-0.000 375 658 488 48 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 375 658 488 48(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 48(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 48| = 0.000 375 658 488 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: 0.000 375 658 488 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) 0.000 375 658 488 48 × 2 = 0 + 0.000 751 316 976 96;
  • 2) 0.000 751 316 976 96 × 2 = 0 + 0.001 502 633 953 92;
  • 3) 0.001 502 633 953 92 × 2 = 0 + 0.003 005 267 907 84;
  • 4) 0.003 005 267 907 84 × 2 = 0 + 0.006 010 535 815 68;
  • 5) 0.006 010 535 815 68 × 2 = 0 + 0.012 021 071 631 36;
  • 6) 0.012 021 071 631 36 × 2 = 0 + 0.024 042 143 262 72;
  • 7) 0.024 042 143 262 72 × 2 = 0 + 0.048 084 286 525 44;
  • 8) 0.048 084 286 525 44 × 2 = 0 + 0.096 168 573 050 88;
  • 9) 0.096 168 573 050 88 × 2 = 0 + 0.192 337 146 101 76;
  • 10) 0.192 337 146 101 76 × 2 = 0 + 0.384 674 292 203 52;
  • 11) 0.384 674 292 203 52 × 2 = 0 + 0.769 348 584 407 04;
  • 12) 0.769 348 584 407 04 × 2 = 1 + 0.538 697 168 814 08;
  • 13) 0.538 697 168 814 08 × 2 = 1 + 0.077 394 337 628 16;
  • 14) 0.077 394 337 628 16 × 2 = 0 + 0.154 788 675 256 32;
  • 15) 0.154 788 675 256 32 × 2 = 0 + 0.309 577 350 512 64;
  • 16) 0.309 577 350 512 64 × 2 = 0 + 0.619 154 701 025 28;
  • 17) 0.619 154 701 025 28 × 2 = 1 + 0.238 309 402 050 56;
  • 18) 0.238 309 402 050 56 × 2 = 0 + 0.476 618 804 101 12;
  • 19) 0.476 618 804 101 12 × 2 = 0 + 0.953 237 608 202 24;
  • 20) 0.953 237 608 202 24 × 2 = 1 + 0.906 475 216 404 48;
  • 21) 0.906 475 216 404 48 × 2 = 1 + 0.812 950 432 808 96;
  • 22) 0.812 950 432 808 96 × 2 = 1 + 0.625 900 865 617 92;
  • 23) 0.625 900 865 617 92 × 2 = 1 + 0.251 801 731 235 84;
  • 24) 0.251 801 731 235 84 × 2 = 0 + 0.503 603 462 471 68;
  • 25) 0.503 603 462 471 68 × 2 = 1 + 0.007 206 924 943 36;
  • 26) 0.007 206 924 943 36 × 2 = 0 + 0.014 413 849 886 72;
  • 27) 0.014 413 849 886 72 × 2 = 0 + 0.028 827 699 773 44;
  • 28) 0.028 827 699 773 44 × 2 = 0 + 0.057 655 399 546 88;
  • 29) 0.057 655 399 546 88 × 2 = 0 + 0.115 310 799 093 76;
  • 30) 0.115 310 799 093 76 × 2 = 0 + 0.230 621 598 187 52;
  • 31) 0.230 621 598 187 52 × 2 = 0 + 0.461 243 196 375 04;
  • 32) 0.461 243 196 375 04 × 2 = 0 + 0.922 486 392 750 08;
  • 33) 0.922 486 392 750 08 × 2 = 1 + 0.844 972 785 500 16;
  • 34) 0.844 972 785 500 16 × 2 = 1 + 0.689 945 571 000 32;
  • 35) 0.689 945 571 000 32 × 2 = 1 + 0.379 891 142 000 64;
  • 36) 0.379 891 142 000 64 × 2 = 0 + 0.759 782 284 001 28;
  • 37) 0.759 782 284 001 28 × 2 = 1 + 0.519 564 568 002 56;
  • 38) 0.519 564 568 002 56 × 2 = 1 + 0.039 129 136 005 12;
  • 39) 0.039 129 136 005 12 × 2 = 0 + 0.078 258 272 010 24;
  • 40) 0.078 258 272 010 24 × 2 = 0 + 0.156 516 544 020 48;
  • 41) 0.156 516 544 020 48 × 2 = 0 + 0.313 033 088 040 96;
  • 42) 0.313 033 088 040 96 × 2 = 0 + 0.626 066 176 081 92;
  • 43) 0.626 066 176 081 92 × 2 = 1 + 0.252 132 352 163 84;
  • 44) 0.252 132 352 163 84 × 2 = 0 + 0.504 264 704 327 68;
  • 45) 0.504 264 704 327 68 × 2 = 1 + 0.008 529 408 655 36;
  • 46) 0.008 529 408 655 36 × 2 = 0 + 0.017 058 817 310 72;
  • 47) 0.017 058 817 310 72 × 2 = 0 + 0.034 117 634 621 44;
  • 48) 0.034 117 634 621 44 × 2 = 0 + 0.068 235 269 242 88;
  • 49) 0.068 235 269 242 88 × 2 = 0 + 0.136 470 538 485 76;
  • 50) 0.136 470 538 485 76 × 2 = 0 + 0.272 941 076 971 52;
  • 51) 0.272 941 076 971 52 × 2 = 0 + 0.545 882 153 943 04;
  • 52) 0.545 882 153 943 04 × 2 = 1 + 0.091 764 307 886 08;
  • 53) 0.091 764 307 886 08 × 2 = 0 + 0.183 528 615 772 16;
  • 54) 0.183 528 615 772 16 × 2 = 0 + 0.367 057 231 544 32;
  • 55) 0.367 057 231 544 32 × 2 = 0 + 0.734 114 463 088 64;
  • 56) 0.734 114 463 088 64 × 2 = 1 + 0.468 228 926 177 28;
  • 57) 0.468 228 926 177 28 × 2 = 0 + 0.936 457 852 354 56;
  • 58) 0.936 457 852 354 56 × 2 = 1 + 0.872 915 704 709 12;
  • 59) 0.872 915 704 709 12 × 2 = 1 + 0.745 831 409 418 24;
  • 60) 0.745 831 409 418 24 × 2 = 1 + 0.491 662 818 836 48;
  • 61) 0.491 662 818 836 48 × 2 = 0 + 0.983 325 637 672 96;
  • 62) 0.983 325 637 672 96 × 2 = 1 + 0.966 651 275 345 92;
  • 63) 0.966 651 275 345 92 × 2 = 1 + 0.933 302 550 691 84;
  • 64) 0.933 302 550 691 84 × 2 = 1 + 0.866 605 101 383 68;

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 48(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1100 0010 1000 0001 0001 0111 0111(2)

6. Positive number before normalization:

0.000 375 658 488 48(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1100 0010 1000 0001 0001 0111 0111(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 48(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1100 0010 1000 0001 0001 0111 0111(2) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1100 0010 1000 0001 0001 0111 0111(2) × 20 =

1.1000 1001 1110 1000 0000 1110 1100 0010 1000 0001 0001 0111 0111(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 0010 1000 0001 0001 0111 0111


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 0010 1000 0001 0001 0111 0111 =

1000 1001 1110 1000 0000 1110 1100 0010 1000 0001 0001 0111 0111


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 0010 1000 0001 0001 0111 0111


Decimal number -0.000 375 658 488 48 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 1000 1001 1110 1000 0000 1110 1100 0010 1000 0001 0001 0111 0111

Share

» Convert decimal -0.000 375 658 489 02 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 Attention Economy - The Battlefield For Your Focus. Gemini Notebook YouTube Video of 7.50 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