-0.000 282 005 917 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 917 2(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
-0.000 282 005 917 2(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:

|-0.000 282 005 917 2| = 0.000 282 005 917 2


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: 0.000 282 005 917 2.

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) 0.000 282 005 917 2 × 2 = 0 + 0.000 564 011 834 4;
  • 2) 0.000 564 011 834 4 × 2 = 0 + 0.001 128 023 668 8;
  • 3) 0.001 128 023 668 8 × 2 = 0 + 0.002 256 047 337 6;
  • 4) 0.002 256 047 337 6 × 2 = 0 + 0.004 512 094 675 2;
  • 5) 0.004 512 094 675 2 × 2 = 0 + 0.009 024 189 350 4;
  • 6) 0.009 024 189 350 4 × 2 = 0 + 0.018 048 378 700 8;
  • 7) 0.018 048 378 700 8 × 2 = 0 + 0.036 096 757 401 6;
  • 8) 0.036 096 757 401 6 × 2 = 0 + 0.072 193 514 803 2;
  • 9) 0.072 193 514 803 2 × 2 = 0 + 0.144 387 029 606 4;
  • 10) 0.144 387 029 606 4 × 2 = 0 + 0.288 774 059 212 8;
  • 11) 0.288 774 059 212 8 × 2 = 0 + 0.577 548 118 425 6;
  • 12) 0.577 548 118 425 6 × 2 = 1 + 0.155 096 236 851 2;
  • 13) 0.155 096 236 851 2 × 2 = 0 + 0.310 192 473 702 4;
  • 14) 0.310 192 473 702 4 × 2 = 0 + 0.620 384 947 404 8;
  • 15) 0.620 384 947 404 8 × 2 = 1 + 0.240 769 894 809 6;
  • 16) 0.240 769 894 809 6 × 2 = 0 + 0.481 539 789 619 2;
  • 17) 0.481 539 789 619 2 × 2 = 0 + 0.963 079 579 238 4;
  • 18) 0.963 079 579 238 4 × 2 = 1 + 0.926 159 158 476 8;
  • 19) 0.926 159 158 476 8 × 2 = 1 + 0.852 318 316 953 6;
  • 20) 0.852 318 316 953 6 × 2 = 1 + 0.704 636 633 907 2;
  • 21) 0.704 636 633 907 2 × 2 = 1 + 0.409 273 267 814 4;
  • 22) 0.409 273 267 814 4 × 2 = 0 + 0.818 546 535 628 8;
  • 23) 0.818 546 535 628 8 × 2 = 1 + 0.637 093 071 257 6;
  • 24) 0.637 093 071 257 6 × 2 = 1 + 0.274 186 142 515 2;
  • 25) 0.274 186 142 515 2 × 2 = 0 + 0.548 372 285 030 4;
  • 26) 0.548 372 285 030 4 × 2 = 1 + 0.096 744 570 060 8;
  • 27) 0.096 744 570 060 8 × 2 = 0 + 0.193 489 140 121 6;
  • 28) 0.193 489 140 121 6 × 2 = 0 + 0.386 978 280 243 2;
  • 29) 0.386 978 280 243 2 × 2 = 0 + 0.773 956 560 486 4;
  • 30) 0.773 956 560 486 4 × 2 = 1 + 0.547 913 120 972 8;
  • 31) 0.547 913 120 972 8 × 2 = 1 + 0.095 826 241 945 6;
  • 32) 0.095 826 241 945 6 × 2 = 0 + 0.191 652 483 891 2;
  • 33) 0.191 652 483 891 2 × 2 = 0 + 0.383 304 967 782 4;
  • 34) 0.383 304 967 782 4 × 2 = 0 + 0.766 609 935 564 8;
  • 35) 0.766 609 935 564 8 × 2 = 1 + 0.533 219 871 129 6;
  • 36) 0.533 219 871 129 6 × 2 = 1 + 0.066 439 742 259 2;
  • 37) 0.066 439 742 259 2 × 2 = 0 + 0.132 879 484 518 4;
  • 38) 0.132 879 484 518 4 × 2 = 0 + 0.265 758 969 036 8;
  • 39) 0.265 758 969 036 8 × 2 = 0 + 0.531 517 938 073 6;
  • 40) 0.531 517 938 073 6 × 2 = 1 + 0.063 035 876 147 2;
  • 41) 0.063 035 876 147 2 × 2 = 0 + 0.126 071 752 294 4;
  • 42) 0.126 071 752 294 4 × 2 = 0 + 0.252 143 504 588 8;
  • 43) 0.252 143 504 588 8 × 2 = 0 + 0.504 287 009 177 6;
  • 44) 0.504 287 009 177 6 × 2 = 1 + 0.008 574 018 355 2;
  • 45) 0.008 574 018 355 2 × 2 = 0 + 0.017 148 036 710 4;
  • 46) 0.017 148 036 710 4 × 2 = 0 + 0.034 296 073 420 8;
  • 47) 0.034 296 073 420 8 × 2 = 0 + 0.068 592 146 841 6;
  • 48) 0.068 592 146 841 6 × 2 = 0 + 0.137 184 293 683 2;
  • 49) 0.137 184 293 683 2 × 2 = 0 + 0.274 368 587 366 4;
  • 50) 0.274 368 587 366 4 × 2 = 0 + 0.548 737 174 732 8;
  • 51) 0.548 737 174 732 8 × 2 = 1 + 0.097 474 349 465 6;
  • 52) 0.097 474 349 465 6 × 2 = 0 + 0.194 948 698 931 2;
  • 53) 0.194 948 698 931 2 × 2 = 0 + 0.389 897 397 862 4;
  • 54) 0.389 897 397 862 4 × 2 = 0 + 0.779 794 795 724 8;
  • 55) 0.779 794 795 724 8 × 2 = 1 + 0.559 589 591 449 6;
  • 56) 0.559 589 591 449 6 × 2 = 1 + 0.119 179 182 899 2;
  • 57) 0.119 179 182 899 2 × 2 = 0 + 0.238 358 365 798 4;
  • 58) 0.238 358 365 798 4 × 2 = 0 + 0.476 716 731 596 8;
  • 59) 0.476 716 731 596 8 × 2 = 0 + 0.953 433 463 193 6;
  • 60) 0.953 433 463 193 6 × 2 = 1 + 0.906 866 926 387 2;
  • 61) 0.906 866 926 387 2 × 2 = 1 + 0.813 733 852 774 4;
  • 62) 0.813 733 852 774 4 × 2 = 1 + 0.627 467 705 548 8;
  • 63) 0.627 467 705 548 8 × 2 = 1 + 0.254 935 411 097 6;
  • 64) 0.254 935 411 097 6 × 2 = 0 + 0.509 870 822 195 2;

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:


0.000 282 005 917 2(10) =

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

6. Positive number before normalization:

0.000 282 005 917 2(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0001 0001 0000 0010 0011 0001 1110(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:


0.000 282 005 917 2(10) =

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

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

1.0010 0111 1011 0100 0110 0011 0001 0001 0000 0010 0011 0001 1110(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 0011 0001 0001 0000 0010 0011 0001 1110


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 0011 0001 0001 0000 0010 0011 0001 1110 =

0010 0111 1011 0100 0110 0011 0001 0001 0000 0010 0011 0001 1110


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 0011 0001 0001 0000 0010 0011 0001 1110


Decimal number -0.000 282 005 917 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0011 0001 0001 0000 0010 0011 0001 1110

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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