-0.008 788 423 612 796 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 612 796(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
-0.008 788 423 612 796(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:

|-0.008 788 423 612 796| = 0.008 788 423 612 796


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: 0.008 788 423 612 796.

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) 0.008 788 423 612 796 × 2 = 0 + 0.017 576 847 225 592;
  • 2) 0.017 576 847 225 592 × 2 = 0 + 0.035 153 694 451 184;
  • 3) 0.035 153 694 451 184 × 2 = 0 + 0.070 307 388 902 368;
  • 4) 0.070 307 388 902 368 × 2 = 0 + 0.140 614 777 804 736;
  • 5) 0.140 614 777 804 736 × 2 = 0 + 0.281 229 555 609 472;
  • 6) 0.281 229 555 609 472 × 2 = 0 + 0.562 459 111 218 944;
  • 7) 0.562 459 111 218 944 × 2 = 1 + 0.124 918 222 437 888;
  • 8) 0.124 918 222 437 888 × 2 = 0 + 0.249 836 444 875 776;
  • 9) 0.249 836 444 875 776 × 2 = 0 + 0.499 672 889 751 552;
  • 10) 0.499 672 889 751 552 × 2 = 0 + 0.999 345 779 503 104;
  • 11) 0.999 345 779 503 104 × 2 = 1 + 0.998 691 559 006 208;
  • 12) 0.998 691 559 006 208 × 2 = 1 + 0.997 383 118 012 416;
  • 13) 0.997 383 118 012 416 × 2 = 1 + 0.994 766 236 024 832;
  • 14) 0.994 766 236 024 832 × 2 = 1 + 0.989 532 472 049 664;
  • 15) 0.989 532 472 049 664 × 2 = 1 + 0.979 064 944 099 328;
  • 16) 0.979 064 944 099 328 × 2 = 1 + 0.958 129 888 198 656;
  • 17) 0.958 129 888 198 656 × 2 = 1 + 0.916 259 776 397 312;
  • 18) 0.916 259 776 397 312 × 2 = 1 + 0.832 519 552 794 624;
  • 19) 0.832 519 552 794 624 × 2 = 1 + 0.665 039 105 589 248;
  • 20) 0.665 039 105 589 248 × 2 = 1 + 0.330 078 211 178 496;
  • 21) 0.330 078 211 178 496 × 2 = 0 + 0.660 156 422 356 992;
  • 22) 0.660 156 422 356 992 × 2 = 1 + 0.320 312 844 713 984;
  • 23) 0.320 312 844 713 984 × 2 = 0 + 0.640 625 689 427 968;
  • 24) 0.640 625 689 427 968 × 2 = 1 + 0.281 251 378 855 936;
  • 25) 0.281 251 378 855 936 × 2 = 0 + 0.562 502 757 711 872;
  • 26) 0.562 502 757 711 872 × 2 = 1 + 0.125 005 515 423 744;
  • 27) 0.125 005 515 423 744 × 2 = 0 + 0.250 011 030 847 488;
  • 28) 0.250 011 030 847 488 × 2 = 0 + 0.500 022 061 694 976;
  • 29) 0.500 022 061 694 976 × 2 = 1 + 0.000 044 123 389 952;
  • 30) 0.000 044 123 389 952 × 2 = 0 + 0.000 088 246 779 904;
  • 31) 0.000 088 246 779 904 × 2 = 0 + 0.000 176 493 559 808;
  • 32) 0.000 176 493 559 808 × 2 = 0 + 0.000 352 987 119 616;
  • 33) 0.000 352 987 119 616 × 2 = 0 + 0.000 705 974 239 232;
  • 34) 0.000 705 974 239 232 × 2 = 0 + 0.001 411 948 478 464;
  • 35) 0.001 411 948 478 464 × 2 = 0 + 0.002 823 896 956 928;
  • 36) 0.002 823 896 956 928 × 2 = 0 + 0.005 647 793 913 856;
  • 37) 0.005 647 793 913 856 × 2 = 0 + 0.011 295 587 827 712;
  • 38) 0.011 295 587 827 712 × 2 = 0 + 0.022 591 175 655 424;
  • 39) 0.022 591 175 655 424 × 2 = 0 + 0.045 182 351 310 848;
  • 40) 0.045 182 351 310 848 × 2 = 0 + 0.090 364 702 621 696;
  • 41) 0.090 364 702 621 696 × 2 = 0 + 0.180 729 405 243 392;
  • 42) 0.180 729 405 243 392 × 2 = 0 + 0.361 458 810 486 784;
  • 43) 0.361 458 810 486 784 × 2 = 0 + 0.722 917 620 973 568;
  • 44) 0.722 917 620 973 568 × 2 = 1 + 0.445 835 241 947 136;
  • 45) 0.445 835 241 947 136 × 2 = 0 + 0.891 670 483 894 272;
  • 46) 0.891 670 483 894 272 × 2 = 1 + 0.783 340 967 788 544;
  • 47) 0.783 340 967 788 544 × 2 = 1 + 0.566 681 935 577 088;
  • 48) 0.566 681 935 577 088 × 2 = 1 + 0.133 363 871 154 176;
  • 49) 0.133 363 871 154 176 × 2 = 0 + 0.266 727 742 308 352;
  • 50) 0.266 727 742 308 352 × 2 = 0 + 0.533 455 484 616 704;
  • 51) 0.533 455 484 616 704 × 2 = 1 + 0.066 910 969 233 408;
  • 52) 0.066 910 969 233 408 × 2 = 0 + 0.133 821 938 466 816;
  • 53) 0.133 821 938 466 816 × 2 = 0 + 0.267 643 876 933 632;
  • 54) 0.267 643 876 933 632 × 2 = 0 + 0.535 287 753 867 264;
  • 55) 0.535 287 753 867 264 × 2 = 1 + 0.070 575 507 734 528;
  • 56) 0.070 575 507 734 528 × 2 = 0 + 0.141 151 015 469 056;
  • 57) 0.141 151 015 469 056 × 2 = 0 + 0.282 302 030 938 112;
  • 58) 0.282 302 030 938 112 × 2 = 0 + 0.564 604 061 876 224;
  • 59) 0.564 604 061 876 224 × 2 = 1 + 0.129 208 123 752 448;

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:


0.008 788 423 612 796(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0001 0111 0010 0010 001(2)

6. Positive number before normalization:

0.008 788 423 612 796(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0001 0111 0010 0010 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:


0.008 788 423 612 796(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0001 0111 0010 0010 001(2) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0001 0111 0010 0010 001(2) × 20 =

1.0001 1111 1111 1010 1010 0100 0000 0000 0000 1011 1001 0001 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 0000 0000 1011 1001 0001 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 0000 0000 1011 1001 0001 0001 =

0001 1111 1111 1010 1010 0100 0000 0000 0000 1011 1001 0001 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 0000 0000 1011 1001 0001 0001


Decimal number -0.008 788 423 612 796 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0001 1111 1111 1010 1010 0100 0000 0000 0000 1011 1001 0001 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