 000 000 160 880 247 439 36;
  • 38) 0.000 000 000 160 880 247 439 36 × 2 = 0 + 0.000 000 000 321 760 494 878 72;
  • 39) 0.000 000 000 321 760 494 878 72 × 2 = 0 + 0.000 000 000 643 520 989 757 44;
  • 40) 0.000 000 000 643 520 989 757 44 × 2 = 0 + 0.000 000 001 287 041 979 514 88;
  • 41) 0.000 000 001 287 041 979 514 88 × 2 = 0 + 0.000 000 002 574 083 959 029 76;
  • 42) 0.000 000 002 574 083 959 029 76 × 2 = 0 + 0.000 000 005 148 167 918 059 52;
  • 43) 0.000 000 005 148 167 918 059 52 × 2 = 0 + 0.000 000 010 296 335 836 119 04;
  • 44) 0.000 000 010 296 335 836 119 04 × 2 = 0 + 0.000 000 020 592 671 672 238 08;
  • 45) 0.000 000 020 592 671 672 238 08 × 2 = 0 + 0.000 000 041 185 343 344 476 16;
  • 46) 0.000 000 041 185 343 344 476 16 × 2 = 0 + 0.000 000 082 370 686 688 952 32;
  • 47) 0.000 000 082 370 686 688 952 32 × 2 = 0 + 0.000 000 164 741 373 377 904 64;
  • 48) 0.000 000 164 741 373 377 904 64 × 2 = 0 + 0.000 000 329 482 746 755 809 28;
  • 49) 0.000 000 329 482 746 755 809 28 × 2 = 0 + 0.000 000 658 965 493 511 618 56;
  • 50) 0.000 000 658 965 493 511 618 56 × 2 = 0 + 0.000 001 317 930 987 023 237 12;
  • 51) 0.000 001 317 930 987 023 237 12 × 2 = 0 + 0.000 002 635 861 974 046 474 24;
  • 52) 0.000 002 635 861 974 046 474 24 × 2 = 0 + 0.000 005 271 723 948 092 948 48;
  • 53) 0.000 005 271 723 948 092 948 48 × 2 = 0 + 0.000 010 543 447 896 185 896 96;
  • 54) 0.000 010 543 447 896 185 896 96 × 2 = 0 + 0.000 021 086 895 792 371 793 92;

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:


0.000 000 000 742 147 676 647 88(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 647 88(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 647 88(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 647 88 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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