-0.000 282 009 86 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 009 86(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
-0.000 282 009 86(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:

|-0.000 282 009 86| = 0.000 282 009 86


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: 0.000 282 009 86.

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) 0.000 282 009 86 × 2 = 0 + 0.000 564 019 72;
  • 2) 0.000 564 019 72 × 2 = 0 + 0.001 128 039 44;
  • 3) 0.001 128 039 44 × 2 = 0 + 0.002 256 078 88;
  • 4) 0.002 256 078 88 × 2 = 0 + 0.004 512 157 76;
  • 5) 0.004 512 157 76 × 2 = 0 + 0.009 024 315 52;
  • 6) 0.009 024 315 52 × 2 = 0 + 0.018 048 631 04;
  • 7) 0.018 048 631 04 × 2 = 0 + 0.036 097 262 08;
  • 8) 0.036 097 262 08 × 2 = 0 + 0.072 194 524 16;
  • 9) 0.072 194 524 16 × 2 = 0 + 0.144 389 048 32;
  • 10) 0.144 389 048 32 × 2 = 0 + 0.288 778 096 64;
  • 11) 0.288 778 096 64 × 2 = 0 + 0.577 556 193 28;
  • 12) 0.577 556 193 28 × 2 = 1 + 0.155 112 386 56;
  • 13) 0.155 112 386 56 × 2 = 0 + 0.310 224 773 12;
  • 14) 0.310 224 773 12 × 2 = 0 + 0.620 449 546 24;
  • 15) 0.620 449 546 24 × 2 = 1 + 0.240 899 092 48;
  • 16) 0.240 899 092 48 × 2 = 0 + 0.481 798 184 96;
  • 17) 0.481 798 184 96 × 2 = 0 + 0.963 596 369 92;
  • 18) 0.963 596 369 92 × 2 = 1 + 0.927 192 739 84;
  • 19) 0.927 192 739 84 × 2 = 1 + 0.854 385 479 68;
  • 20) 0.854 385 479 68 × 2 = 1 + 0.708 770 959 36;
  • 21) 0.708 770 959 36 × 2 = 1 + 0.417 541 918 72;
  • 22) 0.417 541 918 72 × 2 = 0 + 0.835 083 837 44;
  • 23) 0.835 083 837 44 × 2 = 1 + 0.670 167 674 88;
  • 24) 0.670 167 674 88 × 2 = 1 + 0.340 335 349 76;
  • 25) 0.340 335 349 76 × 2 = 0 + 0.680 670 699 52;
  • 26) 0.680 670 699 52 × 2 = 1 + 0.361 341 399 04;
  • 27) 0.361 341 399 04 × 2 = 0 + 0.722 682 798 08;
  • 28) 0.722 682 798 08 × 2 = 1 + 0.445 365 596 16;
  • 29) 0.445 365 596 16 × 2 = 0 + 0.890 731 192 32;
  • 30) 0.890 731 192 32 × 2 = 1 + 0.781 462 384 64;
  • 31) 0.781 462 384 64 × 2 = 1 + 0.562 924 769 28;
  • 32) 0.562 924 769 28 × 2 = 1 + 0.125 849 538 56;
  • 33) 0.125 849 538 56 × 2 = 0 + 0.251 699 077 12;
  • 34) 0.251 699 077 12 × 2 = 0 + 0.503 398 154 24;
  • 35) 0.503 398 154 24 × 2 = 1 + 0.006 796 308 48;
  • 36) 0.006 796 308 48 × 2 = 0 + 0.013 592 616 96;
  • 37) 0.013 592 616 96 × 2 = 0 + 0.027 185 233 92;
  • 38) 0.027 185 233 92 × 2 = 0 + 0.054 370 467 84;
  • 39) 0.054 370 467 84 × 2 = 0 + 0.108 740 935 68;
  • 40) 0.108 740 935 68 × 2 = 0 + 0.217 481 871 36;
  • 41) 0.217 481 871 36 × 2 = 0 + 0.434 963 742 72;
  • 42) 0.434 963 742 72 × 2 = 0 + 0.869 927 485 44;
  • 43) 0.869 927 485 44 × 2 = 1 + 0.739 854 970 88;
  • 44) 0.739 854 970 88 × 2 = 1 + 0.479 709 941 76;
  • 45) 0.479 709 941 76 × 2 = 0 + 0.959 419 883 52;
  • 46) 0.959 419 883 52 × 2 = 1 + 0.918 839 767 04;
  • 47) 0.918 839 767 04 × 2 = 1 + 0.837 679 534 08;
  • 48) 0.837 679 534 08 × 2 = 1 + 0.675 359 068 16;
  • 49) 0.675 359 068 16 × 2 = 1 + 0.350 718 136 32;
  • 50) 0.350 718 136 32 × 2 = 0 + 0.701 436 272 64;
  • 51) 0.701 436 272 64 × 2 = 1 + 0.402 872 545 28;
  • 52) 0.402 872 545 28 × 2 = 0 + 0.805 745 090 56;
  • 53) 0.805 745 090 56 × 2 = 1 + 0.611 490 181 12;
  • 54) 0.611 490 181 12 × 2 = 1 + 0.222 980 362 24;
  • 55) 0.222 980 362 24 × 2 = 0 + 0.445 960 724 48;
  • 56) 0.445 960 724 48 × 2 = 0 + 0.891 921 448 96;
  • 57) 0.891 921 448 96 × 2 = 1 + 0.783 842 897 92;
  • 58) 0.783 842 897 92 × 2 = 1 + 0.567 685 795 84;
  • 59) 0.567 685 795 84 × 2 = 1 + 0.135 371 591 68;
  • 60) 0.135 371 591 68 × 2 = 0 + 0.270 743 183 36;
  • 61) 0.270 743 183 36 × 2 = 0 + 0.541 486 366 72;
  • 62) 0.541 486 366 72 × 2 = 1 + 0.082 972 733 44;
  • 63) 0.082 972 733 44 × 2 = 0 + 0.165 945 466 88;
  • 64) 0.165 945 466 88 × 2 = 0 + 0.331 890 933 76;

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:


0.000 282 009 86(10) =

0.0000 0000 0001 0010 0111 1011 0101 0111 0010 0000 0011 0111 1010 1100 1110 0100(2)

6. Positive number before normalization:

0.000 282 009 86(10) =

0.0000 0000 0001 0010 0111 1011 0101 0111 0010 0000 0011 0111 1010 1100 1110 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:


0.000 282 009 86(10) =

0.0000 0000 0001 0010 0111 1011 0101 0111 0010 0000 0011 0111 1010 1100 1110 0100(2) =

0.0000 0000 0001 0010 0111 1011 0101 0111 0010 0000 0011 0111 1010 1100 1110 0100(2) × 20 =

1.0010 0111 1011 0101 0111 0010 0000 0011 0111 1010 1100 1110 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 0111 0010 0000 0011 0111 1010 1100 1110 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 0111 0010 0000 0011 0111 1010 1100 1110 0100 =

0010 0111 1011 0101 0111 0010 0000 0011 0111 1010 1100 1110 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 0111 0010 0000 0011 0111 1010 1100 1110 0100


Decimal number -0.000 282 009 86 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 0111 0010 0000 0011 0111 1010 1100 1110 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