-0.000 282 005 866 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 866 4(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
-0.000 282 005 866 4(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:

|-0.000 282 005 866 4| = 0.000 282 005 866 4


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: 0.000 282 005 866 4.

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) 0.000 282 005 866 4 × 2 = 0 + 0.000 564 011 732 8;
  • 2) 0.000 564 011 732 8 × 2 = 0 + 0.001 128 023 465 6;
  • 3) 0.001 128 023 465 6 × 2 = 0 + 0.002 256 046 931 2;
  • 4) 0.002 256 046 931 2 × 2 = 0 + 0.004 512 093 862 4;
  • 5) 0.004 512 093 862 4 × 2 = 0 + 0.009 024 187 724 8;
  • 6) 0.009 024 187 724 8 × 2 = 0 + 0.018 048 375 449 6;
  • 7) 0.018 048 375 449 6 × 2 = 0 + 0.036 096 750 899 2;
  • 8) 0.036 096 750 899 2 × 2 = 0 + 0.072 193 501 798 4;
  • 9) 0.072 193 501 798 4 × 2 = 0 + 0.144 387 003 596 8;
  • 10) 0.144 387 003 596 8 × 2 = 0 + 0.288 774 007 193 6;
  • 11) 0.288 774 007 193 6 × 2 = 0 + 0.577 548 014 387 2;
  • 12) 0.577 548 014 387 2 × 2 = 1 + 0.155 096 028 774 4;
  • 13) 0.155 096 028 774 4 × 2 = 0 + 0.310 192 057 548 8;
  • 14) 0.310 192 057 548 8 × 2 = 0 + 0.620 384 115 097 6;
  • 15) 0.620 384 115 097 6 × 2 = 1 + 0.240 768 230 195 2;
  • 16) 0.240 768 230 195 2 × 2 = 0 + 0.481 536 460 390 4;
  • 17) 0.481 536 460 390 4 × 2 = 0 + 0.963 072 920 780 8;
  • 18) 0.963 072 920 780 8 × 2 = 1 + 0.926 145 841 561 6;
  • 19) 0.926 145 841 561 6 × 2 = 1 + 0.852 291 683 123 2;
  • 20) 0.852 291 683 123 2 × 2 = 1 + 0.704 583 366 246 4;
  • 21) 0.704 583 366 246 4 × 2 = 1 + 0.409 166 732 492 8;
  • 22) 0.409 166 732 492 8 × 2 = 0 + 0.818 333 464 985 6;
  • 23) 0.818 333 464 985 6 × 2 = 1 + 0.636 666 929 971 2;
  • 24) 0.636 666 929 971 2 × 2 = 1 + 0.273 333 859 942 4;
  • 25) 0.273 333 859 942 4 × 2 = 0 + 0.546 667 719 884 8;
  • 26) 0.546 667 719 884 8 × 2 = 1 + 0.093 335 439 769 6;
  • 27) 0.093 335 439 769 6 × 2 = 0 + 0.186 670 879 539 2;
  • 28) 0.186 670 879 539 2 × 2 = 0 + 0.373 341 759 078 4;
  • 29) 0.373 341 759 078 4 × 2 = 0 + 0.746 683 518 156 8;
  • 30) 0.746 683 518 156 8 × 2 = 1 + 0.493 367 036 313 6;
  • 31) 0.493 367 036 313 6 × 2 = 0 + 0.986 734 072 627 2;
  • 32) 0.986 734 072 627 2 × 2 = 1 + 0.973 468 145 254 4;
  • 33) 0.973 468 145 254 4 × 2 = 1 + 0.946 936 290 508 8;
  • 34) 0.946 936 290 508 8 × 2 = 1 + 0.893 872 581 017 6;
  • 35) 0.893 872 581 017 6 × 2 = 1 + 0.787 745 162 035 2;
  • 36) 0.787 745 162 035 2 × 2 = 1 + 0.575 490 324 070 4;
  • 37) 0.575 490 324 070 4 × 2 = 1 + 0.150 980 648 140 8;
  • 38) 0.150 980 648 140 8 × 2 = 0 + 0.301 961 296 281 6;
  • 39) 0.301 961 296 281 6 × 2 = 0 + 0.603 922 592 563 2;
  • 40) 0.603 922 592 563 2 × 2 = 1 + 0.207 845 185 126 4;
  • 41) 0.207 845 185 126 4 × 2 = 0 + 0.415 690 370 252 8;
  • 42) 0.415 690 370 252 8 × 2 = 0 + 0.831 380 740 505 6;
  • 43) 0.831 380 740 505 6 × 2 = 1 + 0.662 761 481 011 2;
  • 44) 0.662 761 481 011 2 × 2 = 1 + 0.325 522 962 022 4;
  • 45) 0.325 522 962 022 4 × 2 = 0 + 0.651 045 924 044 8;
  • 46) 0.651 045 924 044 8 × 2 = 1 + 0.302 091 848 089 6;
  • 47) 0.302 091 848 089 6 × 2 = 0 + 0.604 183 696 179 2;
  • 48) 0.604 183 696 179 2 × 2 = 1 + 0.208 367 392 358 4;
  • 49) 0.208 367 392 358 4 × 2 = 0 + 0.416 734 784 716 8;
  • 50) 0.416 734 784 716 8 × 2 = 0 + 0.833 469 569 433 6;
  • 51) 0.833 469 569 433 6 × 2 = 1 + 0.666 939 138 867 2;
  • 52) 0.666 939 138 867 2 × 2 = 1 + 0.333 878 277 734 4;
  • 53) 0.333 878 277 734 4 × 2 = 0 + 0.667 756 555 468 8;
  • 54) 0.667 756 555 468 8 × 2 = 1 + 0.335 513 110 937 6;
  • 55) 0.335 513 110 937 6 × 2 = 0 + 0.671 026 221 875 2;
  • 56) 0.671 026 221 875 2 × 2 = 1 + 0.342 052 443 750 4;
  • 57) 0.342 052 443 750 4 × 2 = 0 + 0.684 104 887 500 8;
  • 58) 0.684 104 887 500 8 × 2 = 1 + 0.368 209 775 001 6;
  • 59) 0.368 209 775 001 6 × 2 = 0 + 0.736 419 550 003 2;
  • 60) 0.736 419 550 003 2 × 2 = 1 + 0.472 839 100 006 4;
  • 61) 0.472 839 100 006 4 × 2 = 0 + 0.945 678 200 012 8;
  • 62) 0.945 678 200 012 8 × 2 = 1 + 0.891 356 400 025 6;
  • 63) 0.891 356 400 025 6 × 2 = 1 + 0.782 712 800 051 2;
  • 64) 0.782 712 800 051 2 × 2 = 1 + 0.565 425 600 102 4;

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:


0.000 282 005 866 4(10) =

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

6. Positive number before normalization:

0.000 282 005 866 4(10) =

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


0.000 282 005 866 4(10) =

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

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

1.0010 0111 1011 0100 0101 1111 1001 0011 0101 0011 0101 0101 0111(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 0101 1111 1001 0011 0101 0011 0101 0101 0111


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 0101 1111 1001 0011 0101 0011 0101 0101 0111 =

0010 0111 1011 0100 0101 1111 1001 0011 0101 0011 0101 0101 0111


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 0101 1111 1001 0011 0101 0011 0101 0101 0111


Decimal number -0.000 282 005 866 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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