-0.008 788 423 612 981 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 612 981(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
-0.008 788 423 612 981(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:

|-0.008 788 423 612 981| = 0.008 788 423 612 981


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: 0.008 788 423 612 981.

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) 0.008 788 423 612 981 × 2 = 0 + 0.017 576 847 225 962;
  • 2) 0.017 576 847 225 962 × 2 = 0 + 0.035 153 694 451 924;
  • 3) 0.035 153 694 451 924 × 2 = 0 + 0.070 307 388 903 848;
  • 4) 0.070 307 388 903 848 × 2 = 0 + 0.140 614 777 807 696;
  • 5) 0.140 614 777 807 696 × 2 = 0 + 0.281 229 555 615 392;
  • 6) 0.281 229 555 615 392 × 2 = 0 + 0.562 459 111 230 784;
  • 7) 0.562 459 111 230 784 × 2 = 1 + 0.124 918 222 461 568;
  • 8) 0.124 918 222 461 568 × 2 = 0 + 0.249 836 444 923 136;
  • 9) 0.249 836 444 923 136 × 2 = 0 + 0.499 672 889 846 272;
  • 10) 0.499 672 889 846 272 × 2 = 0 + 0.999 345 779 692 544;
  • 11) 0.999 345 779 692 544 × 2 = 1 + 0.998 691 559 385 088;
  • 12) 0.998 691 559 385 088 × 2 = 1 + 0.997 383 118 770 176;
  • 13) 0.997 383 118 770 176 × 2 = 1 + 0.994 766 237 540 352;
  • 14) 0.994 766 237 540 352 × 2 = 1 + 0.989 532 475 080 704;
  • 15) 0.989 532 475 080 704 × 2 = 1 + 0.979 064 950 161 408;
  • 16) 0.979 064 950 161 408 × 2 = 1 + 0.958 129 900 322 816;
  • 17) 0.958 129 900 322 816 × 2 = 1 + 0.916 259 800 645 632;
  • 18) 0.916 259 800 645 632 × 2 = 1 + 0.832 519 601 291 264;
  • 19) 0.832 519 601 291 264 × 2 = 1 + 0.665 039 202 582 528;
  • 20) 0.665 039 202 582 528 × 2 = 1 + 0.330 078 405 165 056;
  • 21) 0.330 078 405 165 056 × 2 = 0 + 0.660 156 810 330 112;
  • 22) 0.660 156 810 330 112 × 2 = 1 + 0.320 313 620 660 224;
  • 23) 0.320 313 620 660 224 × 2 = 0 + 0.640 627 241 320 448;
  • 24) 0.640 627 241 320 448 × 2 = 1 + 0.281 254 482 640 896;
  • 25) 0.281 254 482 640 896 × 2 = 0 + 0.562 508 965 281 792;
  • 26) 0.562 508 965 281 792 × 2 = 1 + 0.125 017 930 563 584;
  • 27) 0.125 017 930 563 584 × 2 = 0 + 0.250 035 861 127 168;
  • 28) 0.250 035 861 127 168 × 2 = 0 + 0.500 071 722 254 336;
  • 29) 0.500 071 722 254 336 × 2 = 1 + 0.000 143 444 508 672;
  • 30) 0.000 143 444 508 672 × 2 = 0 + 0.000 286 889 017 344;
  • 31) 0.000 286 889 017 344 × 2 = 0 + 0.000 573 778 034 688;
  • 32) 0.000 573 778 034 688 × 2 = 0 + 0.001 147 556 069 376;
  • 33) 0.001 147 556 069 376 × 2 = 0 + 0.002 295 112 138 752;
  • 34) 0.002 295 112 138 752 × 2 = 0 + 0.004 590 224 277 504;
  • 35) 0.004 590 224 277 504 × 2 = 0 + 0.009 180 448 555 008;
  • 36) 0.009 180 448 555 008 × 2 = 0 + 0.018 360 897 110 016;
  • 37) 0.018 360 897 110 016 × 2 = 0 + 0.036 721 794 220 032;
  • 38) 0.036 721 794 220 032 × 2 = 0 + 0.073 443 588 440 064;
  • 39) 0.073 443 588 440 064 × 2 = 0 + 0.146 887 176 880 128;
  • 40) 0.146 887 176 880 128 × 2 = 0 + 0.293 774 353 760 256;
  • 41) 0.293 774 353 760 256 × 2 = 0 + 0.587 548 707 520 512;
  • 42) 0.587 548 707 520 512 × 2 = 1 + 0.175 097 415 041 024;
  • 43) 0.175 097 415 041 024 × 2 = 0 + 0.350 194 830 082 048;
  • 44) 0.350 194 830 082 048 × 2 = 0 + 0.700 389 660 164 096;
  • 45) 0.700 389 660 164 096 × 2 = 1 + 0.400 779 320 328 192;
  • 46) 0.400 779 320 328 192 × 2 = 0 + 0.801 558 640 656 384;
  • 47) 0.801 558 640 656 384 × 2 = 1 + 0.603 117 281 312 768;
  • 48) 0.603 117 281 312 768 × 2 = 1 + 0.206 234 562 625 536;
  • 49) 0.206 234 562 625 536 × 2 = 0 + 0.412 469 125 251 072;
  • 50) 0.412 469 125 251 072 × 2 = 0 + 0.824 938 250 502 144;
  • 51) 0.824 938 250 502 144 × 2 = 1 + 0.649 876 501 004 288;
  • 52) 0.649 876 501 004 288 × 2 = 1 + 0.299 753 002 008 576;
  • 53) 0.299 753 002 008 576 × 2 = 0 + 0.599 506 004 017 152;
  • 54) 0.599 506 004 017 152 × 2 = 1 + 0.199 012 008 034 304;
  • 55) 0.199 012 008 034 304 × 2 = 0 + 0.398 024 016 068 608;
  • 56) 0.398 024 016 068 608 × 2 = 0 + 0.796 048 032 137 216;
  • 57) 0.796 048 032 137 216 × 2 = 1 + 0.592 096 064 274 432;
  • 58) 0.592 096 064 274 432 × 2 = 1 + 0.184 192 128 548 864;
  • 59) 0.184 192 128 548 864 × 2 = 0 + 0.368 384 257 097 728;

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:


0.008 788 423 612 981(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0100 1011 0011 0100 110(2)

6. Positive number before normalization:

0.008 788 423 612 981(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0100 1011 0011 0100 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:


0.008 788 423 612 981(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0100 1011 0011 0100 110(2) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0100 1011 0011 0100 110(2) × 20 =

1.0001 1111 1111 1010 1010 0100 0000 0000 0010 0101 1001 1010 0110(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 0010 0101 1001 1010 0110


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 0010 0101 1001 1010 0110 =

0001 1111 1111 1010 1010 0100 0000 0000 0010 0101 1001 1010 0110


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 0010 0101 1001 1010 0110


Decimal number -0.008 788 423 612 981 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0001 1111 1111 1010 1010 0100 0000 0000 0010 0101 1001 1010 0110

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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