-0.000 282 005 913 27 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 913 27(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 27(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 27| = 0.000 282 005 913 27


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 27.

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 27 × 2 = 0 + 0.000 564 011 826 54;
  • 2) 0.000 564 011 826 54 × 2 = 0 + 0.001 128 023 653 08;
  • 3) 0.001 128 023 653 08 × 2 = 0 + 0.002 256 047 306 16;
  • 4) 0.002 256 047 306 16 × 2 = 0 + 0.004 512 094 612 32;
  • 5) 0.004 512 094 612 32 × 2 = 0 + 0.009 024 189 224 64;
  • 6) 0.009 024 189 224 64 × 2 = 0 + 0.018 048 378 449 28;
  • 7) 0.018 048 378 449 28 × 2 = 0 + 0.036 096 756 898 56;
  • 8) 0.036 096 756 898 56 × 2 = 0 + 0.072 193 513 797 12;
  • 9) 0.072 193 513 797 12 × 2 = 0 + 0.144 387 027 594 24;
  • 10) 0.144 387 027 594 24 × 2 = 0 + 0.288 774 055 188 48;
  • 11) 0.288 774 055 188 48 × 2 = 0 + 0.577 548 110 376 96;
  • 12) 0.577 548 110 376 96 × 2 = 1 + 0.155 096 220 753 92;
  • 13) 0.155 096 220 753 92 × 2 = 0 + 0.310 192 441 507 84;
  • 14) 0.310 192 441 507 84 × 2 = 0 + 0.620 384 883 015 68;
  • 15) 0.620 384 883 015 68 × 2 = 1 + 0.240 769 766 031 36;
  • 16) 0.240 769 766 031 36 × 2 = 0 + 0.481 539 532 062 72;
  • 17) 0.481 539 532 062 72 × 2 = 0 + 0.963 079 064 125 44;
  • 18) 0.963 079 064 125 44 × 2 = 1 + 0.926 158 128 250 88;
  • 19) 0.926 158 128 250 88 × 2 = 1 + 0.852 316 256 501 76;
  • 20) 0.852 316 256 501 76 × 2 = 1 + 0.704 632 513 003 52;
  • 21) 0.704 632 513 003 52 × 2 = 1 + 0.409 265 026 007 04;
  • 22) 0.409 265 026 007 04 × 2 = 0 + 0.818 530 052 014 08;
  • 23) 0.818 530 052 014 08 × 2 = 1 + 0.637 060 104 028 16;
  • 24) 0.637 060 104 028 16 × 2 = 1 + 0.274 120 208 056 32;
  • 25) 0.274 120 208 056 32 × 2 = 0 + 0.548 240 416 112 64;
  • 26) 0.548 240 416 112 64 × 2 = 1 + 0.096 480 832 225 28;
  • 27) 0.096 480 832 225 28 × 2 = 0 + 0.192 961 664 450 56;
  • 28) 0.192 961 664 450 56 × 2 = 0 + 0.385 923 328 901 12;
  • 29) 0.385 923 328 901 12 × 2 = 0 + 0.771 846 657 802 24;
  • 30) 0.771 846 657 802 24 × 2 = 1 + 0.543 693 315 604 48;
  • 31) 0.543 693 315 604 48 × 2 = 1 + 0.087 386 631 208 96;
  • 32) 0.087 386 631 208 96 × 2 = 0 + 0.174 773 262 417 92;
  • 33) 0.174 773 262 417 92 × 2 = 0 + 0.349 546 524 835 84;
  • 34) 0.349 546 524 835 84 × 2 = 0 + 0.699 093 049 671 68;
  • 35) 0.699 093 049 671 68 × 2 = 1 + 0.398 186 099 343 36;
  • 36) 0.398 186 099 343 36 × 2 = 0 + 0.796 372 198 686 72;
  • 37) 0.796 372 198 686 72 × 2 = 1 + 0.592 744 397 373 44;
  • 38) 0.592 744 397 373 44 × 2 = 1 + 0.185 488 794 746 88;
  • 39) 0.185 488 794 746 88 × 2 = 0 + 0.370 977 589 493 76;
  • 40) 0.370 977 589 493 76 × 2 = 0 + 0.741 955 178 987 52;
  • 41) 0.741 955 178 987 52 × 2 = 1 + 0.483 910 357 975 04;
  • 42) 0.483 910 357 975 04 × 2 = 0 + 0.967 820 715 950 08;
  • 43) 0.967 820 715 950 08 × 2 = 1 + 0.935 641 431 900 16;
  • 44) 0.935 641 431 900 16 × 2 = 1 + 0.871 282 863 800 32;
  • 45) 0.871 282 863 800 32 × 2 = 1 + 0.742 565 727 600 64;
  • 46) 0.742 565 727 600 64 × 2 = 1 + 0.485 131 455 201 28;
  • 47) 0.485 131 455 201 28 × 2 = 0 + 0.970 262 910 402 56;
  • 48) 0.970 262 910 402 56 × 2 = 1 + 0.940 525 820 805 12;
  • 49) 0.940 525 820 805 12 × 2 = 1 + 0.881 051 641 610 24;
  • 50) 0.881 051 641 610 24 × 2 = 1 + 0.762 103 283 220 48;
  • 51) 0.762 103 283 220 48 × 2 = 1 + 0.524 206 566 440 96;
  • 52) 0.524 206 566 440 96 × 2 = 1 + 0.048 413 132 881 92;
  • 53) 0.048 413 132 881 92 × 2 = 0 + 0.096 826 265 763 84;
  • 54) 0.096 826 265 763 84 × 2 = 0 + 0.193 652 531 527 68;
  • 55) 0.193 652 531 527 68 × 2 = 0 + 0.387 305 063 055 36;
  • 56) 0.387 305 063 055 36 × 2 = 0 + 0.774 610 126 110 72;
  • 57) 0.774 610 126 110 72 × 2 = 1 + 0.549 220 252 221 44;
  • 58) 0.549 220 252 221 44 × 2 = 1 + 0.098 440 504 442 88;
  • 59) 0.098 440 504 442 88 × 2 = 0 + 0.196 881 008 885 76;
  • 60) 0.196 881 008 885 76 × 2 = 0 + 0.393 762 017 771 52;
  • 61) 0.393 762 017 771 52 × 2 = 0 + 0.787 524 035 543 04;
  • 62) 0.787 524 035 543 04 × 2 = 1 + 0.575 048 071 086 08;
  • 63) 0.575 048 071 086 08 × 2 = 1 + 0.150 096 142 172 16;
  • 64) 0.150 096 142 172 16 × 2 = 0 + 0.300 192 284 344 32;

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 27(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1011 1101 1111 0000 1100 0110(2)

6. Positive number before normalization:

0.000 282 005 913 27(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1011 1101 1111 0000 1100 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:


0.000 282 005 913 27(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1011 1101 1111 0000 1100 0110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1011 1101 1111 0000 1100 0110(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1100 1011 1101 1111 0000 1100 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 1100 1011 1101 1111 0000 1100 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 1100 1011 1101 1111 0000 1100 0110 =

0010 0111 1011 0100 0110 0010 1100 1011 1101 1111 0000 1100 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 1100 1011 1101 1111 0000 1100 0110


Decimal number -0.000 282 005 913 27 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1100 1011 1101 1111 0000 1100 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