-0.000 000 000 742 147 676 646 48 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 48(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
-0.000 000 000 742 147 676 646 48(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:

|-0.000 000 000 742 147 676 646 48| = 0.000 000 000 742 147 676 646 48


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: 0.000 000 000 742 147 676 646 48.

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) 0.000 000 000 742 147 676 646 48 × 2 = 0 + 0.000 000 001 484 295 353 292 96;
  • 2) 0.000 000 001 484 295 353 292 96 × 2 = 0 + 0.000 000 002 968 590 706 585 92;
  • 3) 0.000 000 002 968 590 706 585 92 × 2 = 0 + 0.000 000 005 937 181 413 171 84;
  • 4) 0.000 000 005 937 181 413 171 84 × 2 = 0 + 0.000 000 011 874 362 826 343 68;
  • 5) 0.000 000 011 874 362 826 343 68 × 2 = 0 + 0.000 000 023 748 725 652 687 36;
  • 6) 0.000 000 023 748 725 652 687 36 × 2 = 0 + 0.000 000 047 497 451 305 374 72;
  • 7) 0.000 000 047 497 451 305 374 72 × 2 = 0 + 0.000 000 094 994 902 610 749 44;
  • 8) 0.000 000 094 994 902 610 749 44 × 2 = 0 + 0.000 000 189 989 805 221 498 88;
  • 9) 0.000 000 189 989 805 221 498 88 × 2 = 0 + 0.000 000 379 979 610 442 997 76;
  • 10) 0.000 000 379 979 610 442 997 76 × 2 = 0 + 0.000 000 759 959 220 885 995 52;
  • 11) 0.000 000 759 959 220 885 995 52 × 2 = 0 + 0.000 001 519 918 441 771 991 04;
  • 12) 0.000 001 519 918 441 771 991 04 × 2 = 0 + 0.000 003 039 836 883 543 982 08;
  • 13) 0.000 003 039 836 883 543 982 08 × 2 = 0 + 0.000 006 079 673 767 087 964 16;
  • 14) 0.000 006 079 673 767 087 964 16 × 2 = 0 + 0.000 012 159 347 534 175 928 32;
  • 15) 0.000 012 159 347 534 175 928 32 × 2 = 0 + 0.000 024 318 695 068 351 856 64;
  • 16) 0.000 024 318 695 068 351 856 64 × 2 = 0 + 0.000 048 637 390 136 703 713 28;
  • 17) 0.000 048 637 390 136 703 713 28 × 2 = 0 + 0.000 097 274 780 273 407 426 56;
  • 18) 0.000 097 274 780 273 407 426 56 × 2 = 0 + 0.000 194 549 560 546 814 853 12;
  • 19) 0.000 194 549 560 546 814 853 12 × 2 = 0 + 0.000 389 099 121 093 629 706 24;
  • 20) 0.000 389 099 121 093 629 706 24 × 2 = 0 + 0.000 778 198 242 187 259 412 48;
  • 21) 0.000 778 198 242 187 259 412 48 × 2 = 0 + 0.001 556 396 484 374 518 824 96;
  • 22) 0.001 556 396 484 374 518 824 96 × 2 = 0 + 0.003 112 792 968 749 037 649 92;
  • 23) 0.003 112 792 968 749 037 649 92 × 2 = 0 + 0.006 225 585 937 498 075 299 84;
  • 24) 0.006 225 585 937 498 075 299 84 × 2 = 0 + 0.012 451 171 874 996 150 599 68;
  • 25) 0.012 451 171 874 996 150 599 68 × 2 = 0 + 0.024 902 343 749 992 301 199 36;
  • 26) 0.024 902 343 749 992 301 199 36 × 2 = 0 + 0.049 804 687 499 984 602 398 72;
  • 27) 0.049 804 687 499 984 602 398 72 × 2 = 0 + 0.099 609 374 999 969 204 797 44;
  • 28) 0.099 609 374 999 969 204 797 44 × 2 = 0 + 0.199 218 749 999 938 409 594 88;
  • 29) 0.199 218 749 999 938 409 594 88 × 2 = 0 + 0.398 437 499 999 876 819 189 76;
  • 30) 0.398 437 499 999 876 819 189 76 × 2 = 0 + 0.796 874 999 999 753 638 379 52;
  • 31) 0.796 874 999 999 753 638 379 52 × 2 = 1 + 0.593 749 999 999 507 276 759 04;
  • 32) 0.593 749 999 999 507 276 759 04 × 2 = 1 + 0.187 499 999 999 014 553 518 08;
  • 33) 0.187 499 999 999 014 553 518 08 × 2 = 0 + 0.374 999 999 998 029 107 036 16;
  • 34) 0.374 999 999 998 029 107 036 16 × 2 = 0 + 0.749 999 999 996 058 214 072 32;
  • 35) 0.749 999 999 996 058 214 072 32 × 2 = 1 + 0.499 999 999 992 116 428 144 64;
  • 36) 0.499 999 999 992 116 428 144 64 × 2 = 0 + 0.999 999 999 984 232 856 289 28;
  • 37) 0.999 999 999 984 232 856 289 28 × 2 = 1 + 0.999 999 999 968 465 712 578 56;
  • 38) 0.999 999 999 968 465 712 578 56 × 2 = 1 + 0.999 999 999 936 931 425 157 12;
  • 39) 0.999 999 999 936 931 425 157 12 × 2 = 1 + 0.999 999 999 873 862 850 314 24;
  • 40) 0.999 999 999 873 862 850 314 24 × 2 = 1 + 0.999 999 999 747 725 700 628 48;
  • 41) 0.999 999 999 747 725 700 628 48 × 2 = 1 + 0.999 999 999 495 451 401 256 96;
  • 42) 0.999 999 999 495 451 401 256 96 × 2 = 1 + 0.999 999 998 990 902 802 513 92;
  • 43) 0.999 999 998 990 902 802 513 92 × 2 = 1 + 0.999 999 997 981 805 605 027 84;
  • 44) 0.999 999 997 981 805 605 027 84 × 2 = 1 + 0.999 999 995 963 611 210 055 68;
  • 45) 0.999 999 995 963 611 210 055 68 × 2 = 1 + 0.999 999 991 927 222 420 111 36;
  • 46) 0.999 999 991 927 222 420 111 36 × 2 = 1 + 0.999 999 983 854 444 840 222 72;
  • 47) 0.999 999 983 854 444 840 222 72 × 2 = 1 + 0.999 999 967 708 889 680 445 44;
  • 48) 0.999 999 967 708 889 680 445 44 × 2 = 1 + 0.999 999 935 417 779 360 890 88;
  • 49) 0.999 999 935 417 779 360 890 88 × 2 = 1 + 0.999 999 870 835 558 721 781 76;
  • 50) 0.999 999 870 835 558 721 781 76 × 2 = 1 + 0.999 999 741 671 117 443 563 52;
  • 51) 0.999 999 741 671 117 443 563 52 × 2 = 1 + 0.999 999 483 342 234 887 127 04;
  • 52) 0.999 999 483 342 234 887 127 04 × 2 = 1 + 0.999 998 966 684 469 774 254 08;
  • 53) 0.999 998 966 684 469 774 254 08 × 2 = 1 + 0.999 997 933 368 939 548 508 16;
  • 54) 0.999 997 933 368 939 548 508 16 × 2 = 1 + 0.999 995 866 737 879 097 016 32;

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:


0.000 000 000 742 147 676 646 48(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 48(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:


0.000 000 000 742 147 676 646 48(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 -0.000 000 000 742 147 676 646 48 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