-0.000 282 005 914 476 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 476(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
-0.000 282 005 914 476(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:

|-0.000 282 005 914 476| = 0.000 282 005 914 476


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: 0.000 282 005 914 476.

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) 0.000 282 005 914 476 × 2 = 0 + 0.000 564 011 828 952;
  • 2) 0.000 564 011 828 952 × 2 = 0 + 0.001 128 023 657 904;
  • 3) 0.001 128 023 657 904 × 2 = 0 + 0.002 256 047 315 808;
  • 4) 0.002 256 047 315 808 × 2 = 0 + 0.004 512 094 631 616;
  • 5) 0.004 512 094 631 616 × 2 = 0 + 0.009 024 189 263 232;
  • 6) 0.009 024 189 263 232 × 2 = 0 + 0.018 048 378 526 464;
  • 7) 0.018 048 378 526 464 × 2 = 0 + 0.036 096 757 052 928;
  • 8) 0.036 096 757 052 928 × 2 = 0 + 0.072 193 514 105 856;
  • 9) 0.072 193 514 105 856 × 2 = 0 + 0.144 387 028 211 712;
  • 10) 0.144 387 028 211 712 × 2 = 0 + 0.288 774 056 423 424;
  • 11) 0.288 774 056 423 424 × 2 = 0 + 0.577 548 112 846 848;
  • 12) 0.577 548 112 846 848 × 2 = 1 + 0.155 096 225 693 696;
  • 13) 0.155 096 225 693 696 × 2 = 0 + 0.310 192 451 387 392;
  • 14) 0.310 192 451 387 392 × 2 = 0 + 0.620 384 902 774 784;
  • 15) 0.620 384 902 774 784 × 2 = 1 + 0.240 769 805 549 568;
  • 16) 0.240 769 805 549 568 × 2 = 0 + 0.481 539 611 099 136;
  • 17) 0.481 539 611 099 136 × 2 = 0 + 0.963 079 222 198 272;
  • 18) 0.963 079 222 198 272 × 2 = 1 + 0.926 158 444 396 544;
  • 19) 0.926 158 444 396 544 × 2 = 1 + 0.852 316 888 793 088;
  • 20) 0.852 316 888 793 088 × 2 = 1 + 0.704 633 777 586 176;
  • 21) 0.704 633 777 586 176 × 2 = 1 + 0.409 267 555 172 352;
  • 22) 0.409 267 555 172 352 × 2 = 0 + 0.818 535 110 344 704;
  • 23) 0.818 535 110 344 704 × 2 = 1 + 0.637 070 220 689 408;
  • 24) 0.637 070 220 689 408 × 2 = 1 + 0.274 140 441 378 816;
  • 25) 0.274 140 441 378 816 × 2 = 0 + 0.548 280 882 757 632;
  • 26) 0.548 280 882 757 632 × 2 = 1 + 0.096 561 765 515 264;
  • 27) 0.096 561 765 515 264 × 2 = 0 + 0.193 123 531 030 528;
  • 28) 0.193 123 531 030 528 × 2 = 0 + 0.386 247 062 061 056;
  • 29) 0.386 247 062 061 056 × 2 = 0 + 0.772 494 124 122 112;
  • 30) 0.772 494 124 122 112 × 2 = 1 + 0.544 988 248 244 224;
  • 31) 0.544 988 248 244 224 × 2 = 1 + 0.089 976 496 488 448;
  • 32) 0.089 976 496 488 448 × 2 = 0 + 0.179 952 992 976 896;
  • 33) 0.179 952 992 976 896 × 2 = 0 + 0.359 905 985 953 792;
  • 34) 0.359 905 985 953 792 × 2 = 0 + 0.719 811 971 907 584;
  • 35) 0.719 811 971 907 584 × 2 = 1 + 0.439 623 943 815 168;
  • 36) 0.439 623 943 815 168 × 2 = 0 + 0.879 247 887 630 336;
  • 37) 0.879 247 887 630 336 × 2 = 1 + 0.758 495 775 260 672;
  • 38) 0.758 495 775 260 672 × 2 = 1 + 0.516 991 550 521 344;
  • 39) 0.516 991 550 521 344 × 2 = 1 + 0.033 983 101 042 688;
  • 40) 0.033 983 101 042 688 × 2 = 0 + 0.067 966 202 085 376;
  • 41) 0.067 966 202 085 376 × 2 = 0 + 0.135 932 404 170 752;
  • 42) 0.135 932 404 170 752 × 2 = 0 + 0.271 864 808 341 504;
  • 43) 0.271 864 808 341 504 × 2 = 0 + 0.543 729 616 683 008;
  • 44) 0.543 729 616 683 008 × 2 = 1 + 0.087 459 233 366 016;
  • 45) 0.087 459 233 366 016 × 2 = 0 + 0.174 918 466 732 032;
  • 46) 0.174 918 466 732 032 × 2 = 0 + 0.349 836 933 464 064;
  • 47) 0.349 836 933 464 064 × 2 = 0 + 0.699 673 866 928 128;
  • 48) 0.699 673 866 928 128 × 2 = 1 + 0.399 347 733 856 256;
  • 49) 0.399 347 733 856 256 × 2 = 0 + 0.798 695 467 712 512;
  • 50) 0.798 695 467 712 512 × 2 = 1 + 0.597 390 935 425 024;
  • 51) 0.597 390 935 425 024 × 2 = 1 + 0.194 781 870 850 048;
  • 52) 0.194 781 870 850 048 × 2 = 0 + 0.389 563 741 700 096;
  • 53) 0.389 563 741 700 096 × 2 = 0 + 0.779 127 483 400 192;
  • 54) 0.779 127 483 400 192 × 2 = 1 + 0.558 254 966 800 384;
  • 55) 0.558 254 966 800 384 × 2 = 1 + 0.116 509 933 600 768;
  • 56) 0.116 509 933 600 768 × 2 = 0 + 0.233 019 867 201 536;
  • 57) 0.233 019 867 201 536 × 2 = 0 + 0.466 039 734 403 072;
  • 58) 0.466 039 734 403 072 × 2 = 0 + 0.932 079 468 806 144;
  • 59) 0.932 079 468 806 144 × 2 = 1 + 0.864 158 937 612 288;
  • 60) 0.864 158 937 612 288 × 2 = 1 + 0.728 317 875 224 576;
  • 61) 0.728 317 875 224 576 × 2 = 1 + 0.456 635 750 449 152;
  • 62) 0.456 635 750 449 152 × 2 = 0 + 0.913 271 500 898 304;
  • 63) 0.913 271 500 898 304 × 2 = 1 + 0.826 543 001 796 608;
  • 64) 0.826 543 001 796 608 × 2 = 1 + 0.653 086 003 593 216;

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:


0.000 282 005 914 476(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 0001 0110 0110 0011 1011(2)

6. Positive number before normalization:

0.000 282 005 914 476(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 0001 0110 0110 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:


0.000 282 005 914 476(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 0001 0110 0110 0011 1011(2) =

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

1.0010 0111 1011 0100 0110 0010 1110 0001 0001 0110 0110 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):
1.0010 0111 1011 0100 0110 0010 1110 0001 0001 0110 0110 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. 0010 0111 1011 0100 0110 0010 1110 0001 0001 0110 0110 0011 1011 =

0010 0111 1011 0100 0110 0010 1110 0001 0001 0110 0110 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) =
0010 0111 1011 0100 0110 0010 1110 0001 0001 0110 0110 0011 1011


Decimal number -0.000 282 005 914 476 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0001 0001 0110 0110 0011 1011

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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