-0.000 282 009 91 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 009 91(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 91(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 91| = 0.000 282 009 91


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 91.

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 91 × 2 = 0 + 0.000 564 019 82;
  • 2) 0.000 564 019 82 × 2 = 0 + 0.001 128 039 64;
  • 3) 0.001 128 039 64 × 2 = 0 + 0.002 256 079 28;
  • 4) 0.002 256 079 28 × 2 = 0 + 0.004 512 158 56;
  • 5) 0.004 512 158 56 × 2 = 0 + 0.009 024 317 12;
  • 6) 0.009 024 317 12 × 2 = 0 + 0.018 048 634 24;
  • 7) 0.018 048 634 24 × 2 = 0 + 0.036 097 268 48;
  • 8) 0.036 097 268 48 × 2 = 0 + 0.072 194 536 96;
  • 9) 0.072 194 536 96 × 2 = 0 + 0.144 389 073 92;
  • 10) 0.144 389 073 92 × 2 = 0 + 0.288 778 147 84;
  • 11) 0.288 778 147 84 × 2 = 0 + 0.577 556 295 68;
  • 12) 0.577 556 295 68 × 2 = 1 + 0.155 112 591 36;
  • 13) 0.155 112 591 36 × 2 = 0 + 0.310 225 182 72;
  • 14) 0.310 225 182 72 × 2 = 0 + 0.620 450 365 44;
  • 15) 0.620 450 365 44 × 2 = 1 + 0.240 900 730 88;
  • 16) 0.240 900 730 88 × 2 = 0 + 0.481 801 461 76;
  • 17) 0.481 801 461 76 × 2 = 0 + 0.963 602 923 52;
  • 18) 0.963 602 923 52 × 2 = 1 + 0.927 205 847 04;
  • 19) 0.927 205 847 04 × 2 = 1 + 0.854 411 694 08;
  • 20) 0.854 411 694 08 × 2 = 1 + 0.708 823 388 16;
  • 21) 0.708 823 388 16 × 2 = 1 + 0.417 646 776 32;
  • 22) 0.417 646 776 32 × 2 = 0 + 0.835 293 552 64;
  • 23) 0.835 293 552 64 × 2 = 1 + 0.670 587 105 28;
  • 24) 0.670 587 105 28 × 2 = 1 + 0.341 174 210 56;
  • 25) 0.341 174 210 56 × 2 = 0 + 0.682 348 421 12;
  • 26) 0.682 348 421 12 × 2 = 1 + 0.364 696 842 24;
  • 27) 0.364 696 842 24 × 2 = 0 + 0.729 393 684 48;
  • 28) 0.729 393 684 48 × 2 = 1 + 0.458 787 368 96;
  • 29) 0.458 787 368 96 × 2 = 0 + 0.917 574 737 92;
  • 30) 0.917 574 737 92 × 2 = 1 + 0.835 149 475 84;
  • 31) 0.835 149 475 84 × 2 = 1 + 0.670 298 951 68;
  • 32) 0.670 298 951 68 × 2 = 1 + 0.340 597 903 36;
  • 33) 0.340 597 903 36 × 2 = 0 + 0.681 195 806 72;
  • 34) 0.681 195 806 72 × 2 = 1 + 0.362 391 613 44;
  • 35) 0.362 391 613 44 × 2 = 0 + 0.724 783 226 88;
  • 36) 0.724 783 226 88 × 2 = 1 + 0.449 566 453 76;
  • 37) 0.449 566 453 76 × 2 = 0 + 0.899 132 907 52;
  • 38) 0.899 132 907 52 × 2 = 1 + 0.798 265 815 04;
  • 39) 0.798 265 815 04 × 2 = 1 + 0.596 531 630 08;
  • 40) 0.596 531 630 08 × 2 = 1 + 0.193 063 260 16;
  • 41) 0.193 063 260 16 × 2 = 0 + 0.386 126 520 32;
  • 42) 0.386 126 520 32 × 2 = 0 + 0.772 253 040 64;
  • 43) 0.772 253 040 64 × 2 = 1 + 0.544 506 081 28;
  • 44) 0.544 506 081 28 × 2 = 1 + 0.089 012 162 56;
  • 45) 0.089 012 162 56 × 2 = 0 + 0.178 024 325 12;
  • 46) 0.178 024 325 12 × 2 = 0 + 0.356 048 650 24;
  • 47) 0.356 048 650 24 × 2 = 0 + 0.712 097 300 48;
  • 48) 0.712 097 300 48 × 2 = 1 + 0.424 194 600 96;
  • 49) 0.424 194 600 96 × 2 = 0 + 0.848 389 201 92;
  • 50) 0.848 389 201 92 × 2 = 1 + 0.696 778 403 84;
  • 51) 0.696 778 403 84 × 2 = 1 + 0.393 556 807 68;
  • 52) 0.393 556 807 68 × 2 = 0 + 0.787 113 615 36;
  • 53) 0.787 113 615 36 × 2 = 1 + 0.574 227 230 72;
  • 54) 0.574 227 230 72 × 2 = 1 + 0.148 454 461 44;
  • 55) 0.148 454 461 44 × 2 = 0 + 0.296 908 922 88;
  • 56) 0.296 908 922 88 × 2 = 0 + 0.593 817 845 76;
  • 57) 0.593 817 845 76 × 2 = 1 + 0.187 635 691 52;
  • 58) 0.187 635 691 52 × 2 = 0 + 0.375 271 383 04;
  • 59) 0.375 271 383 04 × 2 = 0 + 0.750 542 766 08;
  • 60) 0.750 542 766 08 × 2 = 1 + 0.501 085 532 16;
  • 61) 0.501 085 532 16 × 2 = 1 + 0.002 171 064 32;
  • 62) 0.002 171 064 32 × 2 = 0 + 0.004 342 128 64;
  • 63) 0.004 342 128 64 × 2 = 0 + 0.008 684 257 28;
  • 64) 0.008 684 257 28 × 2 = 0 + 0.017 368 514 56;

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 91(10) =

0.0000 0000 0001 0010 0111 1011 0101 0111 0101 0111 0011 0001 0110 1100 1001 1000(2)

6. Positive number before normalization:

0.000 282 009 91(10) =

0.0000 0000 0001 0010 0111 1011 0101 0111 0101 0111 0011 0001 0110 1100 1001 1000(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 91(10) =

0.0000 0000 0001 0010 0111 1011 0101 0111 0101 0111 0011 0001 0110 1100 1001 1000(2) =

0.0000 0000 0001 0010 0111 1011 0101 0111 0101 0111 0011 0001 0110 1100 1001 1000(2) × 20 =

1.0010 0111 1011 0101 0111 0101 0111 0011 0001 0110 1100 1001 1000(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 0101 0111 0011 0001 0110 1100 1001 1000


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 0101 0111 0011 0001 0110 1100 1001 1000 =

0010 0111 1011 0101 0111 0101 0111 0011 0001 0110 1100 1001 1000


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 0101 0111 0011 0001 0110 1100 1001 1000


Decimal number -0.000 282 009 91 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 0111 0101 0111 0011 0001 0110 1100 1001 1000

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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