-0.000 282 005 891 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 891 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
-0.000 282 005 891 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:

|-0.000 282 005 891 2| = 0.000 282 005 891 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: 0.000 282 005 891 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) 0.000 282 005 891 2 × 2 = 0 + 0.000 564 011 782 4;
  • 2) 0.000 564 011 782 4 × 2 = 0 + 0.001 128 023 564 8;
  • 3) 0.001 128 023 564 8 × 2 = 0 + 0.002 256 047 129 6;
  • 4) 0.002 256 047 129 6 × 2 = 0 + 0.004 512 094 259 2;
  • 5) 0.004 512 094 259 2 × 2 = 0 + 0.009 024 188 518 4;
  • 6) 0.009 024 188 518 4 × 2 = 0 + 0.018 048 377 036 8;
  • 7) 0.018 048 377 036 8 × 2 = 0 + 0.036 096 754 073 6;
  • 8) 0.036 096 754 073 6 × 2 = 0 + 0.072 193 508 147 2;
  • 9) 0.072 193 508 147 2 × 2 = 0 + 0.144 387 016 294 4;
  • 10) 0.144 387 016 294 4 × 2 = 0 + 0.288 774 032 588 8;
  • 11) 0.288 774 032 588 8 × 2 = 0 + 0.577 548 065 177 6;
  • 12) 0.577 548 065 177 6 × 2 = 1 + 0.155 096 130 355 2;
  • 13) 0.155 096 130 355 2 × 2 = 0 + 0.310 192 260 710 4;
  • 14) 0.310 192 260 710 4 × 2 = 0 + 0.620 384 521 420 8;
  • 15) 0.620 384 521 420 8 × 2 = 1 + 0.240 769 042 841 6;
  • 16) 0.240 769 042 841 6 × 2 = 0 + 0.481 538 085 683 2;
  • 17) 0.481 538 085 683 2 × 2 = 0 + 0.963 076 171 366 4;
  • 18) 0.963 076 171 366 4 × 2 = 1 + 0.926 152 342 732 8;
  • 19) 0.926 152 342 732 8 × 2 = 1 + 0.852 304 685 465 6;
  • 20) 0.852 304 685 465 6 × 2 = 1 + 0.704 609 370 931 2;
  • 21) 0.704 609 370 931 2 × 2 = 1 + 0.409 218 741 862 4;
  • 22) 0.409 218 741 862 4 × 2 = 0 + 0.818 437 483 724 8;
  • 23) 0.818 437 483 724 8 × 2 = 1 + 0.636 874 967 449 6;
  • 24) 0.636 874 967 449 6 × 2 = 1 + 0.273 749 934 899 2;
  • 25) 0.273 749 934 899 2 × 2 = 0 + 0.547 499 869 798 4;
  • 26) 0.547 499 869 798 4 × 2 = 1 + 0.094 999 739 596 8;
  • 27) 0.094 999 739 596 8 × 2 = 0 + 0.189 999 479 193 6;
  • 28) 0.189 999 479 193 6 × 2 = 0 + 0.379 998 958 387 2;
  • 29) 0.379 998 958 387 2 × 2 = 0 + 0.759 997 916 774 4;
  • 30) 0.759 997 916 774 4 × 2 = 1 + 0.519 995 833 548 8;
  • 31) 0.519 995 833 548 8 × 2 = 1 + 0.039 991 667 097 6;
  • 32) 0.039 991 667 097 6 × 2 = 0 + 0.079 983 334 195 2;
  • 33) 0.079 983 334 195 2 × 2 = 0 + 0.159 966 668 390 4;
  • 34) 0.159 966 668 390 4 × 2 = 0 + 0.319 933 336 780 8;
  • 35) 0.319 933 336 780 8 × 2 = 0 + 0.639 866 673 561 6;
  • 36) 0.639 866 673 561 6 × 2 = 1 + 0.279 733 347 123 2;
  • 37) 0.279 733 347 123 2 × 2 = 0 + 0.559 466 694 246 4;
  • 38) 0.559 466 694 246 4 × 2 = 1 + 0.118 933 388 492 8;
  • 39) 0.118 933 388 492 8 × 2 = 0 + 0.237 866 776 985 6;
  • 40) 0.237 866 776 985 6 × 2 = 0 + 0.475 733 553 971 2;
  • 41) 0.475 733 553 971 2 × 2 = 0 + 0.951 467 107 942 4;
  • 42) 0.951 467 107 942 4 × 2 = 1 + 0.902 934 215 884 8;
  • 43) 0.902 934 215 884 8 × 2 = 1 + 0.805 868 431 769 6;
  • 44) 0.805 868 431 769 6 × 2 = 1 + 0.611 736 863 539 2;
  • 45) 0.611 736 863 539 2 × 2 = 1 + 0.223 473 727 078 4;
  • 46) 0.223 473 727 078 4 × 2 = 0 + 0.446 947 454 156 8;
  • 47) 0.446 947 454 156 8 × 2 = 0 + 0.893 894 908 313 6;
  • 48) 0.893 894 908 313 6 × 2 = 1 + 0.787 789 816 627 2;
  • 49) 0.787 789 816 627 2 × 2 = 1 + 0.575 579 633 254 4;
  • 50) 0.575 579 633 254 4 × 2 = 1 + 0.151 159 266 508 8;
  • 51) 0.151 159 266 508 8 × 2 = 0 + 0.302 318 533 017 6;
  • 52) 0.302 318 533 017 6 × 2 = 0 + 0.604 637 066 035 2;
  • 53) 0.604 637 066 035 2 × 2 = 1 + 0.209 274 132 070 4;
  • 54) 0.209 274 132 070 4 × 2 = 0 + 0.418 548 264 140 8;
  • 55) 0.418 548 264 140 8 × 2 = 0 + 0.837 096 528 281 6;
  • 56) 0.837 096 528 281 6 × 2 = 1 + 0.674 193 056 563 2;
  • 57) 0.674 193 056 563 2 × 2 = 1 + 0.348 386 113 126 4;
  • 58) 0.348 386 113 126 4 × 2 = 0 + 0.696 772 226 252 8;
  • 59) 0.696 772 226 252 8 × 2 = 1 + 0.393 544 452 505 6;
  • 60) 0.393 544 452 505 6 × 2 = 0 + 0.787 088 905 011 2;
  • 61) 0.787 088 905 011 2 × 2 = 1 + 0.574 177 810 022 4;
  • 62) 0.574 177 810 022 4 × 2 = 1 + 0.148 355 620 044 8;
  • 63) 0.148 355 620 044 8 × 2 = 0 + 0.296 711 240 089 6;
  • 64) 0.296 711 240 089 6 × 2 = 0 + 0.593 422 480 179 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:


0.000 282 005 891 2(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0100 0111 1001 1100 1001 1010 1100(2)

6. Positive number before normalization:

0.000 282 005 891 2(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0100 0111 1001 1100 1001 1010 1100(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:


0.000 282 005 891 2(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0100 0111 1001 1100 1001 1010 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0100 0111 1001 1100 1001 1010 1100(2) × 20 =

1.0010 0111 1011 0100 0110 0001 0100 0111 1001 1100 1001 1010 1100(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 0001 0100 0111 1001 1100 1001 1010 1100


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 0001 0100 0111 1001 1100 1001 1010 1100 =

0010 0111 1011 0100 0110 0001 0100 0111 1001 1100 1001 1010 1100


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 0001 0100 0111 1001 1100 1001 1010 1100


Decimal number -0.000 282 005 891 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0001 0100 0111 1001 1100 1001 1010 1100

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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