-0.000 282 005 908 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 908 7(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
-0.000 282 005 908 7(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:

|-0.000 282 005 908 7| = 0.000 282 005 908 7


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: 0.000 282 005 908 7.

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) 0.000 282 005 908 7 × 2 = 0 + 0.000 564 011 817 4;
  • 2) 0.000 564 011 817 4 × 2 = 0 + 0.001 128 023 634 8;
  • 3) 0.001 128 023 634 8 × 2 = 0 + 0.002 256 047 269 6;
  • 4) 0.002 256 047 269 6 × 2 = 0 + 0.004 512 094 539 2;
  • 5) 0.004 512 094 539 2 × 2 = 0 + 0.009 024 189 078 4;
  • 6) 0.009 024 189 078 4 × 2 = 0 + 0.018 048 378 156 8;
  • 7) 0.018 048 378 156 8 × 2 = 0 + 0.036 096 756 313 6;
  • 8) 0.036 096 756 313 6 × 2 = 0 + 0.072 193 512 627 2;
  • 9) 0.072 193 512 627 2 × 2 = 0 + 0.144 387 025 254 4;
  • 10) 0.144 387 025 254 4 × 2 = 0 + 0.288 774 050 508 8;
  • 11) 0.288 774 050 508 8 × 2 = 0 + 0.577 548 101 017 6;
  • 12) 0.577 548 101 017 6 × 2 = 1 + 0.155 096 202 035 2;
  • 13) 0.155 096 202 035 2 × 2 = 0 + 0.310 192 404 070 4;
  • 14) 0.310 192 404 070 4 × 2 = 0 + 0.620 384 808 140 8;
  • 15) 0.620 384 808 140 8 × 2 = 1 + 0.240 769 616 281 6;
  • 16) 0.240 769 616 281 6 × 2 = 0 + 0.481 539 232 563 2;
  • 17) 0.481 539 232 563 2 × 2 = 0 + 0.963 078 465 126 4;
  • 18) 0.963 078 465 126 4 × 2 = 1 + 0.926 156 930 252 8;
  • 19) 0.926 156 930 252 8 × 2 = 1 + 0.852 313 860 505 6;
  • 20) 0.852 313 860 505 6 × 2 = 1 + 0.704 627 721 011 2;
  • 21) 0.704 627 721 011 2 × 2 = 1 + 0.409 255 442 022 4;
  • 22) 0.409 255 442 022 4 × 2 = 0 + 0.818 510 884 044 8;
  • 23) 0.818 510 884 044 8 × 2 = 1 + 0.637 021 768 089 6;
  • 24) 0.637 021 768 089 6 × 2 = 1 + 0.274 043 536 179 2;
  • 25) 0.274 043 536 179 2 × 2 = 0 + 0.548 087 072 358 4;
  • 26) 0.548 087 072 358 4 × 2 = 1 + 0.096 174 144 716 8;
  • 27) 0.096 174 144 716 8 × 2 = 0 + 0.192 348 289 433 6;
  • 28) 0.192 348 289 433 6 × 2 = 0 + 0.384 696 578 867 2;
  • 29) 0.384 696 578 867 2 × 2 = 0 + 0.769 393 157 734 4;
  • 30) 0.769 393 157 734 4 × 2 = 1 + 0.538 786 315 468 8;
  • 31) 0.538 786 315 468 8 × 2 = 1 + 0.077 572 630 937 6;
  • 32) 0.077 572 630 937 6 × 2 = 0 + 0.155 145 261 875 2;
  • 33) 0.155 145 261 875 2 × 2 = 0 + 0.310 290 523 750 4;
  • 34) 0.310 290 523 750 4 × 2 = 0 + 0.620 581 047 500 8;
  • 35) 0.620 581 047 500 8 × 2 = 1 + 0.241 162 095 001 6;
  • 36) 0.241 162 095 001 6 × 2 = 0 + 0.482 324 190 003 2;
  • 37) 0.482 324 190 003 2 × 2 = 0 + 0.964 648 380 006 4;
  • 38) 0.964 648 380 006 4 × 2 = 1 + 0.929 296 760 012 8;
  • 39) 0.929 296 760 012 8 × 2 = 1 + 0.858 593 520 025 6;
  • 40) 0.858 593 520 025 6 × 2 = 1 + 0.717 187 040 051 2;
  • 41) 0.717 187 040 051 2 × 2 = 1 + 0.434 374 080 102 4;
  • 42) 0.434 374 080 102 4 × 2 = 0 + 0.868 748 160 204 8;
  • 43) 0.868 748 160 204 8 × 2 = 1 + 0.737 496 320 409 6;
  • 44) 0.737 496 320 409 6 × 2 = 1 + 0.474 992 640 819 2;
  • 45) 0.474 992 640 819 2 × 2 = 0 + 0.949 985 281 638 4;
  • 46) 0.949 985 281 638 4 × 2 = 1 + 0.899 970 563 276 8;
  • 47) 0.899 970 563 276 8 × 2 = 1 + 0.799 941 126 553 6;
  • 48) 0.799 941 126 553 6 × 2 = 1 + 0.599 882 253 107 2;
  • 49) 0.599 882 253 107 2 × 2 = 1 + 0.199 764 506 214 4;
  • 50) 0.199 764 506 214 4 × 2 = 0 + 0.399 529 012 428 8;
  • 51) 0.399 529 012 428 8 × 2 = 0 + 0.799 058 024 857 6;
  • 52) 0.799 058 024 857 6 × 2 = 1 + 0.598 116 049 715 2;
  • 53) 0.598 116 049 715 2 × 2 = 1 + 0.196 232 099 430 4;
  • 54) 0.196 232 099 430 4 × 2 = 0 + 0.392 464 198 860 8;
  • 55) 0.392 464 198 860 8 × 2 = 0 + 0.784 928 397 721 6;
  • 56) 0.784 928 397 721 6 × 2 = 1 + 0.569 856 795 443 2;
  • 57) 0.569 856 795 443 2 × 2 = 1 + 0.139 713 590 886 4;
  • 58) 0.139 713 590 886 4 × 2 = 0 + 0.279 427 181 772 8;
  • 59) 0.279 427 181 772 8 × 2 = 0 + 0.558 854 363 545 6;
  • 60) 0.558 854 363 545 6 × 2 = 1 + 0.117 708 727 091 2;
  • 61) 0.117 708 727 091 2 × 2 = 0 + 0.235 417 454 182 4;
  • 62) 0.235 417 454 182 4 × 2 = 0 + 0.470 834 908 364 8;
  • 63) 0.470 834 908 364 8 × 2 = 0 + 0.941 669 816 729 6;
  • 64) 0.941 669 816 729 6 × 2 = 1 + 0.883 339 633 459 2;

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:


0.000 282 005 908 7(10) =

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

6. Positive number before normalization:

0.000 282 005 908 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0111 1011 0111 1001 1001 1001 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:


0.000 282 005 908 7(10) =

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

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

1.0010 0111 1011 0100 0110 0010 0111 1011 0111 1001 1001 1001 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 0111 1011 0111 1001 1001 1001 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 0111 1011 0111 1001 1001 1001 0001 =

0010 0111 1011 0100 0110 0010 0111 1011 0111 1001 1001 1001 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 0111 1011 0111 1001 1001 1001 0001


Decimal number -0.000 282 005 908 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 0111 1011 0111 1001 1001 1001 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