-0.000 282 005 914 384 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 384(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
-0.000 282 005 914 384(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:

|-0.000 282 005 914 384| = 0.000 282 005 914 384


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: 0.000 282 005 914 384.

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) 0.000 282 005 914 384 × 2 = 0 + 0.000 564 011 828 768;
  • 2) 0.000 564 011 828 768 × 2 = 0 + 0.001 128 023 657 536;
  • 3) 0.001 128 023 657 536 × 2 = 0 + 0.002 256 047 315 072;
  • 4) 0.002 256 047 315 072 × 2 = 0 + 0.004 512 094 630 144;
  • 5) 0.004 512 094 630 144 × 2 = 0 + 0.009 024 189 260 288;
  • 6) 0.009 024 189 260 288 × 2 = 0 + 0.018 048 378 520 576;
  • 7) 0.018 048 378 520 576 × 2 = 0 + 0.036 096 757 041 152;
  • 8) 0.036 096 757 041 152 × 2 = 0 + 0.072 193 514 082 304;
  • 9) 0.072 193 514 082 304 × 2 = 0 + 0.144 387 028 164 608;
  • 10) 0.144 387 028 164 608 × 2 = 0 + 0.288 774 056 329 216;
  • 11) 0.288 774 056 329 216 × 2 = 0 + 0.577 548 112 658 432;
  • 12) 0.577 548 112 658 432 × 2 = 1 + 0.155 096 225 316 864;
  • 13) 0.155 096 225 316 864 × 2 = 0 + 0.310 192 450 633 728;
  • 14) 0.310 192 450 633 728 × 2 = 0 + 0.620 384 901 267 456;
  • 15) 0.620 384 901 267 456 × 2 = 1 + 0.240 769 802 534 912;
  • 16) 0.240 769 802 534 912 × 2 = 0 + 0.481 539 605 069 824;
  • 17) 0.481 539 605 069 824 × 2 = 0 + 0.963 079 210 139 648;
  • 18) 0.963 079 210 139 648 × 2 = 1 + 0.926 158 420 279 296;
  • 19) 0.926 158 420 279 296 × 2 = 1 + 0.852 316 840 558 592;
  • 20) 0.852 316 840 558 592 × 2 = 1 + 0.704 633 681 117 184;
  • 21) 0.704 633 681 117 184 × 2 = 1 + 0.409 267 362 234 368;
  • 22) 0.409 267 362 234 368 × 2 = 0 + 0.818 534 724 468 736;
  • 23) 0.818 534 724 468 736 × 2 = 1 + 0.637 069 448 937 472;
  • 24) 0.637 069 448 937 472 × 2 = 1 + 0.274 138 897 874 944;
  • 25) 0.274 138 897 874 944 × 2 = 0 + 0.548 277 795 749 888;
  • 26) 0.548 277 795 749 888 × 2 = 1 + 0.096 555 591 499 776;
  • 27) 0.096 555 591 499 776 × 2 = 0 + 0.193 111 182 999 552;
  • 28) 0.193 111 182 999 552 × 2 = 0 + 0.386 222 365 999 104;
  • 29) 0.386 222 365 999 104 × 2 = 0 + 0.772 444 731 998 208;
  • 30) 0.772 444 731 998 208 × 2 = 1 + 0.544 889 463 996 416;
  • 31) 0.544 889 463 996 416 × 2 = 1 + 0.089 778 927 992 832;
  • 32) 0.089 778 927 992 832 × 2 = 0 + 0.179 557 855 985 664;
  • 33) 0.179 557 855 985 664 × 2 = 0 + 0.359 115 711 971 328;
  • 34) 0.359 115 711 971 328 × 2 = 0 + 0.718 231 423 942 656;
  • 35) 0.718 231 423 942 656 × 2 = 1 + 0.436 462 847 885 312;
  • 36) 0.436 462 847 885 312 × 2 = 0 + 0.872 925 695 770 624;
  • 37) 0.872 925 695 770 624 × 2 = 1 + 0.745 851 391 541 248;
  • 38) 0.745 851 391 541 248 × 2 = 1 + 0.491 702 783 082 496;
  • 39) 0.491 702 783 082 496 × 2 = 0 + 0.983 405 566 164 992;
  • 40) 0.983 405 566 164 992 × 2 = 1 + 0.966 811 132 329 984;
  • 41) 0.966 811 132 329 984 × 2 = 1 + 0.933 622 264 659 968;
  • 42) 0.933 622 264 659 968 × 2 = 1 + 0.867 244 529 319 936;
  • 43) 0.867 244 529 319 936 × 2 = 1 + 0.734 489 058 639 872;
  • 44) 0.734 489 058 639 872 × 2 = 1 + 0.468 978 117 279 744;
  • 45) 0.468 978 117 279 744 × 2 = 0 + 0.937 956 234 559 488;
  • 46) 0.937 956 234 559 488 × 2 = 1 + 0.875 912 469 118 976;
  • 47) 0.875 912 469 118 976 × 2 = 1 + 0.751 824 938 237 952;
  • 48) 0.751 824 938 237 952 × 2 = 1 + 0.503 649 876 475 904;
  • 49) 0.503 649 876 475 904 × 2 = 1 + 0.007 299 752 951 808;
  • 50) 0.007 299 752 951 808 × 2 = 0 + 0.014 599 505 903 616;
  • 51) 0.014 599 505 903 616 × 2 = 0 + 0.029 199 011 807 232;
  • 52) 0.029 199 011 807 232 × 2 = 0 + 0.058 398 023 614 464;
  • 53) 0.058 398 023 614 464 × 2 = 0 + 0.116 796 047 228 928;
  • 54) 0.116 796 047 228 928 × 2 = 0 + 0.233 592 094 457 856;
  • 55) 0.233 592 094 457 856 × 2 = 0 + 0.467 184 188 915 712;
  • 56) 0.467 184 188 915 712 × 2 = 0 + 0.934 368 377 831 424;
  • 57) 0.934 368 377 831 424 × 2 = 1 + 0.868 736 755 662 848;
  • 58) 0.868 736 755 662 848 × 2 = 1 + 0.737 473 511 325 696;
  • 59) 0.737 473 511 325 696 × 2 = 1 + 0.474 947 022 651 392;
  • 60) 0.474 947 022 651 392 × 2 = 0 + 0.949 894 045 302 784;
  • 61) 0.949 894 045 302 784 × 2 = 1 + 0.899 788 090 605 568;
  • 62) 0.899 788 090 605 568 × 2 = 1 + 0.799 576 181 211 136;
  • 63) 0.799 576 181 211 136 × 2 = 1 + 0.599 152 362 422 272;
  • 64) 0.599 152 362 422 272 × 2 = 1 + 0.198 304 724 844 544;

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:


0.000 282 005 914 384(10) =

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

6. Positive number before normalization:

0.000 282 005 914 384(10) =

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


0.000 282 005 914 384(10) =

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

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

1.0010 0111 1011 0100 0110 0010 1101 1111 0111 1000 0000 1110 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 1101 1111 0111 1000 0000 1110 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 1101 1111 0111 1000 0000 1110 1111 =

0010 0111 1011 0100 0110 0010 1101 1111 0111 1000 0000 1110 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 1101 1111 0111 1000 0000 1110 1111


Decimal number -0.000 282 005 914 384 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1111 0111 1000 0000 1110 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