-0.000 282 005 914 469 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 469(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
-0.000 282 005 914 469(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:

|-0.000 282 005 914 469| = 0.000 282 005 914 469


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: 0.000 282 005 914 469.

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) 0.000 282 005 914 469 × 2 = 0 + 0.000 564 011 828 938;
  • 2) 0.000 564 011 828 938 × 2 = 0 + 0.001 128 023 657 876;
  • 3) 0.001 128 023 657 876 × 2 = 0 + 0.002 256 047 315 752;
  • 4) 0.002 256 047 315 752 × 2 = 0 + 0.004 512 094 631 504;
  • 5) 0.004 512 094 631 504 × 2 = 0 + 0.009 024 189 263 008;
  • 6) 0.009 024 189 263 008 × 2 = 0 + 0.018 048 378 526 016;
  • 7) 0.018 048 378 526 016 × 2 = 0 + 0.036 096 757 052 032;
  • 8) 0.036 096 757 052 032 × 2 = 0 + 0.072 193 514 104 064;
  • 9) 0.072 193 514 104 064 × 2 = 0 + 0.144 387 028 208 128;
  • 10) 0.144 387 028 208 128 × 2 = 0 + 0.288 774 056 416 256;
  • 11) 0.288 774 056 416 256 × 2 = 0 + 0.577 548 112 832 512;
  • 12) 0.577 548 112 832 512 × 2 = 1 + 0.155 096 225 665 024;
  • 13) 0.155 096 225 665 024 × 2 = 0 + 0.310 192 451 330 048;
  • 14) 0.310 192 451 330 048 × 2 = 0 + 0.620 384 902 660 096;
  • 15) 0.620 384 902 660 096 × 2 = 1 + 0.240 769 805 320 192;
  • 16) 0.240 769 805 320 192 × 2 = 0 + 0.481 539 610 640 384;
  • 17) 0.481 539 610 640 384 × 2 = 0 + 0.963 079 221 280 768;
  • 18) 0.963 079 221 280 768 × 2 = 1 + 0.926 158 442 561 536;
  • 19) 0.926 158 442 561 536 × 2 = 1 + 0.852 316 885 123 072;
  • 20) 0.852 316 885 123 072 × 2 = 1 + 0.704 633 770 246 144;
  • 21) 0.704 633 770 246 144 × 2 = 1 + 0.409 267 540 492 288;
  • 22) 0.409 267 540 492 288 × 2 = 0 + 0.818 535 080 984 576;
  • 23) 0.818 535 080 984 576 × 2 = 1 + 0.637 070 161 969 152;
  • 24) 0.637 070 161 969 152 × 2 = 1 + 0.274 140 323 938 304;
  • 25) 0.274 140 323 938 304 × 2 = 0 + 0.548 280 647 876 608;
  • 26) 0.548 280 647 876 608 × 2 = 1 + 0.096 561 295 753 216;
  • 27) 0.096 561 295 753 216 × 2 = 0 + 0.193 122 591 506 432;
  • 28) 0.193 122 591 506 432 × 2 = 0 + 0.386 245 183 012 864;
  • 29) 0.386 245 183 012 864 × 2 = 0 + 0.772 490 366 025 728;
  • 30) 0.772 490 366 025 728 × 2 = 1 + 0.544 980 732 051 456;
  • 31) 0.544 980 732 051 456 × 2 = 1 + 0.089 961 464 102 912;
  • 32) 0.089 961 464 102 912 × 2 = 0 + 0.179 922 928 205 824;
  • 33) 0.179 922 928 205 824 × 2 = 0 + 0.359 845 856 411 648;
  • 34) 0.359 845 856 411 648 × 2 = 0 + 0.719 691 712 823 296;
  • 35) 0.719 691 712 823 296 × 2 = 1 + 0.439 383 425 646 592;
  • 36) 0.439 383 425 646 592 × 2 = 0 + 0.878 766 851 293 184;
  • 37) 0.878 766 851 293 184 × 2 = 1 + 0.757 533 702 586 368;
  • 38) 0.757 533 702 586 368 × 2 = 1 + 0.515 067 405 172 736;
  • 39) 0.515 067 405 172 736 × 2 = 1 + 0.030 134 810 345 472;
  • 40) 0.030 134 810 345 472 × 2 = 0 + 0.060 269 620 690 944;
  • 41) 0.060 269 620 690 944 × 2 = 0 + 0.120 539 241 381 888;
  • 42) 0.120 539 241 381 888 × 2 = 0 + 0.241 078 482 763 776;
  • 43) 0.241 078 482 763 776 × 2 = 0 + 0.482 156 965 527 552;
  • 44) 0.482 156 965 527 552 × 2 = 0 + 0.964 313 931 055 104;
  • 45) 0.964 313 931 055 104 × 2 = 1 + 0.928 627 862 110 208;
  • 46) 0.928 627 862 110 208 × 2 = 1 + 0.857 255 724 220 416;
  • 47) 0.857 255 724 220 416 × 2 = 1 + 0.714 511 448 440 832;
  • 48) 0.714 511 448 440 832 × 2 = 1 + 0.429 022 896 881 664;
  • 49) 0.429 022 896 881 664 × 2 = 0 + 0.858 045 793 763 328;
  • 50) 0.858 045 793 763 328 × 2 = 1 + 0.716 091 587 526 656;
  • 51) 0.716 091 587 526 656 × 2 = 1 + 0.432 183 175 053 312;
  • 52) 0.432 183 175 053 312 × 2 = 0 + 0.864 366 350 106 624;
  • 53) 0.864 366 350 106 624 × 2 = 1 + 0.728 732 700 213 248;
  • 54) 0.728 732 700 213 248 × 2 = 1 + 0.457 465 400 426 496;
  • 55) 0.457 465 400 426 496 × 2 = 0 + 0.914 930 800 852 992;
  • 56) 0.914 930 800 852 992 × 2 = 1 + 0.829 861 601 705 984;
  • 57) 0.829 861 601 705 984 × 2 = 1 + 0.659 723 203 411 968;
  • 58) 0.659 723 203 411 968 × 2 = 1 + 0.319 446 406 823 936;
  • 59) 0.319 446 406 823 936 × 2 = 0 + 0.638 892 813 647 872;
  • 60) 0.638 892 813 647 872 × 2 = 1 + 0.277 785 627 295 744;
  • 61) 0.277 785 627 295 744 × 2 = 0 + 0.555 571 254 591 488;
  • 62) 0.555 571 254 591 488 × 2 = 1 + 0.111 142 509 182 976;
  • 63) 0.111 142 509 182 976 × 2 = 0 + 0.222 285 018 365 952;
  • 64) 0.222 285 018 365 952 × 2 = 0 + 0.444 570 036 731 904;

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:


0.000 282 005 914 469(10) =

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

6. Positive number before normalization:

0.000 282 005 914 469(10) =

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


0.000 282 005 914 469(10) =

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

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

1.0010 0111 1011 0100 0110 0010 1110 0000 1111 0110 1101 1101 0100(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 1110 0000 1111 0110 1101 1101 0100


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 1110 0000 1111 0110 1101 1101 0100 =

0010 0111 1011 0100 0110 0010 1110 0000 1111 0110 1101 1101 0100


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 1110 0000 1111 0110 1101 1101 0100


Decimal number -0.000 282 005 914 469 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0000 1111 0110 1101 1101 0100

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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