-0.008 788 423 612 69 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 612 69(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
-0.008 788 423 612 69(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:

|-0.008 788 423 612 69| = 0.008 788 423 612 69


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: 0.008 788 423 612 69.

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) 0.008 788 423 612 69 × 2 = 0 + 0.017 576 847 225 38;
  • 2) 0.017 576 847 225 38 × 2 = 0 + 0.035 153 694 450 76;
  • 3) 0.035 153 694 450 76 × 2 = 0 + 0.070 307 388 901 52;
  • 4) 0.070 307 388 901 52 × 2 = 0 + 0.140 614 777 803 04;
  • 5) 0.140 614 777 803 04 × 2 = 0 + 0.281 229 555 606 08;
  • 6) 0.281 229 555 606 08 × 2 = 0 + 0.562 459 111 212 16;
  • 7) 0.562 459 111 212 16 × 2 = 1 + 0.124 918 222 424 32;
  • 8) 0.124 918 222 424 32 × 2 = 0 + 0.249 836 444 848 64;
  • 9) 0.249 836 444 848 64 × 2 = 0 + 0.499 672 889 697 28;
  • 10) 0.499 672 889 697 28 × 2 = 0 + 0.999 345 779 394 56;
  • 11) 0.999 345 779 394 56 × 2 = 1 + 0.998 691 558 789 12;
  • 12) 0.998 691 558 789 12 × 2 = 1 + 0.997 383 117 578 24;
  • 13) 0.997 383 117 578 24 × 2 = 1 + 0.994 766 235 156 48;
  • 14) 0.994 766 235 156 48 × 2 = 1 + 0.989 532 470 312 96;
  • 15) 0.989 532 470 312 96 × 2 = 1 + 0.979 064 940 625 92;
  • 16) 0.979 064 940 625 92 × 2 = 1 + 0.958 129 881 251 84;
  • 17) 0.958 129 881 251 84 × 2 = 1 + 0.916 259 762 503 68;
  • 18) 0.916 259 762 503 68 × 2 = 1 + 0.832 519 525 007 36;
  • 19) 0.832 519 525 007 36 × 2 = 1 + 0.665 039 050 014 72;
  • 20) 0.665 039 050 014 72 × 2 = 1 + 0.330 078 100 029 44;
  • 21) 0.330 078 100 029 44 × 2 = 0 + 0.660 156 200 058 88;
  • 22) 0.660 156 200 058 88 × 2 = 1 + 0.320 312 400 117 76;
  • 23) 0.320 312 400 117 76 × 2 = 0 + 0.640 624 800 235 52;
  • 24) 0.640 624 800 235 52 × 2 = 1 + 0.281 249 600 471 04;
  • 25) 0.281 249 600 471 04 × 2 = 0 + 0.562 499 200 942 08;
  • 26) 0.562 499 200 942 08 × 2 = 1 + 0.124 998 401 884 16;
  • 27) 0.124 998 401 884 16 × 2 = 0 + 0.249 996 803 768 32;
  • 28) 0.249 996 803 768 32 × 2 = 0 + 0.499 993 607 536 64;
  • 29) 0.499 993 607 536 64 × 2 = 0 + 0.999 987 215 073 28;
  • 30) 0.999 987 215 073 28 × 2 = 1 + 0.999 974 430 146 56;
  • 31) 0.999 974 430 146 56 × 2 = 1 + 0.999 948 860 293 12;
  • 32) 0.999 948 860 293 12 × 2 = 1 + 0.999 897 720 586 24;
  • 33) 0.999 897 720 586 24 × 2 = 1 + 0.999 795 441 172 48;
  • 34) 0.999 795 441 172 48 × 2 = 1 + 0.999 590 882 344 96;
  • 35) 0.999 590 882 344 96 × 2 = 1 + 0.999 181 764 689 92;
  • 36) 0.999 181 764 689 92 × 2 = 1 + 0.998 363 529 379 84;
  • 37) 0.998 363 529 379 84 × 2 = 1 + 0.996 727 058 759 68;
  • 38) 0.996 727 058 759 68 × 2 = 1 + 0.993 454 117 519 36;
  • 39) 0.993 454 117 519 36 × 2 = 1 + 0.986 908 235 038 72;
  • 40) 0.986 908 235 038 72 × 2 = 1 + 0.973 816 470 077 44;
  • 41) 0.973 816 470 077 44 × 2 = 1 + 0.947 632 940 154 88;
  • 42) 0.947 632 940 154 88 × 2 = 1 + 0.895 265 880 309 76;
  • 43) 0.895 265 880 309 76 × 2 = 1 + 0.790 531 760 619 52;
  • 44) 0.790 531 760 619 52 × 2 = 1 + 0.581 063 521 239 04;
  • 45) 0.581 063 521 239 04 × 2 = 1 + 0.162 127 042 478 08;
  • 46) 0.162 127 042 478 08 × 2 = 0 + 0.324 254 084 956 16;
  • 47) 0.324 254 084 956 16 × 2 = 0 + 0.648 508 169 912 32;
  • 48) 0.648 508 169 912 32 × 2 = 1 + 0.297 016 339 824 64;
  • 49) 0.297 016 339 824 64 × 2 = 0 + 0.594 032 679 649 28;
  • 50) 0.594 032 679 649 28 × 2 = 1 + 0.188 065 359 298 56;
  • 51) 0.188 065 359 298 56 × 2 = 0 + 0.376 130 718 597 12;
  • 52) 0.376 130 718 597 12 × 2 = 0 + 0.752 261 437 194 24;
  • 53) 0.752 261 437 194 24 × 2 = 1 + 0.504 522 874 388 48;
  • 54) 0.504 522 874 388 48 × 2 = 1 + 0.009 045 748 776 96;
  • 55) 0.009 045 748 776 96 × 2 = 0 + 0.018 091 497 553 92;
  • 56) 0.018 091 497 553 92 × 2 = 0 + 0.036 182 995 107 84;
  • 57) 0.036 182 995 107 84 × 2 = 0 + 0.072 365 990 215 68;
  • 58) 0.072 365 990 215 68 × 2 = 0 + 0.144 731 980 431 36;
  • 59) 0.144 731 980 431 36 × 2 = 0 + 0.289 463 960 862 72;

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:


0.008 788 423 612 69(10) =

0.0000 0010 0011 1111 1111 0101 0100 0111 1111 1111 1111 1001 0100 1100 000(2)

6. Positive number before normalization:

0.008 788 423 612 69(10) =

0.0000 0010 0011 1111 1111 0101 0100 0111 1111 1111 1111 1001 0100 1100 000(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:


0.008 788 423 612 69(10) =

0.0000 0010 0011 1111 1111 0101 0100 0111 1111 1111 1111 1001 0100 1100 000(2) =

0.0000 0010 0011 1111 1111 0101 0100 0111 1111 1111 1111 1001 0100 1100 000(2) × 20 =

1.0001 1111 1111 1010 1010 0011 1111 1111 1111 1100 1010 0110 0000(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 0011 1111 1111 1111 1100 1010 0110 0000


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 0011 1111 1111 1111 1100 1010 0110 0000 =

0001 1111 1111 1010 1010 0011 1111 1111 1111 1100 1010 0110 0000


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 0011 1111 1111 1111 1100 1010 0110 0000


Decimal number -0.008 788 423 612 69 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0001 1111 1111 1010 1010 0011 1111 1111 1111 1100 1010 0110 0000

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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