-0.000 282 005 914 382 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 382(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
-0.000 282 005 914 382(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:

|-0.000 282 005 914 382| = 0.000 282 005 914 382


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: 0.000 282 005 914 382.

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) 0.000 282 005 914 382 × 2 = 0 + 0.000 564 011 828 764;
  • 2) 0.000 564 011 828 764 × 2 = 0 + 0.001 128 023 657 528;
  • 3) 0.001 128 023 657 528 × 2 = 0 + 0.002 256 047 315 056;
  • 4) 0.002 256 047 315 056 × 2 = 0 + 0.004 512 094 630 112;
  • 5) 0.004 512 094 630 112 × 2 = 0 + 0.009 024 189 260 224;
  • 6) 0.009 024 189 260 224 × 2 = 0 + 0.018 048 378 520 448;
  • 7) 0.018 048 378 520 448 × 2 = 0 + 0.036 096 757 040 896;
  • 8) 0.036 096 757 040 896 × 2 = 0 + 0.072 193 514 081 792;
  • 9) 0.072 193 514 081 792 × 2 = 0 + 0.144 387 028 163 584;
  • 10) 0.144 387 028 163 584 × 2 = 0 + 0.288 774 056 327 168;
  • 11) 0.288 774 056 327 168 × 2 = 0 + 0.577 548 112 654 336;
  • 12) 0.577 548 112 654 336 × 2 = 1 + 0.155 096 225 308 672;
  • 13) 0.155 096 225 308 672 × 2 = 0 + 0.310 192 450 617 344;
  • 14) 0.310 192 450 617 344 × 2 = 0 + 0.620 384 901 234 688;
  • 15) 0.620 384 901 234 688 × 2 = 1 + 0.240 769 802 469 376;
  • 16) 0.240 769 802 469 376 × 2 = 0 + 0.481 539 604 938 752;
  • 17) 0.481 539 604 938 752 × 2 = 0 + 0.963 079 209 877 504;
  • 18) 0.963 079 209 877 504 × 2 = 1 + 0.926 158 419 755 008;
  • 19) 0.926 158 419 755 008 × 2 = 1 + 0.852 316 839 510 016;
  • 20) 0.852 316 839 510 016 × 2 = 1 + 0.704 633 679 020 032;
  • 21) 0.704 633 679 020 032 × 2 = 1 + 0.409 267 358 040 064;
  • 22) 0.409 267 358 040 064 × 2 = 0 + 0.818 534 716 080 128;
  • 23) 0.818 534 716 080 128 × 2 = 1 + 0.637 069 432 160 256;
  • 24) 0.637 069 432 160 256 × 2 = 1 + 0.274 138 864 320 512;
  • 25) 0.274 138 864 320 512 × 2 = 0 + 0.548 277 728 641 024;
  • 26) 0.548 277 728 641 024 × 2 = 1 + 0.096 555 457 282 048;
  • 27) 0.096 555 457 282 048 × 2 = 0 + 0.193 110 914 564 096;
  • 28) 0.193 110 914 564 096 × 2 = 0 + 0.386 221 829 128 192;
  • 29) 0.386 221 829 128 192 × 2 = 0 + 0.772 443 658 256 384;
  • 30) 0.772 443 658 256 384 × 2 = 1 + 0.544 887 316 512 768;
  • 31) 0.544 887 316 512 768 × 2 = 1 + 0.089 774 633 025 536;
  • 32) 0.089 774 633 025 536 × 2 = 0 + 0.179 549 266 051 072;
  • 33) 0.179 549 266 051 072 × 2 = 0 + 0.359 098 532 102 144;
  • 34) 0.359 098 532 102 144 × 2 = 0 + 0.718 197 064 204 288;
  • 35) 0.718 197 064 204 288 × 2 = 1 + 0.436 394 128 408 576;
  • 36) 0.436 394 128 408 576 × 2 = 0 + 0.872 788 256 817 152;
  • 37) 0.872 788 256 817 152 × 2 = 1 + 0.745 576 513 634 304;
  • 38) 0.745 576 513 634 304 × 2 = 1 + 0.491 153 027 268 608;
  • 39) 0.491 153 027 268 608 × 2 = 0 + 0.982 306 054 537 216;
  • 40) 0.982 306 054 537 216 × 2 = 1 + 0.964 612 109 074 432;
  • 41) 0.964 612 109 074 432 × 2 = 1 + 0.929 224 218 148 864;
  • 42) 0.929 224 218 148 864 × 2 = 1 + 0.858 448 436 297 728;
  • 43) 0.858 448 436 297 728 × 2 = 1 + 0.716 896 872 595 456;
  • 44) 0.716 896 872 595 456 × 2 = 1 + 0.433 793 745 190 912;
  • 45) 0.433 793 745 190 912 × 2 = 0 + 0.867 587 490 381 824;
  • 46) 0.867 587 490 381 824 × 2 = 1 + 0.735 174 980 763 648;
  • 47) 0.735 174 980 763 648 × 2 = 1 + 0.470 349 961 527 296;
  • 48) 0.470 349 961 527 296 × 2 = 0 + 0.940 699 923 054 592;
  • 49) 0.940 699 923 054 592 × 2 = 1 + 0.881 399 846 109 184;
  • 50) 0.881 399 846 109 184 × 2 = 1 + 0.762 799 692 218 368;
  • 51) 0.762 799 692 218 368 × 2 = 1 + 0.525 599 384 436 736;
  • 52) 0.525 599 384 436 736 × 2 = 1 + 0.051 198 768 873 472;
  • 53) 0.051 198 768 873 472 × 2 = 0 + 0.102 397 537 746 944;
  • 54) 0.102 397 537 746 944 × 2 = 0 + 0.204 795 075 493 888;
  • 55) 0.204 795 075 493 888 × 2 = 0 + 0.409 590 150 987 776;
  • 56) 0.409 590 150 987 776 × 2 = 0 + 0.819 180 301 975 552;
  • 57) 0.819 180 301 975 552 × 2 = 1 + 0.638 360 603 951 104;
  • 58) 0.638 360 603 951 104 × 2 = 1 + 0.276 721 207 902 208;
  • 59) 0.276 721 207 902 208 × 2 = 0 + 0.553 442 415 804 416;
  • 60) 0.553 442 415 804 416 × 2 = 1 + 0.106 884 831 608 832;
  • 61) 0.106 884 831 608 832 × 2 = 0 + 0.213 769 663 217 664;
  • 62) 0.213 769 663 217 664 × 2 = 0 + 0.427 539 326 435 328;
  • 63) 0.427 539 326 435 328 × 2 = 0 + 0.855 078 652 870 656;
  • 64) 0.855 078 652 870 656 × 2 = 1 + 0.710 157 305 741 312;

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:


0.000 282 005 914 382(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 0110 1111 0000 1101 0001(2)

6. Positive number before normalization:

0.000 282 005 914 382(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 0110 1111 0000 1101 0001(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:


0.000 282 005 914 382(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 0110 1111 0000 1101 0001(2) =

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

1.0010 0111 1011 0100 0110 0010 1101 1111 0110 1111 0000 1101 0001(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 0110 1111 0000 1101 0001


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 0110 1111 0000 1101 0001 =

0010 0111 1011 0100 0110 0010 1101 1111 0110 1111 0000 1101 0001


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 0110 1111 0000 1101 0001


Decimal number -0.000 282 005 914 382 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1111 0110 1111 0000 1101 0001

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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