-0.000 282 005 914 62 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 62(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 62(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 62| = 0.000 282 005 914 62


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 62.

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 62 × 2 = 0 + 0.000 564 011 829 24;
  • 2) 0.000 564 011 829 24 × 2 = 0 + 0.001 128 023 658 48;
  • 3) 0.001 128 023 658 48 × 2 = 0 + 0.002 256 047 316 96;
  • 4) 0.002 256 047 316 96 × 2 = 0 + 0.004 512 094 633 92;
  • 5) 0.004 512 094 633 92 × 2 = 0 + 0.009 024 189 267 84;
  • 6) 0.009 024 189 267 84 × 2 = 0 + 0.018 048 378 535 68;
  • 7) 0.018 048 378 535 68 × 2 = 0 + 0.036 096 757 071 36;
  • 8) 0.036 096 757 071 36 × 2 = 0 + 0.072 193 514 142 72;
  • 9) 0.072 193 514 142 72 × 2 = 0 + 0.144 387 028 285 44;
  • 10) 0.144 387 028 285 44 × 2 = 0 + 0.288 774 056 570 88;
  • 11) 0.288 774 056 570 88 × 2 = 0 + 0.577 548 113 141 76;
  • 12) 0.577 548 113 141 76 × 2 = 1 + 0.155 096 226 283 52;
  • 13) 0.155 096 226 283 52 × 2 = 0 + 0.310 192 452 567 04;
  • 14) 0.310 192 452 567 04 × 2 = 0 + 0.620 384 905 134 08;
  • 15) 0.620 384 905 134 08 × 2 = 1 + 0.240 769 810 268 16;
  • 16) 0.240 769 810 268 16 × 2 = 0 + 0.481 539 620 536 32;
  • 17) 0.481 539 620 536 32 × 2 = 0 + 0.963 079 241 072 64;
  • 18) 0.963 079 241 072 64 × 2 = 1 + 0.926 158 482 145 28;
  • 19) 0.926 158 482 145 28 × 2 = 1 + 0.852 316 964 290 56;
  • 20) 0.852 316 964 290 56 × 2 = 1 + 0.704 633 928 581 12;
  • 21) 0.704 633 928 581 12 × 2 = 1 + 0.409 267 857 162 24;
  • 22) 0.409 267 857 162 24 × 2 = 0 + 0.818 535 714 324 48;
  • 23) 0.818 535 714 324 48 × 2 = 1 + 0.637 071 428 648 96;
  • 24) 0.637 071 428 648 96 × 2 = 1 + 0.274 142 857 297 92;
  • 25) 0.274 142 857 297 92 × 2 = 0 + 0.548 285 714 595 84;
  • 26) 0.548 285 714 595 84 × 2 = 1 + 0.096 571 429 191 68;
  • 27) 0.096 571 429 191 68 × 2 = 0 + 0.193 142 858 383 36;
  • 28) 0.193 142 858 383 36 × 2 = 0 + 0.386 285 716 766 72;
  • 29) 0.386 285 716 766 72 × 2 = 0 + 0.772 571 433 533 44;
  • 30) 0.772 571 433 533 44 × 2 = 1 + 0.545 142 867 066 88;
  • 31) 0.545 142 867 066 88 × 2 = 1 + 0.090 285 734 133 76;
  • 32) 0.090 285 734 133 76 × 2 = 0 + 0.180 571 468 267 52;
  • 33) 0.180 571 468 267 52 × 2 = 0 + 0.361 142 936 535 04;
  • 34) 0.361 142 936 535 04 × 2 = 0 + 0.722 285 873 070 08;
  • 35) 0.722 285 873 070 08 × 2 = 1 + 0.444 571 746 140 16;
  • 36) 0.444 571 746 140 16 × 2 = 0 + 0.889 143 492 280 32;
  • 37) 0.889 143 492 280 32 × 2 = 1 + 0.778 286 984 560 64;
  • 38) 0.778 286 984 560 64 × 2 = 1 + 0.556 573 969 121 28;
  • 39) 0.556 573 969 121 28 × 2 = 1 + 0.113 147 938 242 56;
  • 40) 0.113 147 938 242 56 × 2 = 0 + 0.226 295 876 485 12;
  • 41) 0.226 295 876 485 12 × 2 = 0 + 0.452 591 752 970 24;
  • 42) 0.452 591 752 970 24 × 2 = 0 + 0.905 183 505 940 48;
  • 43) 0.905 183 505 940 48 × 2 = 1 + 0.810 367 011 880 96;
  • 44) 0.810 367 011 880 96 × 2 = 1 + 0.620 734 023 761 92;
  • 45) 0.620 734 023 761 92 × 2 = 1 + 0.241 468 047 523 84;
  • 46) 0.241 468 047 523 84 × 2 = 0 + 0.482 936 095 047 68;
  • 47) 0.482 936 095 047 68 × 2 = 0 + 0.965 872 190 095 36;
  • 48) 0.965 872 190 095 36 × 2 = 1 + 0.931 744 380 190 72;
  • 49) 0.931 744 380 190 72 × 2 = 1 + 0.863 488 760 381 44;
  • 50) 0.863 488 760 381 44 × 2 = 1 + 0.726 977 520 762 88;
  • 51) 0.726 977 520 762 88 × 2 = 1 + 0.453 955 041 525 76;
  • 52) 0.453 955 041 525 76 × 2 = 0 + 0.907 910 083 051 52;
  • 53) 0.907 910 083 051 52 × 2 = 1 + 0.815 820 166 103 04;
  • 54) 0.815 820 166 103 04 × 2 = 1 + 0.631 640 332 206 08;
  • 55) 0.631 640 332 206 08 × 2 = 1 + 0.263 280 664 412 16;
  • 56) 0.263 280 664 412 16 × 2 = 0 + 0.526 561 328 824 32;
  • 57) 0.526 561 328 824 32 × 2 = 1 + 0.053 122 657 648 64;
  • 58) 0.053 122 657 648 64 × 2 = 0 + 0.106 245 315 297 28;
  • 59) 0.106 245 315 297 28 × 2 = 0 + 0.212 490 630 594 56;
  • 60) 0.212 490 630 594 56 × 2 = 0 + 0.424 981 261 189 12;
  • 61) 0.424 981 261 189 12 × 2 = 0 + 0.849 962 522 378 24;
  • 62) 0.849 962 522 378 24 × 2 = 1 + 0.699 925 044 756 48;
  • 63) 0.699 925 044 756 48 × 2 = 1 + 0.399 850 089 512 96;
  • 64) 0.399 850 089 512 96 × 2 = 0 + 0.799 700 179 025 92;

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 62(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0011 1001 1110 1110 1000 0110(2)

6. Positive number before normalization:

0.000 282 005 914 62(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0011 1001 1110 1110 1000 0110(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 62(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0011 1001 1110 1110 1000 0110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0011 1001 1110 1110 1000 0110(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0011 1001 1110 1110 1000 0110(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 0011 1001 1110 1110 1000 0110


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 0011 1001 1110 1110 1000 0110 =

0010 0111 1011 0100 0110 0010 1110 0011 1001 1110 1110 1000 0110


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 0011 1001 1110 1110 1000 0110


Decimal number -0.000 282 005 914 62 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0011 1001 1110 1110 1000 0110

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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