-0.008 788 423 614 81 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 614 81(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 81(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 81| = 0.008 788 423 614 81


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 81.

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 81 × 2 = 0 + 0.017 576 847 229 62;
  • 2) 0.017 576 847 229 62 × 2 = 0 + 0.035 153 694 459 24;
  • 3) 0.035 153 694 459 24 × 2 = 0 + 0.070 307 388 918 48;
  • 4) 0.070 307 388 918 48 × 2 = 0 + 0.140 614 777 836 96;
  • 5) 0.140 614 777 836 96 × 2 = 0 + 0.281 229 555 673 92;
  • 6) 0.281 229 555 673 92 × 2 = 0 + 0.562 459 111 347 84;
  • 7) 0.562 459 111 347 84 × 2 = 1 + 0.124 918 222 695 68;
  • 8) 0.124 918 222 695 68 × 2 = 0 + 0.249 836 445 391 36;
  • 9) 0.249 836 445 391 36 × 2 = 0 + 0.499 672 890 782 72;
  • 10) 0.499 672 890 782 72 × 2 = 0 + 0.999 345 781 565 44;
  • 11) 0.999 345 781 565 44 × 2 = 1 + 0.998 691 563 130 88;
  • 12) 0.998 691 563 130 88 × 2 = 1 + 0.997 383 126 261 76;
  • 13) 0.997 383 126 261 76 × 2 = 1 + 0.994 766 252 523 52;
  • 14) 0.994 766 252 523 52 × 2 = 1 + 0.989 532 505 047 04;
  • 15) 0.989 532 505 047 04 × 2 = 1 + 0.979 065 010 094 08;
  • 16) 0.979 065 010 094 08 × 2 = 1 + 0.958 130 020 188 16;
  • 17) 0.958 130 020 188 16 × 2 = 1 + 0.916 260 040 376 32;
  • 18) 0.916 260 040 376 32 × 2 = 1 + 0.832 520 080 752 64;
  • 19) 0.832 520 080 752 64 × 2 = 1 + 0.665 040 161 505 28;
  • 20) 0.665 040 161 505 28 × 2 = 1 + 0.330 080 323 010 56;
  • 21) 0.330 080 323 010 56 × 2 = 0 + 0.660 160 646 021 12;
  • 22) 0.660 160 646 021 12 × 2 = 1 + 0.320 321 292 042 24;
  • 23) 0.320 321 292 042 24 × 2 = 0 + 0.640 642 584 084 48;
  • 24) 0.640 642 584 084 48 × 2 = 1 + 0.281 285 168 168 96;
  • 25) 0.281 285 168 168 96 × 2 = 0 + 0.562 570 336 337 92;
  • 26) 0.562 570 336 337 92 × 2 = 1 + 0.125 140 672 675 84;
  • 27) 0.125 140 672 675 84 × 2 = 0 + 0.250 281 345 351 68;
  • 28) 0.250 281 345 351 68 × 2 = 0 + 0.500 562 690 703 36;
  • 29) 0.500 562 690 703 36 × 2 = 1 + 0.001 125 381 406 72;
  • 30) 0.001 125 381 406 72 × 2 = 0 + 0.002 250 762 813 44;
  • 31) 0.002 250 762 813 44 × 2 = 0 + 0.004 501 525 626 88;
  • 32) 0.004 501 525 626 88 × 2 = 0 + 0.009 003 051 253 76;
  • 33) 0.009 003 051 253 76 × 2 = 0 + 0.018 006 102 507 52;
  • 34) 0.018 006 102 507 52 × 2 = 0 + 0.036 012 205 015 04;
  • 35) 0.036 012 205 015 04 × 2 = 0 + 0.072 024 410 030 08;
  • 36) 0.072 024 410 030 08 × 2 = 0 + 0.144 048 820 060 16;
  • 37) 0.144 048 820 060 16 × 2 = 0 + 0.288 097 640 120 32;
  • 38) 0.288 097 640 120 32 × 2 = 0 + 0.576 195 280 240 64;
  • 39) 0.576 195 280 240 64 × 2 = 1 + 0.152 390 560 481 28;
  • 40) 0.152 390 560 481 28 × 2 = 0 + 0.304 781 120 962 56;
  • 41) 0.304 781 120 962 56 × 2 = 0 + 0.609 562 241 925 12;
  • 42) 0.609 562 241 925 12 × 2 = 1 + 0.219 124 483 850 24;
  • 43) 0.219 124 483 850 24 × 2 = 0 + 0.438 248 967 700 48;
  • 44) 0.438 248 967 700 48 × 2 = 0 + 0.876 497 935 400 96;
  • 45) 0.876 497 935 400 96 × 2 = 1 + 0.752 995 870 801 92;
  • 46) 0.752 995 870 801 92 × 2 = 1 + 0.505 991 741 603 84;
  • 47) 0.505 991 741 603 84 × 2 = 1 + 0.011 983 483 207 68;
  • 48) 0.011 983 483 207 68 × 2 = 0 + 0.023 966 966 415 36;
  • 49) 0.023 966 966 415 36 × 2 = 0 + 0.047 933 932 830 72;
  • 50) 0.047 933 932 830 72 × 2 = 0 + 0.095 867 865 661 44;
  • 51) 0.095 867 865 661 44 × 2 = 0 + 0.191 735 731 322 88;
  • 52) 0.191 735 731 322 88 × 2 = 0 + 0.383 471 462 645 76;
  • 53) 0.383 471 462 645 76 × 2 = 0 + 0.766 942 925 291 52;
  • 54) 0.766 942 925 291 52 × 2 = 1 + 0.533 885 850 583 04;
  • 55) 0.533 885 850 583 04 × 2 = 1 + 0.067 771 701 166 08;
  • 56) 0.067 771 701 166 08 × 2 = 0 + 0.135 543 402 332 16;
  • 57) 0.135 543 402 332 16 × 2 = 0 + 0.271 086 804 664 32;
  • 58) 0.271 086 804 664 32 × 2 = 0 + 0.542 173 609 328 64;
  • 59) 0.542 173 609 328 64 × 2 = 1 + 0.084 347 218 657 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:


0.008 788 423 614 81(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0010 0100 1110 0000 0110 001(2)

6. Positive number before normalization:

0.008 788 423 614 81(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0010 0100 1110 0000 0110 001(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 81(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0010 0100 1110 0000 0110 001(2) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0010 0100 1110 0000 0110 001(2) × 20 =

1.0001 1111 1111 1010 1010 0100 0000 0001 0010 0111 0000 0011 0001(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 0111 0000 0011 0001


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 0111 0000 0011 0001 =

0001 1111 1111 1010 1010 0100 0000 0001 0010 0111 0000 0011 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 1000

Mantissa (52 bits) =
0001 1111 1111 1010 1010 0100 0000 0001 0010 0111 0000 0011 0001


Decimal number -0.008 788 423 614 81 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 0111 0000 0011 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