-0.000 282 005 991 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 991(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
-0.000 282 005 991(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:

|-0.000 282 005 991| = 0.000 282 005 991


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: 0.000 282 005 991.

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) 0.000 282 005 991 × 2 = 0 + 0.000 564 011 982;
  • 2) 0.000 564 011 982 × 2 = 0 + 0.001 128 023 964;
  • 3) 0.001 128 023 964 × 2 = 0 + 0.002 256 047 928;
  • 4) 0.002 256 047 928 × 2 = 0 + 0.004 512 095 856;
  • 5) 0.004 512 095 856 × 2 = 0 + 0.009 024 191 712;
  • 6) 0.009 024 191 712 × 2 = 0 + 0.018 048 383 424;
  • 7) 0.018 048 383 424 × 2 = 0 + 0.036 096 766 848;
  • 8) 0.036 096 766 848 × 2 = 0 + 0.072 193 533 696;
  • 9) 0.072 193 533 696 × 2 = 0 + 0.144 387 067 392;
  • 10) 0.144 387 067 392 × 2 = 0 + 0.288 774 134 784;
  • 11) 0.288 774 134 784 × 2 = 0 + 0.577 548 269 568;
  • 12) 0.577 548 269 568 × 2 = 1 + 0.155 096 539 136;
  • 13) 0.155 096 539 136 × 2 = 0 + 0.310 193 078 272;
  • 14) 0.310 193 078 272 × 2 = 0 + 0.620 386 156 544;
  • 15) 0.620 386 156 544 × 2 = 1 + 0.240 772 313 088;
  • 16) 0.240 772 313 088 × 2 = 0 + 0.481 544 626 176;
  • 17) 0.481 544 626 176 × 2 = 0 + 0.963 089 252 352;
  • 18) 0.963 089 252 352 × 2 = 1 + 0.926 178 504 704;
  • 19) 0.926 178 504 704 × 2 = 1 + 0.852 357 009 408;
  • 20) 0.852 357 009 408 × 2 = 1 + 0.704 714 018 816;
  • 21) 0.704 714 018 816 × 2 = 1 + 0.409 428 037 632;
  • 22) 0.409 428 037 632 × 2 = 0 + 0.818 856 075 264;
  • 23) 0.818 856 075 264 × 2 = 1 + 0.637 712 150 528;
  • 24) 0.637 712 150 528 × 2 = 1 + 0.275 424 301 056;
  • 25) 0.275 424 301 056 × 2 = 0 + 0.550 848 602 112;
  • 26) 0.550 848 602 112 × 2 = 1 + 0.101 697 204 224;
  • 27) 0.101 697 204 224 × 2 = 0 + 0.203 394 408 448;
  • 28) 0.203 394 408 448 × 2 = 0 + 0.406 788 816 896;
  • 29) 0.406 788 816 896 × 2 = 0 + 0.813 577 633 792;
  • 30) 0.813 577 633 792 × 2 = 1 + 0.627 155 267 584;
  • 31) 0.627 155 267 584 × 2 = 1 + 0.254 310 535 168;
  • 32) 0.254 310 535 168 × 2 = 0 + 0.508 621 070 336;
  • 33) 0.508 621 070 336 × 2 = 1 + 0.017 242 140 672;
  • 34) 0.017 242 140 672 × 2 = 0 + 0.034 484 281 344;
  • 35) 0.034 484 281 344 × 2 = 0 + 0.068 968 562 688;
  • 36) 0.068 968 562 688 × 2 = 0 + 0.137 937 125 376;
  • 37) 0.137 937 125 376 × 2 = 0 + 0.275 874 250 752;
  • 38) 0.275 874 250 752 × 2 = 0 + 0.551 748 501 504;
  • 39) 0.551 748 501 504 × 2 = 1 + 0.103 497 003 008;
  • 40) 0.103 497 003 008 × 2 = 0 + 0.206 994 006 016;
  • 41) 0.206 994 006 016 × 2 = 0 + 0.413 988 012 032;
  • 42) 0.413 988 012 032 × 2 = 0 + 0.827 976 024 064;
  • 43) 0.827 976 024 064 × 2 = 1 + 0.655 952 048 128;
  • 44) 0.655 952 048 128 × 2 = 1 + 0.311 904 096 256;
  • 45) 0.311 904 096 256 × 2 = 0 + 0.623 808 192 512;
  • 46) 0.623 808 192 512 × 2 = 1 + 0.247 616 385 024;
  • 47) 0.247 616 385 024 × 2 = 0 + 0.495 232 770 048;
  • 48) 0.495 232 770 048 × 2 = 0 + 0.990 465 540 096;
  • 49) 0.990 465 540 096 × 2 = 1 + 0.980 931 080 192;
  • 50) 0.980 931 080 192 × 2 = 1 + 0.961 862 160 384;
  • 51) 0.961 862 160 384 × 2 = 1 + 0.923 724 320 768;
  • 52) 0.923 724 320 768 × 2 = 1 + 0.847 448 641 536;
  • 53) 0.847 448 641 536 × 2 = 1 + 0.694 897 283 072;
  • 54) 0.694 897 283 072 × 2 = 1 + 0.389 794 566 144;
  • 55) 0.389 794 566 144 × 2 = 0 + 0.779 589 132 288;
  • 56) 0.779 589 132 288 × 2 = 1 + 0.559 178 264 576;
  • 57) 0.559 178 264 576 × 2 = 1 + 0.118 356 529 152;
  • 58) 0.118 356 529 152 × 2 = 0 + 0.236 713 058 304;
  • 59) 0.236 713 058 304 × 2 = 0 + 0.473 426 116 608;
  • 60) 0.473 426 116 608 × 2 = 0 + 0.946 852 233 216;
  • 61) 0.946 852 233 216 × 2 = 1 + 0.893 704 466 432;
  • 62) 0.893 704 466 432 × 2 = 1 + 0.787 408 932 864;
  • 63) 0.787 408 932 864 × 2 = 1 + 0.574 817 865 728;
  • 64) 0.574 817 865 728 × 2 = 1 + 0.149 635 731 456;

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:


0.000 282 005 991(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1000 0010 0011 0100 1111 1101 1000 1111(2)

6. Positive number before normalization:

0.000 282 005 991(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1000 0010 0011 0100 1111 1101 1000 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:


0.000 282 005 991(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1000 0010 0011 0100 1111 1101 1000 1111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1000 0010 0011 0100 1111 1101 1000 1111(2) × 20 =

1.0010 0111 1011 0100 0110 1000 0010 0011 0100 1111 1101 1000 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 1000 0010 0011 0100 1111 1101 1000 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 1000 0010 0011 0100 1111 1101 1000 1111 =

0010 0111 1011 0100 0110 1000 0010 0011 0100 1111 1101 1000 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 1000 0010 0011 0100 1111 1101 1000 1111


Decimal number -0.000 282 005 991 converted to 64 bit double precision IEEE 754 binary floating point representation:

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