-0.000 282 005 719 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 719(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
-0.000 282 005 719(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:

|-0.000 282 005 719| = 0.000 282 005 719


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: 0.000 282 005 719.

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) 0.000 282 005 719 × 2 = 0 + 0.000 564 011 438;
  • 2) 0.000 564 011 438 × 2 = 0 + 0.001 128 022 876;
  • 3) 0.001 128 022 876 × 2 = 0 + 0.002 256 045 752;
  • 4) 0.002 256 045 752 × 2 = 0 + 0.004 512 091 504;
  • 5) 0.004 512 091 504 × 2 = 0 + 0.009 024 183 008;
  • 6) 0.009 024 183 008 × 2 = 0 + 0.018 048 366 016;
  • 7) 0.018 048 366 016 × 2 = 0 + 0.036 096 732 032;
  • 8) 0.036 096 732 032 × 2 = 0 + 0.072 193 464 064;
  • 9) 0.072 193 464 064 × 2 = 0 + 0.144 386 928 128;
  • 10) 0.144 386 928 128 × 2 = 0 + 0.288 773 856 256;
  • 11) 0.288 773 856 256 × 2 = 0 + 0.577 547 712 512;
  • 12) 0.577 547 712 512 × 2 = 1 + 0.155 095 425 024;
  • 13) 0.155 095 425 024 × 2 = 0 + 0.310 190 850 048;
  • 14) 0.310 190 850 048 × 2 = 0 + 0.620 381 700 096;
  • 15) 0.620 381 700 096 × 2 = 1 + 0.240 763 400 192;
  • 16) 0.240 763 400 192 × 2 = 0 + 0.481 526 800 384;
  • 17) 0.481 526 800 384 × 2 = 0 + 0.963 053 600 768;
  • 18) 0.963 053 600 768 × 2 = 1 + 0.926 107 201 536;
  • 19) 0.926 107 201 536 × 2 = 1 + 0.852 214 403 072;
  • 20) 0.852 214 403 072 × 2 = 1 + 0.704 428 806 144;
  • 21) 0.704 428 806 144 × 2 = 1 + 0.408 857 612 288;
  • 22) 0.408 857 612 288 × 2 = 0 + 0.817 715 224 576;
  • 23) 0.817 715 224 576 × 2 = 1 + 0.635 430 449 152;
  • 24) 0.635 430 449 152 × 2 = 1 + 0.270 860 898 304;
  • 25) 0.270 860 898 304 × 2 = 0 + 0.541 721 796 608;
  • 26) 0.541 721 796 608 × 2 = 1 + 0.083 443 593 216;
  • 27) 0.083 443 593 216 × 2 = 0 + 0.166 887 186 432;
  • 28) 0.166 887 186 432 × 2 = 0 + 0.333 774 372 864;
  • 29) 0.333 774 372 864 × 2 = 0 + 0.667 548 745 728;
  • 30) 0.667 548 745 728 × 2 = 1 + 0.335 097 491 456;
  • 31) 0.335 097 491 456 × 2 = 0 + 0.670 194 982 912;
  • 32) 0.670 194 982 912 × 2 = 1 + 0.340 389 965 824;
  • 33) 0.340 389 965 824 × 2 = 0 + 0.680 779 931 648;
  • 34) 0.680 779 931 648 × 2 = 1 + 0.361 559 863 296;
  • 35) 0.361 559 863 296 × 2 = 0 + 0.723 119 726 592;
  • 36) 0.723 119 726 592 × 2 = 1 + 0.446 239 453 184;
  • 37) 0.446 239 453 184 × 2 = 0 + 0.892 478 906 368;
  • 38) 0.892 478 906 368 × 2 = 1 + 0.784 957 812 736;
  • 39) 0.784 957 812 736 × 2 = 1 + 0.569 915 625 472;
  • 40) 0.569 915 625 472 × 2 = 1 + 0.139 831 250 944;
  • 41) 0.139 831 250 944 × 2 = 0 + 0.279 662 501 888;
  • 42) 0.279 662 501 888 × 2 = 0 + 0.559 325 003 776;
  • 43) 0.559 325 003 776 × 2 = 1 + 0.118 650 007 552;
  • 44) 0.118 650 007 552 × 2 = 0 + 0.237 300 015 104;
  • 45) 0.237 300 015 104 × 2 = 0 + 0.474 600 030 208;
  • 46) 0.474 600 030 208 × 2 = 0 + 0.949 200 060 416;
  • 47) 0.949 200 060 416 × 2 = 1 + 0.898 400 120 832;
  • 48) 0.898 400 120 832 × 2 = 1 + 0.796 800 241 664;
  • 49) 0.796 800 241 664 × 2 = 1 + 0.593 600 483 328;
  • 50) 0.593 600 483 328 × 2 = 1 + 0.187 200 966 656;
  • 51) 0.187 200 966 656 × 2 = 0 + 0.374 401 933 312;
  • 52) 0.374 401 933 312 × 2 = 0 + 0.748 803 866 624;
  • 53) 0.748 803 866 624 × 2 = 1 + 0.497 607 733 248;
  • 54) 0.497 607 733 248 × 2 = 0 + 0.995 215 466 496;
  • 55) 0.995 215 466 496 × 2 = 1 + 0.990 430 932 992;
  • 56) 0.990 430 932 992 × 2 = 1 + 0.980 861 865 984;
  • 57) 0.980 861 865 984 × 2 = 1 + 0.961 723 731 968;
  • 58) 0.961 723 731 968 × 2 = 1 + 0.923 447 463 936;
  • 59) 0.923 447 463 936 × 2 = 1 + 0.846 894 927 872;
  • 60) 0.846 894 927 872 × 2 = 1 + 0.693 789 855 744;
  • 61) 0.693 789 855 744 × 2 = 1 + 0.387 579 711 488;
  • 62) 0.387 579 711 488 × 2 = 0 + 0.775 159 422 976;
  • 63) 0.775 159 422 976 × 2 = 1 + 0.550 318 845 952;
  • 64) 0.550 318 845 952 × 2 = 1 + 0.100 637 691 904;

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:


0.000 282 005 719(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0101 0111 0010 0011 1100 1011 1111 1011(2)

6. Positive number before normalization:

0.000 282 005 719(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0101 0111 0010 0011 1100 1011 1111 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:


0.000 282 005 719(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0101 0111 0010 0011 1100 1011 1111 1011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0101 0111 0010 0011 1100 1011 1111 1011(2) × 20 =

1.0010 0111 1011 0100 0101 0101 0111 0010 0011 1100 1011 1111 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 0101 0101 0111 0010 0011 1100 1011 1111 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 0101 0101 0111 0010 0011 1100 1011 1111 1011 =

0010 0111 1011 0100 0101 0101 0111 0010 0011 1100 1011 1111 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 0101 0101 0111 0010 0011 1100 1011 1111 1011


Decimal number -0.000 282 005 719 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 0101 0111 0010 0011 1100 1011 1111 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