-0.000 282 005 913 54 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 913 54(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
-0.000 282 005 913 54(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:

|-0.000 282 005 913 54| = 0.000 282 005 913 54


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: 0.000 282 005 913 54.

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) 0.000 282 005 913 54 × 2 = 0 + 0.000 564 011 827 08;
  • 2) 0.000 564 011 827 08 × 2 = 0 + 0.001 128 023 654 16;
  • 3) 0.001 128 023 654 16 × 2 = 0 + 0.002 256 047 308 32;
  • 4) 0.002 256 047 308 32 × 2 = 0 + 0.004 512 094 616 64;
  • 5) 0.004 512 094 616 64 × 2 = 0 + 0.009 024 189 233 28;
  • 6) 0.009 024 189 233 28 × 2 = 0 + 0.018 048 378 466 56;
  • 7) 0.018 048 378 466 56 × 2 = 0 + 0.036 096 756 933 12;
  • 8) 0.036 096 756 933 12 × 2 = 0 + 0.072 193 513 866 24;
  • 9) 0.072 193 513 866 24 × 2 = 0 + 0.144 387 027 732 48;
  • 10) 0.144 387 027 732 48 × 2 = 0 + 0.288 774 055 464 96;
  • 11) 0.288 774 055 464 96 × 2 = 0 + 0.577 548 110 929 92;
  • 12) 0.577 548 110 929 92 × 2 = 1 + 0.155 096 221 859 84;
  • 13) 0.155 096 221 859 84 × 2 = 0 + 0.310 192 443 719 68;
  • 14) 0.310 192 443 719 68 × 2 = 0 + 0.620 384 887 439 36;
  • 15) 0.620 384 887 439 36 × 2 = 1 + 0.240 769 774 878 72;
  • 16) 0.240 769 774 878 72 × 2 = 0 + 0.481 539 549 757 44;
  • 17) 0.481 539 549 757 44 × 2 = 0 + 0.963 079 099 514 88;
  • 18) 0.963 079 099 514 88 × 2 = 1 + 0.926 158 199 029 76;
  • 19) 0.926 158 199 029 76 × 2 = 1 + 0.852 316 398 059 52;
  • 20) 0.852 316 398 059 52 × 2 = 1 + 0.704 632 796 119 04;
  • 21) 0.704 632 796 119 04 × 2 = 1 + 0.409 265 592 238 08;
  • 22) 0.409 265 592 238 08 × 2 = 0 + 0.818 531 184 476 16;
  • 23) 0.818 531 184 476 16 × 2 = 1 + 0.637 062 368 952 32;
  • 24) 0.637 062 368 952 32 × 2 = 1 + 0.274 124 737 904 64;
  • 25) 0.274 124 737 904 64 × 2 = 0 + 0.548 249 475 809 28;
  • 26) 0.548 249 475 809 28 × 2 = 1 + 0.096 498 951 618 56;
  • 27) 0.096 498 951 618 56 × 2 = 0 + 0.192 997 903 237 12;
  • 28) 0.192 997 903 237 12 × 2 = 0 + 0.385 995 806 474 24;
  • 29) 0.385 995 806 474 24 × 2 = 0 + 0.771 991 612 948 48;
  • 30) 0.771 991 612 948 48 × 2 = 1 + 0.543 983 225 896 96;
  • 31) 0.543 983 225 896 96 × 2 = 1 + 0.087 966 451 793 92;
  • 32) 0.087 966 451 793 92 × 2 = 0 + 0.175 932 903 587 84;
  • 33) 0.175 932 903 587 84 × 2 = 0 + 0.351 865 807 175 68;
  • 34) 0.351 865 807 175 68 × 2 = 0 + 0.703 731 614 351 36;
  • 35) 0.703 731 614 351 36 × 2 = 1 + 0.407 463 228 702 72;
  • 36) 0.407 463 228 702 72 × 2 = 0 + 0.814 926 457 405 44;
  • 37) 0.814 926 457 405 44 × 2 = 1 + 0.629 852 914 810 88;
  • 38) 0.629 852 914 810 88 × 2 = 1 + 0.259 705 829 621 76;
  • 39) 0.259 705 829 621 76 × 2 = 0 + 0.519 411 659 243 52;
  • 40) 0.519 411 659 243 52 × 2 = 1 + 0.038 823 318 487 04;
  • 41) 0.038 823 318 487 04 × 2 = 0 + 0.077 646 636 974 08;
  • 42) 0.077 646 636 974 08 × 2 = 0 + 0.155 293 273 948 16;
  • 43) 0.155 293 273 948 16 × 2 = 0 + 0.310 586 547 896 32;
  • 44) 0.310 586 547 896 32 × 2 = 0 + 0.621 173 095 792 64;
  • 45) 0.621 173 095 792 64 × 2 = 1 + 0.242 346 191 585 28;
  • 46) 0.242 346 191 585 28 × 2 = 0 + 0.484 692 383 170 56;
  • 47) 0.484 692 383 170 56 × 2 = 0 + 0.969 384 766 341 12;
  • 48) 0.969 384 766 341 12 × 2 = 1 + 0.938 769 532 682 24;
  • 49) 0.938 769 532 682 24 × 2 = 1 + 0.877 539 065 364 48;
  • 50) 0.877 539 065 364 48 × 2 = 1 + 0.755 078 130 728 96;
  • 51) 0.755 078 130 728 96 × 2 = 1 + 0.510 156 261 457 92;
  • 52) 0.510 156 261 457 92 × 2 = 1 + 0.020 312 522 915 84;
  • 53) 0.020 312 522 915 84 × 2 = 0 + 0.040 625 045 831 68;
  • 54) 0.040 625 045 831 68 × 2 = 0 + 0.081 250 091 663 36;
  • 55) 0.081 250 091 663 36 × 2 = 0 + 0.162 500 183 326 72;
  • 56) 0.162 500 183 326 72 × 2 = 0 + 0.325 000 366 653 44;
  • 57) 0.325 000 366 653 44 × 2 = 0 + 0.650 000 733 306 88;
  • 58) 0.650 000 733 306 88 × 2 = 1 + 0.300 001 466 613 76;
  • 59) 0.300 001 466 613 76 × 2 = 0 + 0.600 002 933 227 52;
  • 60) 0.600 002 933 227 52 × 2 = 1 + 0.200 005 866 455 04;
  • 61) 0.200 005 866 455 04 × 2 = 0 + 0.400 011 732 910 08;
  • 62) 0.400 011 732 910 08 × 2 = 0 + 0.800 023 465 820 16;
  • 63) 0.800 023 465 820 16 × 2 = 1 + 0.600 046 931 640 32;
  • 64) 0.600 046 931 640 32 × 2 = 1 + 0.200 093 863 280 64;

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:


0.000 282 005 913 54(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0000 1001 1111 0000 0101 0011(2)

6. Positive number before normalization:

0.000 282 005 913 54(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0000 1001 1111 0000 0101 0011(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:


0.000 282 005 913 54(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0000 1001 1111 0000 0101 0011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0000 1001 1111 0000 0101 0011(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 0000 1001 1111 0000 0101 0011(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 0000 1001 1111 0000 0101 0011


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 0000 1001 1111 0000 0101 0011 =

0010 0111 1011 0100 0110 0010 1101 0000 1001 1111 0000 0101 0011


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 0000 1001 1111 0000 0101 0011


Decimal number -0.000 282 005 913 54 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 0000 1001 1111 0000 0101 0011

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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