-0.000 375 658 488 91 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 375 658 488 91(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.000 375 658 488 91(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 375 658 488 91| = 0.000 375 658 488 91


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: 0.000 375 658 488 91.

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) 0.000 375 658 488 91 × 2 = 0 + 0.000 751 316 977 82;
  • 2) 0.000 751 316 977 82 × 2 = 0 + 0.001 502 633 955 64;
  • 3) 0.001 502 633 955 64 × 2 = 0 + 0.003 005 267 911 28;
  • 4) 0.003 005 267 911 28 × 2 = 0 + 0.006 010 535 822 56;
  • 5) 0.006 010 535 822 56 × 2 = 0 + 0.012 021 071 645 12;
  • 6) 0.012 021 071 645 12 × 2 = 0 + 0.024 042 143 290 24;
  • 7) 0.024 042 143 290 24 × 2 = 0 + 0.048 084 286 580 48;
  • 8) 0.048 084 286 580 48 × 2 = 0 + 0.096 168 573 160 96;
  • 9) 0.096 168 573 160 96 × 2 = 0 + 0.192 337 146 321 92;
  • 10) 0.192 337 146 321 92 × 2 = 0 + 0.384 674 292 643 84;
  • 11) 0.384 674 292 643 84 × 2 = 0 + 0.769 348 585 287 68;
  • 12) 0.769 348 585 287 68 × 2 = 1 + 0.538 697 170 575 36;
  • 13) 0.538 697 170 575 36 × 2 = 1 + 0.077 394 341 150 72;
  • 14) 0.077 394 341 150 72 × 2 = 0 + 0.154 788 682 301 44;
  • 15) 0.154 788 682 301 44 × 2 = 0 + 0.309 577 364 602 88;
  • 16) 0.309 577 364 602 88 × 2 = 0 + 0.619 154 729 205 76;
  • 17) 0.619 154 729 205 76 × 2 = 1 + 0.238 309 458 411 52;
  • 18) 0.238 309 458 411 52 × 2 = 0 + 0.476 618 916 823 04;
  • 19) 0.476 618 916 823 04 × 2 = 0 + 0.953 237 833 646 08;
  • 20) 0.953 237 833 646 08 × 2 = 1 + 0.906 475 667 292 16;
  • 21) 0.906 475 667 292 16 × 2 = 1 + 0.812 951 334 584 32;
  • 22) 0.812 951 334 584 32 × 2 = 1 + 0.625 902 669 168 64;
  • 23) 0.625 902 669 168 64 × 2 = 1 + 0.251 805 338 337 28;
  • 24) 0.251 805 338 337 28 × 2 = 0 + 0.503 610 676 674 56;
  • 25) 0.503 610 676 674 56 × 2 = 1 + 0.007 221 353 349 12;
  • 26) 0.007 221 353 349 12 × 2 = 0 + 0.014 442 706 698 24;
  • 27) 0.014 442 706 698 24 × 2 = 0 + 0.028 885 413 396 48;
  • 28) 0.028 885 413 396 48 × 2 = 0 + 0.057 770 826 792 96;
  • 29) 0.057 770 826 792 96 × 2 = 0 + 0.115 541 653 585 92;
  • 30) 0.115 541 653 585 92 × 2 = 0 + 0.231 083 307 171 84;
  • 31) 0.231 083 307 171 84 × 2 = 0 + 0.462 166 614 343 68;
  • 32) 0.462 166 614 343 68 × 2 = 0 + 0.924 333 228 687 36;
  • 33) 0.924 333 228 687 36 × 2 = 1 + 0.848 666 457 374 72;
  • 34) 0.848 666 457 374 72 × 2 = 1 + 0.697 332 914 749 44;
  • 35) 0.697 332 914 749 44 × 2 = 1 + 0.394 665 829 498 88;
  • 36) 0.394 665 829 498 88 × 2 = 0 + 0.789 331 658 997 76;
  • 37) 0.789 331 658 997 76 × 2 = 1 + 0.578 663 317 995 52;
  • 38) 0.578 663 317 995 52 × 2 = 1 + 0.157 326 635 991 04;
  • 39) 0.157 326 635 991 04 × 2 = 0 + 0.314 653 271 982 08;
  • 40) 0.314 653 271 982 08 × 2 = 0 + 0.629 306 543 964 16;
  • 41) 0.629 306 543 964 16 × 2 = 1 + 0.258 613 087 928 32;
  • 42) 0.258 613 087 928 32 × 2 = 0 + 0.517 226 175 856 64;
  • 43) 0.517 226 175 856 64 × 2 = 1 + 0.034 452 351 713 28;
  • 44) 0.034 452 351 713 28 × 2 = 0 + 0.068 904 703 426 56;
  • 45) 0.068 904 703 426 56 × 2 = 0 + 0.137 809 406 853 12;
  • 46) 0.137 809 406 853 12 × 2 = 0 + 0.275 618 813 706 24;
  • 47) 0.275 618 813 706 24 × 2 = 0 + 0.551 237 627 412 48;
  • 48) 0.551 237 627 412 48 × 2 = 1 + 0.102 475 254 824 96;
  • 49) 0.102 475 254 824 96 × 2 = 0 + 0.204 950 509 649 92;
  • 50) 0.204 950 509 649 92 × 2 = 0 + 0.409 901 019 299 84;
  • 51) 0.409 901 019 299 84 × 2 = 0 + 0.819 802 038 599 68;
  • 52) 0.819 802 038 599 68 × 2 = 1 + 0.639 604 077 199 36;
  • 53) 0.639 604 077 199 36 × 2 = 1 + 0.279 208 154 398 72;
  • 54) 0.279 208 154 398 72 × 2 = 0 + 0.558 416 308 797 44;
  • 55) 0.558 416 308 797 44 × 2 = 1 + 0.116 832 617 594 88;
  • 56) 0.116 832 617 594 88 × 2 = 0 + 0.233 665 235 189 76;
  • 57) 0.233 665 235 189 76 × 2 = 0 + 0.467 330 470 379 52;
  • 58) 0.467 330 470 379 52 × 2 = 0 + 0.934 660 940 759 04;
  • 59) 0.934 660 940 759 04 × 2 = 1 + 0.869 321 881 518 08;
  • 60) 0.869 321 881 518 08 × 2 = 1 + 0.738 643 763 036 16;
  • 61) 0.738 643 763 036 16 × 2 = 1 + 0.477 287 526 072 32;
  • 62) 0.477 287 526 072 32 × 2 = 0 + 0.954 575 052 144 64;
  • 63) 0.954 575 052 144 64 × 2 = 1 + 0.909 150 104 289 28;
  • 64) 0.909 150 104 289 28 × 2 = 1 + 0.818 300 208 578 56;

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:


0.000 375 658 488 91(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1100 1010 0001 0001 1010 0011 1011(2)

6. Positive number before normalization:

0.000 375 658 488 91(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1100 1010 0001 0001 1010 0011 1011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 375 658 488 91(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1100 1010 0001 0001 1010 0011 1011(2) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1100 1010 0001 0001 1010 0011 1011(2) × 20 =

1.1000 1001 1110 1000 0000 1110 1100 1010 0001 0001 1010 0011 1011(2) × 2-12


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -12

Mantissa (not normalized):
1.1000 1001 1110 1000 0000 1110 1100 1010 0001 0001 1010 0011 1011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 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) =

1011(10) =

011 1111 0011(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 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 1000 1001 1110 1000 0000 1110 1100 1010 0001 0001 1010 0011 1011 =

1000 1001 1110 1000 0000 1110 1100 1010 0001 0001 1010 0011 1011


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 0011

Mantissa (52 bits) =
1000 1001 1110 1000 0000 1110 1100 1010 0001 0001 1010 0011 1011


Decimal number -0.000 375 658 488 91 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 1000 1001 1110 1000 0000 1110 1100 1010 0001 0001 1010 0011 1011

Share

» Convert decimal -0.000 375 658 489 05 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 09, 2026 [September]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

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

How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double 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 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 from the previous 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 multiplying operations, starting from the top of the list constructed 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, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' 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 -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 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:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. 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) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) 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:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100