-0.000 282 011 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 011 6(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
-0.000 282 011 6(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:

|-0.000 282 011 6| = 0.000 282 011 6


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: 0.000 282 011 6.

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) 0.000 282 011 6 × 2 = 0 + 0.000 564 023 2;
  • 2) 0.000 564 023 2 × 2 = 0 + 0.001 128 046 4;
  • 3) 0.001 128 046 4 × 2 = 0 + 0.002 256 092 8;
  • 4) 0.002 256 092 8 × 2 = 0 + 0.004 512 185 6;
  • 5) 0.004 512 185 6 × 2 = 0 + 0.009 024 371 2;
  • 6) 0.009 024 371 2 × 2 = 0 + 0.018 048 742 4;
  • 7) 0.018 048 742 4 × 2 = 0 + 0.036 097 484 8;
  • 8) 0.036 097 484 8 × 2 = 0 + 0.072 194 969 6;
  • 9) 0.072 194 969 6 × 2 = 0 + 0.144 389 939 2;
  • 10) 0.144 389 939 2 × 2 = 0 + 0.288 779 878 4;
  • 11) 0.288 779 878 4 × 2 = 0 + 0.577 559 756 8;
  • 12) 0.577 559 756 8 × 2 = 1 + 0.155 119 513 6;
  • 13) 0.155 119 513 6 × 2 = 0 + 0.310 239 027 2;
  • 14) 0.310 239 027 2 × 2 = 0 + 0.620 478 054 4;
  • 15) 0.620 478 054 4 × 2 = 1 + 0.240 956 108 8;
  • 16) 0.240 956 108 8 × 2 = 0 + 0.481 912 217 6;
  • 17) 0.481 912 217 6 × 2 = 0 + 0.963 824 435 2;
  • 18) 0.963 824 435 2 × 2 = 1 + 0.927 648 870 4;
  • 19) 0.927 648 870 4 × 2 = 1 + 0.855 297 740 8;
  • 20) 0.855 297 740 8 × 2 = 1 + 0.710 595 481 6;
  • 21) 0.710 595 481 6 × 2 = 1 + 0.421 190 963 2;
  • 22) 0.421 190 963 2 × 2 = 0 + 0.842 381 926 4;
  • 23) 0.842 381 926 4 × 2 = 1 + 0.684 763 852 8;
  • 24) 0.684 763 852 8 × 2 = 1 + 0.369 527 705 6;
  • 25) 0.369 527 705 6 × 2 = 0 + 0.739 055 411 2;
  • 26) 0.739 055 411 2 × 2 = 1 + 0.478 110 822 4;
  • 27) 0.478 110 822 4 × 2 = 0 + 0.956 221 644 8;
  • 28) 0.956 221 644 8 × 2 = 1 + 0.912 443 289 6;
  • 29) 0.912 443 289 6 × 2 = 1 + 0.824 886 579 2;
  • 30) 0.824 886 579 2 × 2 = 1 + 0.649 773 158 4;
  • 31) 0.649 773 158 4 × 2 = 1 + 0.299 546 316 8;
  • 32) 0.299 546 316 8 × 2 = 0 + 0.599 092 633 6;
  • 33) 0.599 092 633 6 × 2 = 1 + 0.198 185 267 2;
  • 34) 0.198 185 267 2 × 2 = 0 + 0.396 370 534 4;
  • 35) 0.396 370 534 4 × 2 = 0 + 0.792 741 068 8;
  • 36) 0.792 741 068 8 × 2 = 1 + 0.585 482 137 6;
  • 37) 0.585 482 137 6 × 2 = 1 + 0.170 964 275 2;
  • 38) 0.170 964 275 2 × 2 = 0 + 0.341 928 550 4;
  • 39) 0.341 928 550 4 × 2 = 0 + 0.683 857 100 8;
  • 40) 0.683 857 100 8 × 2 = 1 + 0.367 714 201 6;
  • 41) 0.367 714 201 6 × 2 = 0 + 0.735 428 403 2;
  • 42) 0.735 428 403 2 × 2 = 1 + 0.470 856 806 4;
  • 43) 0.470 856 806 4 × 2 = 0 + 0.941 713 612 8;
  • 44) 0.941 713 612 8 × 2 = 1 + 0.883 427 225 6;
  • 45) 0.883 427 225 6 × 2 = 1 + 0.766 854 451 2;
  • 46) 0.766 854 451 2 × 2 = 1 + 0.533 708 902 4;
  • 47) 0.533 708 902 4 × 2 = 1 + 0.067 417 804 8;
  • 48) 0.067 417 804 8 × 2 = 0 + 0.134 835 609 6;
  • 49) 0.134 835 609 6 × 2 = 0 + 0.269 671 219 2;
  • 50) 0.269 671 219 2 × 2 = 0 + 0.539 342 438 4;
  • 51) 0.539 342 438 4 × 2 = 1 + 0.078 684 876 8;
  • 52) 0.078 684 876 8 × 2 = 0 + 0.157 369 753 6;
  • 53) 0.157 369 753 6 × 2 = 0 + 0.314 739 507 2;
  • 54) 0.314 739 507 2 × 2 = 0 + 0.629 479 014 4;
  • 55) 0.629 479 014 4 × 2 = 1 + 0.258 958 028 8;
  • 56) 0.258 958 028 8 × 2 = 0 + 0.517 916 057 6;
  • 57) 0.517 916 057 6 × 2 = 1 + 0.035 832 115 2;
  • 58) 0.035 832 115 2 × 2 = 0 + 0.071 664 230 4;
  • 59) 0.071 664 230 4 × 2 = 0 + 0.143 328 460 8;
  • 60) 0.143 328 460 8 × 2 = 0 + 0.286 656 921 6;
  • 61) 0.286 656 921 6 × 2 = 0 + 0.573 313 843 2;
  • 62) 0.573 313 843 2 × 2 = 1 + 0.146 627 686 4;
  • 63) 0.146 627 686 4 × 2 = 0 + 0.293 255 372 8;
  • 64) 0.293 255 372 8 × 2 = 0 + 0.586 510 745 6;

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:


0.000 282 011 6(10) =

0.0000 0000 0001 0010 0111 1011 0101 1110 1001 1001 0101 1110 0010 0010 1000 0100(2)

6. Positive number before normalization:

0.000 282 011 6(10) =

0.0000 0000 0001 0010 0111 1011 0101 1110 1001 1001 0101 1110 0010 0010 1000 0100(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:


0.000 282 011 6(10) =

0.0000 0000 0001 0010 0111 1011 0101 1110 1001 1001 0101 1110 0010 0010 1000 0100(2) =

0.0000 0000 0001 0010 0111 1011 0101 1110 1001 1001 0101 1110 0010 0010 1000 0100(2) × 20 =

1.0010 0111 1011 0101 1110 1001 1001 0101 1110 0010 0010 1000 0100(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 0101 1110 1001 1001 0101 1110 0010 0010 1000 0100


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 0101 1110 1001 1001 0101 1110 0010 0010 1000 0100 =

0010 0111 1011 0101 1110 1001 1001 0101 1110 0010 0010 1000 0100


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 0101 1110 1001 1001 0101 1110 0010 0010 1000 0100


Decimal number -0.000 282 011 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 1110 1001 1001 0101 1110 0010 0010 1000 0100

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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