-0.000 282 005 914 389 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 389(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
-0.000 282 005 914 389(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:

|-0.000 282 005 914 389| = 0.000 282 005 914 389


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: 0.000 282 005 914 389.

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) 0.000 282 005 914 389 × 2 = 0 + 0.000 564 011 828 778;
  • 2) 0.000 564 011 828 778 × 2 = 0 + 0.001 128 023 657 556;
  • 3) 0.001 128 023 657 556 × 2 = 0 + 0.002 256 047 315 112;
  • 4) 0.002 256 047 315 112 × 2 = 0 + 0.004 512 094 630 224;
  • 5) 0.004 512 094 630 224 × 2 = 0 + 0.009 024 189 260 448;
  • 6) 0.009 024 189 260 448 × 2 = 0 + 0.018 048 378 520 896;
  • 7) 0.018 048 378 520 896 × 2 = 0 + 0.036 096 757 041 792;
  • 8) 0.036 096 757 041 792 × 2 = 0 + 0.072 193 514 083 584;
  • 9) 0.072 193 514 083 584 × 2 = 0 + 0.144 387 028 167 168;
  • 10) 0.144 387 028 167 168 × 2 = 0 + 0.288 774 056 334 336;
  • 11) 0.288 774 056 334 336 × 2 = 0 + 0.577 548 112 668 672;
  • 12) 0.577 548 112 668 672 × 2 = 1 + 0.155 096 225 337 344;
  • 13) 0.155 096 225 337 344 × 2 = 0 + 0.310 192 450 674 688;
  • 14) 0.310 192 450 674 688 × 2 = 0 + 0.620 384 901 349 376;
  • 15) 0.620 384 901 349 376 × 2 = 1 + 0.240 769 802 698 752;
  • 16) 0.240 769 802 698 752 × 2 = 0 + 0.481 539 605 397 504;
  • 17) 0.481 539 605 397 504 × 2 = 0 + 0.963 079 210 795 008;
  • 18) 0.963 079 210 795 008 × 2 = 1 + 0.926 158 421 590 016;
  • 19) 0.926 158 421 590 016 × 2 = 1 + 0.852 316 843 180 032;
  • 20) 0.852 316 843 180 032 × 2 = 1 + 0.704 633 686 360 064;
  • 21) 0.704 633 686 360 064 × 2 = 1 + 0.409 267 372 720 128;
  • 22) 0.409 267 372 720 128 × 2 = 0 + 0.818 534 745 440 256;
  • 23) 0.818 534 745 440 256 × 2 = 1 + 0.637 069 490 880 512;
  • 24) 0.637 069 490 880 512 × 2 = 1 + 0.274 138 981 761 024;
  • 25) 0.274 138 981 761 024 × 2 = 0 + 0.548 277 963 522 048;
  • 26) 0.548 277 963 522 048 × 2 = 1 + 0.096 555 927 044 096;
  • 27) 0.096 555 927 044 096 × 2 = 0 + 0.193 111 854 088 192;
  • 28) 0.193 111 854 088 192 × 2 = 0 + 0.386 223 708 176 384;
  • 29) 0.386 223 708 176 384 × 2 = 0 + 0.772 447 416 352 768;
  • 30) 0.772 447 416 352 768 × 2 = 1 + 0.544 894 832 705 536;
  • 31) 0.544 894 832 705 536 × 2 = 1 + 0.089 789 665 411 072;
  • 32) 0.089 789 665 411 072 × 2 = 0 + 0.179 579 330 822 144;
  • 33) 0.179 579 330 822 144 × 2 = 0 + 0.359 158 661 644 288;
  • 34) 0.359 158 661 644 288 × 2 = 0 + 0.718 317 323 288 576;
  • 35) 0.718 317 323 288 576 × 2 = 1 + 0.436 634 646 577 152;
  • 36) 0.436 634 646 577 152 × 2 = 0 + 0.873 269 293 154 304;
  • 37) 0.873 269 293 154 304 × 2 = 1 + 0.746 538 586 308 608;
  • 38) 0.746 538 586 308 608 × 2 = 1 + 0.493 077 172 617 216;
  • 39) 0.493 077 172 617 216 × 2 = 0 + 0.986 154 345 234 432;
  • 40) 0.986 154 345 234 432 × 2 = 1 + 0.972 308 690 468 864;
  • 41) 0.972 308 690 468 864 × 2 = 1 + 0.944 617 380 937 728;
  • 42) 0.944 617 380 937 728 × 2 = 1 + 0.889 234 761 875 456;
  • 43) 0.889 234 761 875 456 × 2 = 1 + 0.778 469 523 750 912;
  • 44) 0.778 469 523 750 912 × 2 = 1 + 0.556 939 047 501 824;
  • 45) 0.556 939 047 501 824 × 2 = 1 + 0.113 878 095 003 648;
  • 46) 0.113 878 095 003 648 × 2 = 0 + 0.227 756 190 007 296;
  • 47) 0.227 756 190 007 296 × 2 = 0 + 0.455 512 380 014 592;
  • 48) 0.455 512 380 014 592 × 2 = 0 + 0.911 024 760 029 184;
  • 49) 0.911 024 760 029 184 × 2 = 1 + 0.822 049 520 058 368;
  • 50) 0.822 049 520 058 368 × 2 = 1 + 0.644 099 040 116 736;
  • 51) 0.644 099 040 116 736 × 2 = 1 + 0.288 198 080 233 472;
  • 52) 0.288 198 080 233 472 × 2 = 0 + 0.576 396 160 466 944;
  • 53) 0.576 396 160 466 944 × 2 = 1 + 0.152 792 320 933 888;
  • 54) 0.152 792 320 933 888 × 2 = 0 + 0.305 584 641 867 776;
  • 55) 0.305 584 641 867 776 × 2 = 0 + 0.611 169 283 735 552;
  • 56) 0.611 169 283 735 552 × 2 = 1 + 0.222 338 567 471 104;
  • 57) 0.222 338 567 471 104 × 2 = 0 + 0.444 677 134 942 208;
  • 58) 0.444 677 134 942 208 × 2 = 0 + 0.889 354 269 884 416;
  • 59) 0.889 354 269 884 416 × 2 = 1 + 0.778 708 539 768 832;
  • 60) 0.778 708 539 768 832 × 2 = 1 + 0.557 417 079 537 664;
  • 61) 0.557 417 079 537 664 × 2 = 1 + 0.114 834 159 075 328;
  • 62) 0.114 834 159 075 328 × 2 = 0 + 0.229 668 318 150 656;
  • 63) 0.229 668 318 150 656 × 2 = 0 + 0.459 336 636 301 312;
  • 64) 0.459 336 636 301 312 × 2 = 0 + 0.918 673 272 602 624;

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:


0.000 282 005 914 389(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1000 1110 1001 0011 1000(2)

6. Positive number before normalization:

0.000 282 005 914 389(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1000 1110 1001 0011 1000(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:


0.000 282 005 914 389(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1000 1110 1001 0011 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1000 1110 1001 0011 1000(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1111 1000 1110 1001 0011 1000(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):
1.0010 0111 1011 0100 0110 0010 1101 1111 1000 1110 1001 0011 1000


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. 0010 0111 1011 0100 0110 0010 1101 1111 1000 1110 1001 0011 1000 =

0010 0111 1011 0100 0110 0010 1101 1111 1000 1110 1001 0011 1000


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) =
0010 0111 1011 0100 0110 0010 1101 1111 1000 1110 1001 0011 1000


Decimal number -0.000 282 005 914 389 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1111 1000 1110 1001 0011 1000

Share

» Convert decimal -0.000 282 005 914 393 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 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 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