-0.000 282 005 914 473 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 473(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
-0.000 282 005 914 473(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:

|-0.000 282 005 914 473| = 0.000 282 005 914 473


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: 0.000 282 005 914 473.

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) 0.000 282 005 914 473 × 2 = 0 + 0.000 564 011 828 946;
  • 2) 0.000 564 011 828 946 × 2 = 0 + 0.001 128 023 657 892;
  • 3) 0.001 128 023 657 892 × 2 = 0 + 0.002 256 047 315 784;
  • 4) 0.002 256 047 315 784 × 2 = 0 + 0.004 512 094 631 568;
  • 5) 0.004 512 094 631 568 × 2 = 0 + 0.009 024 189 263 136;
  • 6) 0.009 024 189 263 136 × 2 = 0 + 0.018 048 378 526 272;
  • 7) 0.018 048 378 526 272 × 2 = 0 + 0.036 096 757 052 544;
  • 8) 0.036 096 757 052 544 × 2 = 0 + 0.072 193 514 105 088;
  • 9) 0.072 193 514 105 088 × 2 = 0 + 0.144 387 028 210 176;
  • 10) 0.144 387 028 210 176 × 2 = 0 + 0.288 774 056 420 352;
  • 11) 0.288 774 056 420 352 × 2 = 0 + 0.577 548 112 840 704;
  • 12) 0.577 548 112 840 704 × 2 = 1 + 0.155 096 225 681 408;
  • 13) 0.155 096 225 681 408 × 2 = 0 + 0.310 192 451 362 816;
  • 14) 0.310 192 451 362 816 × 2 = 0 + 0.620 384 902 725 632;
  • 15) 0.620 384 902 725 632 × 2 = 1 + 0.240 769 805 451 264;
  • 16) 0.240 769 805 451 264 × 2 = 0 + 0.481 539 610 902 528;
  • 17) 0.481 539 610 902 528 × 2 = 0 + 0.963 079 221 805 056;
  • 18) 0.963 079 221 805 056 × 2 = 1 + 0.926 158 443 610 112;
  • 19) 0.926 158 443 610 112 × 2 = 1 + 0.852 316 887 220 224;
  • 20) 0.852 316 887 220 224 × 2 = 1 + 0.704 633 774 440 448;
  • 21) 0.704 633 774 440 448 × 2 = 1 + 0.409 267 548 880 896;
  • 22) 0.409 267 548 880 896 × 2 = 0 + 0.818 535 097 761 792;
  • 23) 0.818 535 097 761 792 × 2 = 1 + 0.637 070 195 523 584;
  • 24) 0.637 070 195 523 584 × 2 = 1 + 0.274 140 391 047 168;
  • 25) 0.274 140 391 047 168 × 2 = 0 + 0.548 280 782 094 336;
  • 26) 0.548 280 782 094 336 × 2 = 1 + 0.096 561 564 188 672;
  • 27) 0.096 561 564 188 672 × 2 = 0 + 0.193 123 128 377 344;
  • 28) 0.193 123 128 377 344 × 2 = 0 + 0.386 246 256 754 688;
  • 29) 0.386 246 256 754 688 × 2 = 0 + 0.772 492 513 509 376;
  • 30) 0.772 492 513 509 376 × 2 = 1 + 0.544 985 027 018 752;
  • 31) 0.544 985 027 018 752 × 2 = 1 + 0.089 970 054 037 504;
  • 32) 0.089 970 054 037 504 × 2 = 0 + 0.179 940 108 075 008;
  • 33) 0.179 940 108 075 008 × 2 = 0 + 0.359 880 216 150 016;
  • 34) 0.359 880 216 150 016 × 2 = 0 + 0.719 760 432 300 032;
  • 35) 0.719 760 432 300 032 × 2 = 1 + 0.439 520 864 600 064;
  • 36) 0.439 520 864 600 064 × 2 = 0 + 0.879 041 729 200 128;
  • 37) 0.879 041 729 200 128 × 2 = 1 + 0.758 083 458 400 256;
  • 38) 0.758 083 458 400 256 × 2 = 1 + 0.516 166 916 800 512;
  • 39) 0.516 166 916 800 512 × 2 = 1 + 0.032 333 833 601 024;
  • 40) 0.032 333 833 601 024 × 2 = 0 + 0.064 667 667 202 048;
  • 41) 0.064 667 667 202 048 × 2 = 0 + 0.129 335 334 404 096;
  • 42) 0.129 335 334 404 096 × 2 = 0 + 0.258 670 668 808 192;
  • 43) 0.258 670 668 808 192 × 2 = 0 + 0.517 341 337 616 384;
  • 44) 0.517 341 337 616 384 × 2 = 1 + 0.034 682 675 232 768;
  • 45) 0.034 682 675 232 768 × 2 = 0 + 0.069 365 350 465 536;
  • 46) 0.069 365 350 465 536 × 2 = 0 + 0.138 730 700 931 072;
  • 47) 0.138 730 700 931 072 × 2 = 0 + 0.277 461 401 862 144;
  • 48) 0.277 461 401 862 144 × 2 = 0 + 0.554 922 803 724 288;
  • 49) 0.554 922 803 724 288 × 2 = 1 + 0.109 845 607 448 576;
  • 50) 0.109 845 607 448 576 × 2 = 0 + 0.219 691 214 897 152;
  • 51) 0.219 691 214 897 152 × 2 = 0 + 0.439 382 429 794 304;
  • 52) 0.439 382 429 794 304 × 2 = 0 + 0.878 764 859 588 608;
  • 53) 0.878 764 859 588 608 × 2 = 1 + 0.757 529 719 177 216;
  • 54) 0.757 529 719 177 216 × 2 = 1 + 0.515 059 438 354 432;
  • 55) 0.515 059 438 354 432 × 2 = 1 + 0.030 118 876 708 864;
  • 56) 0.030 118 876 708 864 × 2 = 0 + 0.060 237 753 417 728;
  • 57) 0.060 237 753 417 728 × 2 = 0 + 0.120 475 506 835 456;
  • 58) 0.120 475 506 835 456 × 2 = 0 + 0.240 951 013 670 912;
  • 59) 0.240 951 013 670 912 × 2 = 0 + 0.481 902 027 341 824;
  • 60) 0.481 902 027 341 824 × 2 = 0 + 0.963 804 054 683 648;
  • 61) 0.963 804 054 683 648 × 2 = 1 + 0.927 608 109 367 296;
  • 62) 0.927 608 109 367 296 × 2 = 1 + 0.855 216 218 734 592;
  • 63) 0.855 216 218 734 592 × 2 = 1 + 0.710 432 437 469 184;
  • 64) 0.710 432 437 469 184 × 2 = 1 + 0.420 864 874 938 368;

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:


0.000 282 005 914 473(10) =

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

6. Positive number before normalization:

0.000 282 005 914 473(10) =

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


0.000 282 005 914 473(10) =

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

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

1.0010 0111 1011 0100 0110 0010 1110 0001 0000 1000 1110 0000 1111(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 0000 1000 1110 0000 1111


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 0000 1000 1110 0000 1111 =

0010 0111 1011 0100 0110 0010 1110 0001 0000 1000 1110 0000 1111


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 0000 1000 1110 0000 1111


Decimal number -0.000 282 005 914 473 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0001 0000 1000 1110 0000 1111

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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