-0.008 788 423 614 84 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 614 84(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
-0.008 788 423 614 84(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:

|-0.008 788 423 614 84| = 0.008 788 423 614 84


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: 0.008 788 423 614 84.

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) 0.008 788 423 614 84 × 2 = 0 + 0.017 576 847 229 68;
  • 2) 0.017 576 847 229 68 × 2 = 0 + 0.035 153 694 459 36;
  • 3) 0.035 153 694 459 36 × 2 = 0 + 0.070 307 388 918 72;
  • 4) 0.070 307 388 918 72 × 2 = 0 + 0.140 614 777 837 44;
  • 5) 0.140 614 777 837 44 × 2 = 0 + 0.281 229 555 674 88;
  • 6) 0.281 229 555 674 88 × 2 = 0 + 0.562 459 111 349 76;
  • 7) 0.562 459 111 349 76 × 2 = 1 + 0.124 918 222 699 52;
  • 8) 0.124 918 222 699 52 × 2 = 0 + 0.249 836 445 399 04;
  • 9) 0.249 836 445 399 04 × 2 = 0 + 0.499 672 890 798 08;
  • 10) 0.499 672 890 798 08 × 2 = 0 + 0.999 345 781 596 16;
  • 11) 0.999 345 781 596 16 × 2 = 1 + 0.998 691 563 192 32;
  • 12) 0.998 691 563 192 32 × 2 = 1 + 0.997 383 126 384 64;
  • 13) 0.997 383 126 384 64 × 2 = 1 + 0.994 766 252 769 28;
  • 14) 0.994 766 252 769 28 × 2 = 1 + 0.989 532 505 538 56;
  • 15) 0.989 532 505 538 56 × 2 = 1 + 0.979 065 011 077 12;
  • 16) 0.979 065 011 077 12 × 2 = 1 + 0.958 130 022 154 24;
  • 17) 0.958 130 022 154 24 × 2 = 1 + 0.916 260 044 308 48;
  • 18) 0.916 260 044 308 48 × 2 = 1 + 0.832 520 088 616 96;
  • 19) 0.832 520 088 616 96 × 2 = 1 + 0.665 040 177 233 92;
  • 20) 0.665 040 177 233 92 × 2 = 1 + 0.330 080 354 467 84;
  • 21) 0.330 080 354 467 84 × 2 = 0 + 0.660 160 708 935 68;
  • 22) 0.660 160 708 935 68 × 2 = 1 + 0.320 321 417 871 36;
  • 23) 0.320 321 417 871 36 × 2 = 0 + 0.640 642 835 742 72;
  • 24) 0.640 642 835 742 72 × 2 = 1 + 0.281 285 671 485 44;
  • 25) 0.281 285 671 485 44 × 2 = 0 + 0.562 571 342 970 88;
  • 26) 0.562 571 342 970 88 × 2 = 1 + 0.125 142 685 941 76;
  • 27) 0.125 142 685 941 76 × 2 = 0 + 0.250 285 371 883 52;
  • 28) 0.250 285 371 883 52 × 2 = 0 + 0.500 570 743 767 04;
  • 29) 0.500 570 743 767 04 × 2 = 1 + 0.001 141 487 534 08;
  • 30) 0.001 141 487 534 08 × 2 = 0 + 0.002 282 975 068 16;
  • 31) 0.002 282 975 068 16 × 2 = 0 + 0.004 565 950 136 32;
  • 32) 0.004 565 950 136 32 × 2 = 0 + 0.009 131 900 272 64;
  • 33) 0.009 131 900 272 64 × 2 = 0 + 0.018 263 800 545 28;
  • 34) 0.018 263 800 545 28 × 2 = 0 + 0.036 527 601 090 56;
  • 35) 0.036 527 601 090 56 × 2 = 0 + 0.073 055 202 181 12;
  • 36) 0.073 055 202 181 12 × 2 = 0 + 0.146 110 404 362 24;
  • 37) 0.146 110 404 362 24 × 2 = 0 + 0.292 220 808 724 48;
  • 38) 0.292 220 808 724 48 × 2 = 0 + 0.584 441 617 448 96;
  • 39) 0.584 441 617 448 96 × 2 = 1 + 0.168 883 234 897 92;
  • 40) 0.168 883 234 897 92 × 2 = 0 + 0.337 766 469 795 84;
  • 41) 0.337 766 469 795 84 × 2 = 0 + 0.675 532 939 591 68;
  • 42) 0.675 532 939 591 68 × 2 = 1 + 0.351 065 879 183 36;
  • 43) 0.351 065 879 183 36 × 2 = 0 + 0.702 131 758 366 72;
  • 44) 0.702 131 758 366 72 × 2 = 1 + 0.404 263 516 733 44;
  • 45) 0.404 263 516 733 44 × 2 = 0 + 0.808 527 033 466 88;
  • 46) 0.808 527 033 466 88 × 2 = 1 + 0.617 054 066 933 76;
  • 47) 0.617 054 066 933 76 × 2 = 1 + 0.234 108 133 867 52;
  • 48) 0.234 108 133 867 52 × 2 = 0 + 0.468 216 267 735 04;
  • 49) 0.468 216 267 735 04 × 2 = 0 + 0.936 432 535 470 08;
  • 50) 0.936 432 535 470 08 × 2 = 1 + 0.872 865 070 940 16;
  • 51) 0.872 865 070 940 16 × 2 = 1 + 0.745 730 141 880 32;
  • 52) 0.745 730 141 880 32 × 2 = 1 + 0.491 460 283 760 64;
  • 53) 0.491 460 283 760 64 × 2 = 0 + 0.982 920 567 521 28;
  • 54) 0.982 920 567 521 28 × 2 = 1 + 0.965 841 135 042 56;
  • 55) 0.965 841 135 042 56 × 2 = 1 + 0.931 682 270 085 12;
  • 56) 0.931 682 270 085 12 × 2 = 1 + 0.863 364 540 170 24;
  • 57) 0.863 364 540 170 24 × 2 = 1 + 0.726 729 080 340 48;
  • 58) 0.726 729 080 340 48 × 2 = 1 + 0.453 458 160 680 96;
  • 59) 0.453 458 160 680 96 × 2 = 0 + 0.906 916 321 361 92;

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:


0.008 788 423 614 84(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0010 0101 0110 0111 0111 110(2)

6. Positive number before normalization:

0.008 788 423 614 84(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0010 0101 0110 0111 0111 110(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:


0.008 788 423 614 84(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0010 0101 0110 0111 0111 110(2) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0010 0101 0110 0111 0111 110(2) × 20 =

1.0001 1111 1111 1010 1010 0100 0000 0001 0010 1011 0011 1011 1110(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 0001 0010 1011 0011 1011 1110


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 0001 0010 1011 0011 1011 1110 =

0001 1111 1111 1010 1010 0100 0000 0001 0010 1011 0011 1011 1110


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 0001 0010 1011 0011 1011 1110


Decimal number -0.008 788 423 614 84 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0001 1111 1111 1010 1010 0100 0000 0001 0010 1011 0011 1011 1110

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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