-0.000 282 005 91 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 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
-0.000 282 005 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:

|-0.000 282 005 91| = 0.000 282 005 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: 0.000 282 005 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) 0.000 282 005 91 × 2 = 0 + 0.000 564 011 82;
  • 2) 0.000 564 011 82 × 2 = 0 + 0.001 128 023 64;
  • 3) 0.001 128 023 64 × 2 = 0 + 0.002 256 047 28;
  • 4) 0.002 256 047 28 × 2 = 0 + 0.004 512 094 56;
  • 5) 0.004 512 094 56 × 2 = 0 + 0.009 024 189 12;
  • 6) 0.009 024 189 12 × 2 = 0 + 0.018 048 378 24;
  • 7) 0.018 048 378 24 × 2 = 0 + 0.036 096 756 48;
  • 8) 0.036 096 756 48 × 2 = 0 + 0.072 193 512 96;
  • 9) 0.072 193 512 96 × 2 = 0 + 0.144 387 025 92;
  • 10) 0.144 387 025 92 × 2 = 0 + 0.288 774 051 84;
  • 11) 0.288 774 051 84 × 2 = 0 + 0.577 548 103 68;
  • 12) 0.577 548 103 68 × 2 = 1 + 0.155 096 207 36;
  • 13) 0.155 096 207 36 × 2 = 0 + 0.310 192 414 72;
  • 14) 0.310 192 414 72 × 2 = 0 + 0.620 384 829 44;
  • 15) 0.620 384 829 44 × 2 = 1 + 0.240 769 658 88;
  • 16) 0.240 769 658 88 × 2 = 0 + 0.481 539 317 76;
  • 17) 0.481 539 317 76 × 2 = 0 + 0.963 078 635 52;
  • 18) 0.963 078 635 52 × 2 = 1 + 0.926 157 271 04;
  • 19) 0.926 157 271 04 × 2 = 1 + 0.852 314 542 08;
  • 20) 0.852 314 542 08 × 2 = 1 + 0.704 629 084 16;
  • 21) 0.704 629 084 16 × 2 = 1 + 0.409 258 168 32;
  • 22) 0.409 258 168 32 × 2 = 0 + 0.818 516 336 64;
  • 23) 0.818 516 336 64 × 2 = 1 + 0.637 032 673 28;
  • 24) 0.637 032 673 28 × 2 = 1 + 0.274 065 346 56;
  • 25) 0.274 065 346 56 × 2 = 0 + 0.548 130 693 12;
  • 26) 0.548 130 693 12 × 2 = 1 + 0.096 261 386 24;
  • 27) 0.096 261 386 24 × 2 = 0 + 0.192 522 772 48;
  • 28) 0.192 522 772 48 × 2 = 0 + 0.385 045 544 96;
  • 29) 0.385 045 544 96 × 2 = 0 + 0.770 091 089 92;
  • 30) 0.770 091 089 92 × 2 = 1 + 0.540 182 179 84;
  • 31) 0.540 182 179 84 × 2 = 1 + 0.080 364 359 68;
  • 32) 0.080 364 359 68 × 2 = 0 + 0.160 728 719 36;
  • 33) 0.160 728 719 36 × 2 = 0 + 0.321 457 438 72;
  • 34) 0.321 457 438 72 × 2 = 0 + 0.642 914 877 44;
  • 35) 0.642 914 877 44 × 2 = 1 + 0.285 829 754 88;
  • 36) 0.285 829 754 88 × 2 = 0 + 0.571 659 509 76;
  • 37) 0.571 659 509 76 × 2 = 1 + 0.143 319 019 52;
  • 38) 0.143 319 019 52 × 2 = 0 + 0.286 638 039 04;
  • 39) 0.286 638 039 04 × 2 = 0 + 0.573 276 078 08;
  • 40) 0.573 276 078 08 × 2 = 1 + 0.146 552 156 16;
  • 41) 0.146 552 156 16 × 2 = 0 + 0.293 104 312 32;
  • 42) 0.293 104 312 32 × 2 = 0 + 0.586 208 624 64;
  • 43) 0.586 208 624 64 × 2 = 1 + 0.172 417 249 28;
  • 44) 0.172 417 249 28 × 2 = 0 + 0.344 834 498 56;
  • 45) 0.344 834 498 56 × 2 = 0 + 0.689 668 997 12;
  • 46) 0.689 668 997 12 × 2 = 1 + 0.379 337 994 24;
  • 47) 0.379 337 994 24 × 2 = 0 + 0.758 675 988 48;
  • 48) 0.758 675 988 48 × 2 = 1 + 0.517 351 976 96;
  • 49) 0.517 351 976 96 × 2 = 1 + 0.034 703 953 92;
  • 50) 0.034 703 953 92 × 2 = 0 + 0.069 407 907 84;
  • 51) 0.069 407 907 84 × 2 = 0 + 0.138 815 815 68;
  • 52) 0.138 815 815 68 × 2 = 0 + 0.277 631 631 36;
  • 53) 0.277 631 631 36 × 2 = 0 + 0.555 263 262 72;
  • 54) 0.555 263 262 72 × 2 = 1 + 0.110 526 525 44;
  • 55) 0.110 526 525 44 × 2 = 0 + 0.221 053 050 88;
  • 56) 0.221 053 050 88 × 2 = 0 + 0.442 106 101 76;
  • 57) 0.442 106 101 76 × 2 = 0 + 0.884 212 203 52;
  • 58) 0.884 212 203 52 × 2 = 1 + 0.768 424 407 04;
  • 59) 0.768 424 407 04 × 2 = 1 + 0.536 848 814 08;
  • 60) 0.536 848 814 08 × 2 = 1 + 0.073 697 628 16;
  • 61) 0.073 697 628 16 × 2 = 0 + 0.147 395 256 32;
  • 62) 0.147 395 256 32 × 2 = 0 + 0.294 790 512 64;
  • 63) 0.294 790 512 64 × 2 = 0 + 0.589 581 025 28;
  • 64) 0.589 581 025 28 × 2 = 1 + 0.179 162 050 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:


0.000 282 005 91(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1001 0010 0101 1000 0100 0111 0001(2)

6. Positive number before normalization:

0.000 282 005 91(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1001 0010 0101 1000 0100 0111 0001(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 91(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1001 0010 0101 1000 0100 0111 0001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1001 0010 0101 1000 0100 0111 0001(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1001 0010 0101 1000 0100 0111 0001(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 0010 1001 0010 0101 1000 0100 0111 0001


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 0010 1001 0010 0101 1000 0100 0111 0001 =

0010 0111 1011 0100 0110 0010 1001 0010 0101 1000 0100 0111 0001


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 0010 1001 0010 0101 1000 0100 0111 0001


Decimal number -0.000 282 005 91 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1001 0010 0101 1000 0100 0111 0001

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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