-0.008 788 423 613 06 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 613 06(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
-0.008 788 423 613 06(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:

|-0.008 788 423 613 06| = 0.008 788 423 613 06


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: 0.008 788 423 613 06.

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) 0.008 788 423 613 06 × 2 = 0 + 0.017 576 847 226 12;
  • 2) 0.017 576 847 226 12 × 2 = 0 + 0.035 153 694 452 24;
  • 3) 0.035 153 694 452 24 × 2 = 0 + 0.070 307 388 904 48;
  • 4) 0.070 307 388 904 48 × 2 = 0 + 0.140 614 777 808 96;
  • 5) 0.140 614 777 808 96 × 2 = 0 + 0.281 229 555 617 92;
  • 6) 0.281 229 555 617 92 × 2 = 0 + 0.562 459 111 235 84;
  • 7) 0.562 459 111 235 84 × 2 = 1 + 0.124 918 222 471 68;
  • 8) 0.124 918 222 471 68 × 2 = 0 + 0.249 836 444 943 36;
  • 9) 0.249 836 444 943 36 × 2 = 0 + 0.499 672 889 886 72;
  • 10) 0.499 672 889 886 72 × 2 = 0 + 0.999 345 779 773 44;
  • 11) 0.999 345 779 773 44 × 2 = 1 + 0.998 691 559 546 88;
  • 12) 0.998 691 559 546 88 × 2 = 1 + 0.997 383 119 093 76;
  • 13) 0.997 383 119 093 76 × 2 = 1 + 0.994 766 238 187 52;
  • 14) 0.994 766 238 187 52 × 2 = 1 + 0.989 532 476 375 04;
  • 15) 0.989 532 476 375 04 × 2 = 1 + 0.979 064 952 750 08;
  • 16) 0.979 064 952 750 08 × 2 = 1 + 0.958 129 905 500 16;
  • 17) 0.958 129 905 500 16 × 2 = 1 + 0.916 259 811 000 32;
  • 18) 0.916 259 811 000 32 × 2 = 1 + 0.832 519 622 000 64;
  • 19) 0.832 519 622 000 64 × 2 = 1 + 0.665 039 244 001 28;
  • 20) 0.665 039 244 001 28 × 2 = 1 + 0.330 078 488 002 56;
  • 21) 0.330 078 488 002 56 × 2 = 0 + 0.660 156 976 005 12;
  • 22) 0.660 156 976 005 12 × 2 = 1 + 0.320 313 952 010 24;
  • 23) 0.320 313 952 010 24 × 2 = 0 + 0.640 627 904 020 48;
  • 24) 0.640 627 904 020 48 × 2 = 1 + 0.281 255 808 040 96;
  • 25) 0.281 255 808 040 96 × 2 = 0 + 0.562 511 616 081 92;
  • 26) 0.562 511 616 081 92 × 2 = 1 + 0.125 023 232 163 84;
  • 27) 0.125 023 232 163 84 × 2 = 0 + 0.250 046 464 327 68;
  • 28) 0.250 046 464 327 68 × 2 = 0 + 0.500 092 928 655 36;
  • 29) 0.500 092 928 655 36 × 2 = 1 + 0.000 185 857 310 72;
  • 30) 0.000 185 857 310 72 × 2 = 0 + 0.000 371 714 621 44;
  • 31) 0.000 371 714 621 44 × 2 = 0 + 0.000 743 429 242 88;
  • 32) 0.000 743 429 242 88 × 2 = 0 + 0.001 486 858 485 76;
  • 33) 0.001 486 858 485 76 × 2 = 0 + 0.002 973 716 971 52;
  • 34) 0.002 973 716 971 52 × 2 = 0 + 0.005 947 433 943 04;
  • 35) 0.005 947 433 943 04 × 2 = 0 + 0.011 894 867 886 08;
  • 36) 0.011 894 867 886 08 × 2 = 0 + 0.023 789 735 772 16;
  • 37) 0.023 789 735 772 16 × 2 = 0 + 0.047 579 471 544 32;
  • 38) 0.047 579 471 544 32 × 2 = 0 + 0.095 158 943 088 64;
  • 39) 0.095 158 943 088 64 × 2 = 0 + 0.190 317 886 177 28;
  • 40) 0.190 317 886 177 28 × 2 = 0 + 0.380 635 772 354 56;
  • 41) 0.380 635 772 354 56 × 2 = 0 + 0.761 271 544 709 12;
  • 42) 0.761 271 544 709 12 × 2 = 1 + 0.522 543 089 418 24;
  • 43) 0.522 543 089 418 24 × 2 = 1 + 0.045 086 178 836 48;
  • 44) 0.045 086 178 836 48 × 2 = 0 + 0.090 172 357 672 96;
  • 45) 0.090 172 357 672 96 × 2 = 0 + 0.180 344 715 345 92;
  • 46) 0.180 344 715 345 92 × 2 = 0 + 0.360 689 430 691 84;
  • 47) 0.360 689 430 691 84 × 2 = 0 + 0.721 378 861 383 68;
  • 48) 0.721 378 861 383 68 × 2 = 1 + 0.442 757 722 767 36;
  • 49) 0.442 757 722 767 36 × 2 = 0 + 0.885 515 445 534 72;
  • 50) 0.885 515 445 534 72 × 2 = 1 + 0.771 030 891 069 44;
  • 51) 0.771 030 891 069 44 × 2 = 1 + 0.542 061 782 138 88;
  • 52) 0.542 061 782 138 88 × 2 = 1 + 0.084 123 564 277 76;
  • 53) 0.084 123 564 277 76 × 2 = 0 + 0.168 247 128 555 52;
  • 54) 0.168 247 128 555 52 × 2 = 0 + 0.336 494 257 111 04;
  • 55) 0.336 494 257 111 04 × 2 = 0 + 0.672 988 514 222 08;
  • 56) 0.672 988 514 222 08 × 2 = 1 + 0.345 977 028 444 16;
  • 57) 0.345 977 028 444 16 × 2 = 0 + 0.691 954 056 888 32;
  • 58) 0.691 954 056 888 32 × 2 = 1 + 0.383 908 113 776 64;
  • 59) 0.383 908 113 776 64 × 2 = 0 + 0.767 816 227 553 28;

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:


0.008 788 423 613 06(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0110 0001 0111 0001 010(2)

6. Positive number before normalization:

0.008 788 423 613 06(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0110 0001 0111 0001 010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right, so that only one non zero digit remains to the left of it:


0.008 788 423 613 06(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0110 0001 0111 0001 010(2) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0110 0001 0111 0001 010(2) × 20 =

1.0001 1111 1111 1010 1010 0100 0000 0000 0011 0000 1011 1000 1010(2) × 2-7


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): -7

Mantissa (not normalized):
1.0001 1111 1111 1010 1010 0100 0000 0000 0011 0000 1011 1000 1010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-7 + 2(11-1) - 1 =

(-7 + 1 023)(10) =

1 016(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;
  • 1 016 ÷ 2 = 508 + 0;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1016(10) =

011 1111 1000(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. 0001 1111 1111 1010 1010 0100 0000 0000 0011 0000 1011 1000 1010 =

0001 1111 1111 1010 1010 0100 0000 0000 0011 0000 1011 1000 1010


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 1000

Mantissa (52 bits) =
0001 1111 1111 1010 1010 0100 0000 0000 0011 0000 1011 1000 1010


Decimal number -0.008 788 423 613 06 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0001 1111 1111 1010 1010 0100 0000 0000 0011 0000 1011 1000 1010

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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