-0.000 282 005 803 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 803(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
-0.000 282 005 803(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:

|-0.000 282 005 803| = 0.000 282 005 803


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: 0.000 282 005 803.

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) 0.000 282 005 803 × 2 = 0 + 0.000 564 011 606;
  • 2) 0.000 564 011 606 × 2 = 0 + 0.001 128 023 212;
  • 3) 0.001 128 023 212 × 2 = 0 + 0.002 256 046 424;
  • 4) 0.002 256 046 424 × 2 = 0 + 0.004 512 092 848;
  • 5) 0.004 512 092 848 × 2 = 0 + 0.009 024 185 696;
  • 6) 0.009 024 185 696 × 2 = 0 + 0.018 048 371 392;
  • 7) 0.018 048 371 392 × 2 = 0 + 0.036 096 742 784;
  • 8) 0.036 096 742 784 × 2 = 0 + 0.072 193 485 568;
  • 9) 0.072 193 485 568 × 2 = 0 + 0.144 386 971 136;
  • 10) 0.144 386 971 136 × 2 = 0 + 0.288 773 942 272;
  • 11) 0.288 773 942 272 × 2 = 0 + 0.577 547 884 544;
  • 12) 0.577 547 884 544 × 2 = 1 + 0.155 095 769 088;
  • 13) 0.155 095 769 088 × 2 = 0 + 0.310 191 538 176;
  • 14) 0.310 191 538 176 × 2 = 0 + 0.620 383 076 352;
  • 15) 0.620 383 076 352 × 2 = 1 + 0.240 766 152 704;
  • 16) 0.240 766 152 704 × 2 = 0 + 0.481 532 305 408;
  • 17) 0.481 532 305 408 × 2 = 0 + 0.963 064 610 816;
  • 18) 0.963 064 610 816 × 2 = 1 + 0.926 129 221 632;
  • 19) 0.926 129 221 632 × 2 = 1 + 0.852 258 443 264;
  • 20) 0.852 258 443 264 × 2 = 1 + 0.704 516 886 528;
  • 21) 0.704 516 886 528 × 2 = 1 + 0.409 033 773 056;
  • 22) 0.409 033 773 056 × 2 = 0 + 0.818 067 546 112;
  • 23) 0.818 067 546 112 × 2 = 1 + 0.636 135 092 224;
  • 24) 0.636 135 092 224 × 2 = 1 + 0.272 270 184 448;
  • 25) 0.272 270 184 448 × 2 = 0 + 0.544 540 368 896;
  • 26) 0.544 540 368 896 × 2 = 1 + 0.089 080 737 792;
  • 27) 0.089 080 737 792 × 2 = 0 + 0.178 161 475 584;
  • 28) 0.178 161 475 584 × 2 = 0 + 0.356 322 951 168;
  • 29) 0.356 322 951 168 × 2 = 0 + 0.712 645 902 336;
  • 30) 0.712 645 902 336 × 2 = 1 + 0.425 291 804 672;
  • 31) 0.425 291 804 672 × 2 = 0 + 0.850 583 609 344;
  • 32) 0.850 583 609 344 × 2 = 1 + 0.701 167 218 688;
  • 33) 0.701 167 218 688 × 2 = 1 + 0.402 334 437 376;
  • 34) 0.402 334 437 376 × 2 = 0 + 0.804 668 874 752;
  • 35) 0.804 668 874 752 × 2 = 1 + 0.609 337 749 504;
  • 36) 0.609 337 749 504 × 2 = 1 + 0.218 675 499 008;
  • 37) 0.218 675 499 008 × 2 = 0 + 0.437 350 998 016;
  • 38) 0.437 350 998 016 × 2 = 0 + 0.874 701 996 032;
  • 39) 0.874 701 996 032 × 2 = 1 + 0.749 403 992 064;
  • 40) 0.749 403 992 064 × 2 = 1 + 0.498 807 984 128;
  • 41) 0.498 807 984 128 × 2 = 0 + 0.997 615 968 256;
  • 42) 0.997 615 968 256 × 2 = 1 + 0.995 231 936 512;
  • 43) 0.995 231 936 512 × 2 = 1 + 0.990 463 873 024;
  • 44) 0.990 463 873 024 × 2 = 1 + 0.980 927 746 048;
  • 45) 0.980 927 746 048 × 2 = 1 + 0.961 855 492 096;
  • 46) 0.961 855 492 096 × 2 = 1 + 0.923 710 984 192;
  • 47) 0.923 710 984 192 × 2 = 1 + 0.847 421 968 384;
  • 48) 0.847 421 968 384 × 2 = 1 + 0.694 843 936 768;
  • 49) 0.694 843 936 768 × 2 = 1 + 0.389 687 873 536;
  • 50) 0.389 687 873 536 × 2 = 0 + 0.779 375 747 072;
  • 51) 0.779 375 747 072 × 2 = 1 + 0.558 751 494 144;
  • 52) 0.558 751 494 144 × 2 = 1 + 0.117 502 988 288;
  • 53) 0.117 502 988 288 × 2 = 0 + 0.235 005 976 576;
  • 54) 0.235 005 976 576 × 2 = 0 + 0.470 011 953 152;
  • 55) 0.470 011 953 152 × 2 = 0 + 0.940 023 906 304;
  • 56) 0.940 023 906 304 × 2 = 1 + 0.880 047 812 608;
  • 57) 0.880 047 812 608 × 2 = 1 + 0.760 095 625 216;
  • 58) 0.760 095 625 216 × 2 = 1 + 0.520 191 250 432;
  • 59) 0.520 191 250 432 × 2 = 1 + 0.040 382 500 864;
  • 60) 0.040 382 500 864 × 2 = 0 + 0.080 765 001 728;
  • 61) 0.080 765 001 728 × 2 = 0 + 0.161 530 003 456;
  • 62) 0.161 530 003 456 × 2 = 0 + 0.323 060 006 912;
  • 63) 0.323 060 006 912 × 2 = 0 + 0.646 120 013 824;
  • 64) 0.646 120 013 824 × 2 = 1 + 0.292 240 027 648;

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:


0.000 282 005 803(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1011 0011 0111 1111 1011 0001 1110 0001(2)

6. Positive number before normalization:

0.000 282 005 803(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1011 0011 0111 1111 1011 0001 1110 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:


0.000 282 005 803(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1011 0011 0111 1111 1011 0001 1110 0001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1011 0011 0111 1111 1011 0001 1110 0001(2) × 20 =

1.0010 0111 1011 0100 0101 1011 0011 0111 1111 1011 0001 1110 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 0101 1011 0011 0111 1111 1011 0001 1110 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 0101 1011 0011 0111 1111 1011 0001 1110 0001 =

0010 0111 1011 0100 0101 1011 0011 0111 1111 1011 0001 1110 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 0101 1011 0011 0111 1111 1011 0001 1110 0001


Decimal number -0.000 282 005 803 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 1011 0011 0111 1111 1011 0001 1110 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