-0.000 282 026 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 026 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
-0.000 282 026 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:

|-0.000 282 026 6| = 0.000 282 026 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: 0.000 282 026 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) 0.000 282 026 6 × 2 = 0 + 0.000 564 053 2;
  • 2) 0.000 564 053 2 × 2 = 0 + 0.001 128 106 4;
  • 3) 0.001 128 106 4 × 2 = 0 + 0.002 256 212 8;
  • 4) 0.002 256 212 8 × 2 = 0 + 0.004 512 425 6;
  • 5) 0.004 512 425 6 × 2 = 0 + 0.009 024 851 2;
  • 6) 0.009 024 851 2 × 2 = 0 + 0.018 049 702 4;
  • 7) 0.018 049 702 4 × 2 = 0 + 0.036 099 404 8;
  • 8) 0.036 099 404 8 × 2 = 0 + 0.072 198 809 6;
  • 9) 0.072 198 809 6 × 2 = 0 + 0.144 397 619 2;
  • 10) 0.144 397 619 2 × 2 = 0 + 0.288 795 238 4;
  • 11) 0.288 795 238 4 × 2 = 0 + 0.577 590 476 8;
  • 12) 0.577 590 476 8 × 2 = 1 + 0.155 180 953 6;
  • 13) 0.155 180 953 6 × 2 = 0 + 0.310 361 907 2;
  • 14) 0.310 361 907 2 × 2 = 0 + 0.620 723 814 4;
  • 15) 0.620 723 814 4 × 2 = 1 + 0.241 447 628 8;
  • 16) 0.241 447 628 8 × 2 = 0 + 0.482 895 257 6;
  • 17) 0.482 895 257 6 × 2 = 0 + 0.965 790 515 2;
  • 18) 0.965 790 515 2 × 2 = 1 + 0.931 581 030 4;
  • 19) 0.931 581 030 4 × 2 = 1 + 0.863 162 060 8;
  • 20) 0.863 162 060 8 × 2 = 1 + 0.726 324 121 6;
  • 21) 0.726 324 121 6 × 2 = 1 + 0.452 648 243 2;
  • 22) 0.452 648 243 2 × 2 = 0 + 0.905 296 486 4;
  • 23) 0.905 296 486 4 × 2 = 1 + 0.810 592 972 8;
  • 24) 0.810 592 972 8 × 2 = 1 + 0.621 185 945 6;
  • 25) 0.621 185 945 6 × 2 = 1 + 0.242 371 891 2;
  • 26) 0.242 371 891 2 × 2 = 0 + 0.484 743 782 4;
  • 27) 0.484 743 782 4 × 2 = 0 + 0.969 487 564 8;
  • 28) 0.969 487 564 8 × 2 = 1 + 0.938 975 129 6;
  • 29) 0.938 975 129 6 × 2 = 1 + 0.877 950 259 2;
  • 30) 0.877 950 259 2 × 2 = 1 + 0.755 900 518 4;
  • 31) 0.755 900 518 4 × 2 = 1 + 0.511 801 036 8;
  • 32) 0.511 801 036 8 × 2 = 1 + 0.023 602 073 6;
  • 33) 0.023 602 073 6 × 2 = 0 + 0.047 204 147 2;
  • 34) 0.047 204 147 2 × 2 = 0 + 0.094 408 294 4;
  • 35) 0.094 408 294 4 × 2 = 0 + 0.188 816 588 8;
  • 36) 0.188 816 588 8 × 2 = 0 + 0.377 633 177 6;
  • 37) 0.377 633 177 6 × 2 = 0 + 0.755 266 355 2;
  • 38) 0.755 266 355 2 × 2 = 1 + 0.510 532 710 4;
  • 39) 0.510 532 710 4 × 2 = 1 + 0.021 065 420 8;
  • 40) 0.021 065 420 8 × 2 = 0 + 0.042 130 841 6;
  • 41) 0.042 130 841 6 × 2 = 0 + 0.084 261 683 2;
  • 42) 0.084 261 683 2 × 2 = 0 + 0.168 523 366 4;
  • 43) 0.168 523 366 4 × 2 = 0 + 0.337 046 732 8;
  • 44) 0.337 046 732 8 × 2 = 0 + 0.674 093 465 6;
  • 45) 0.674 093 465 6 × 2 = 1 + 0.348 186 931 2;
  • 46) 0.348 186 931 2 × 2 = 0 + 0.696 373 862 4;
  • 47) 0.696 373 862 4 × 2 = 1 + 0.392 747 724 8;
  • 48) 0.392 747 724 8 × 2 = 0 + 0.785 495 449 6;
  • 49) 0.785 495 449 6 × 2 = 1 + 0.570 990 899 2;
  • 50) 0.570 990 899 2 × 2 = 1 + 0.141 981 798 4;
  • 51) 0.141 981 798 4 × 2 = 0 + 0.283 963 596 8;
  • 52) 0.283 963 596 8 × 2 = 0 + 0.567 927 193 6;
  • 53) 0.567 927 193 6 × 2 = 1 + 0.135 854 387 2;
  • 54) 0.135 854 387 2 × 2 = 0 + 0.271 708 774 4;
  • 55) 0.271 708 774 4 × 2 = 0 + 0.543 417 548 8;
  • 56) 0.543 417 548 8 × 2 = 1 + 0.086 835 097 6;
  • 57) 0.086 835 097 6 × 2 = 0 + 0.173 670 195 2;
  • 58) 0.173 670 195 2 × 2 = 0 + 0.347 340 390 4;
  • 59) 0.347 340 390 4 × 2 = 0 + 0.694 680 780 8;
  • 60) 0.694 680 780 8 × 2 = 1 + 0.389 361 561 6;
  • 61) 0.389 361 561 6 × 2 = 0 + 0.778 723 123 2;
  • 62) 0.778 723 123 2 × 2 = 1 + 0.557 446 246 4;
  • 63) 0.557 446 246 4 × 2 = 1 + 0.114 892 492 8;
  • 64) 0.114 892 492 8 × 2 = 0 + 0.229 784 985 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:


0.000 282 026 6(10) =

0.0000 0000 0001 0010 0111 1011 1001 1111 0000 0110 0000 1010 1100 1001 0001 0110(2)

6. Positive number before normalization:

0.000 282 026 6(10) =

0.0000 0000 0001 0010 0111 1011 1001 1111 0000 0110 0000 1010 1100 1001 0001 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:


0.000 282 026 6(10) =

0.0000 0000 0001 0010 0111 1011 1001 1111 0000 0110 0000 1010 1100 1001 0001 0110(2) =

0.0000 0000 0001 0010 0111 1011 1001 1111 0000 0110 0000 1010 1100 1001 0001 0110(2) × 20 =

1.0010 0111 1011 1001 1111 0000 0110 0000 1010 1100 1001 0001 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 1001 1111 0000 0110 0000 1010 1100 1001 0001 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 1001 1111 0000 0110 0000 1010 1100 1001 0001 0110 =

0010 0111 1011 1001 1111 0000 0110 0000 1010 1100 1001 0001 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 1001 1111 0000 0110 0000 1010 1100 1001 0001 0110


Decimal number -0.000 282 026 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 1001 1111 0000 0110 0000 1010 1100 1001 0001 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