0.996 194 697 999 999 934 090 02 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.996 194 697 999 999 934 090 02(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
0.996 194 697 999 999 934 090 02(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.996 194 697 999 999 934 090 02.

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) 0.996 194 697 999 999 934 090 02 × 2 = 1 + 0.992 389 395 999 999 868 180 04;
  • 2) 0.992 389 395 999 999 868 180 04 × 2 = 1 + 0.984 778 791 999 999 736 360 08;
  • 3) 0.984 778 791 999 999 736 360 08 × 2 = 1 + 0.969 557 583 999 999 472 720 16;
  • 4) 0.969 557 583 999 999 472 720 16 × 2 = 1 + 0.939 115 167 999 998 945 440 32;
  • 5) 0.939 115 167 999 998 945 440 32 × 2 = 1 + 0.878 230 335 999 997 890 880 64;
  • 6) 0.878 230 335 999 997 890 880 64 × 2 = 1 + 0.756 460 671 999 995 781 761 28;
  • 7) 0.756 460 671 999 995 781 761 28 × 2 = 1 + 0.512 921 343 999 991 563 522 56;
  • 8) 0.512 921 343 999 991 563 522 56 × 2 = 1 + 0.025 842 687 999 983 127 045 12;
  • 9) 0.025 842 687 999 983 127 045 12 × 2 = 0 + 0.051 685 375 999 966 254 090 24;
  • 10) 0.051 685 375 999 966 254 090 24 × 2 = 0 + 0.103 370 751 999 932 508 180 48;
  • 11) 0.103 370 751 999 932 508 180 48 × 2 = 0 + 0.206 741 503 999 865 016 360 96;
  • 12) 0.206 741 503 999 865 016 360 96 × 2 = 0 + 0.413 483 007 999 730 032 721 92;
  • 13) 0.413 483 007 999 730 032 721 92 × 2 = 0 + 0.826 966 015 999 460 065 443 84;
  • 14) 0.826 966 015 999 460 065 443 84 × 2 = 1 + 0.653 932 031 998 920 130 887 68;
  • 15) 0.653 932 031 998 920 130 887 68 × 2 = 1 + 0.307 864 063 997 840 261 775 36;
  • 16) 0.307 864 063 997 840 261 775 36 × 2 = 0 + 0.615 728 127 995 680 523 550 72;
  • 17) 0.615 728 127 995 680 523 550 72 × 2 = 1 + 0.231 456 255 991 361 047 101 44;
  • 18) 0.231 456 255 991 361 047 101 44 × 2 = 0 + 0.462 912 511 982 722 094 202 88;
  • 19) 0.462 912 511 982 722 094 202 88 × 2 = 0 + 0.925 825 023 965 444 188 405 76;
  • 20) 0.925 825 023 965 444 188 405 76 × 2 = 1 + 0.851 650 047 930 888 376 811 52;
  • 21) 0.851 650 047 930 888 376 811 52 × 2 = 1 + 0.703 300 095 861 776 753 623 04;
  • 22) 0.703 300 095 861 776 753 623 04 × 2 = 1 + 0.406 600 191 723 553 507 246 08;
  • 23) 0.406 600 191 723 553 507 246 08 × 2 = 0 + 0.813 200 383 447 107 014 492 16;
  • 24) 0.813 200 383 447 107 014 492 16 × 2 = 1 + 0.626 400 766 894 214 028 984 32;
  • 25) 0.626 400 766 894 214 028 984 32 × 2 = 1 + 0.252 801 533 788 428 057 968 64;
  • 26) 0.252 801 533 788 428 057 968 64 × 2 = 0 + 0.505 603 067 576 856 115 937 28;
  • 27) 0.505 603 067 576 856 115 937 28 × 2 = 1 + 0.011 206 135 153 712 231 874 56;
  • 28) 0.011 206 135 153 712 231 874 56 × 2 = 0 + 0.022 412 270 307 424 463 749 12;
  • 29) 0.022 412 270 307 424 463 749 12 × 2 = 0 + 0.044 824 540 614 848 927 498 24;
  • 30) 0.044 824 540 614 848 927 498 24 × 2 = 0 + 0.089 649 081 229 697 854 996 48;
  • 31) 0.089 649 081 229 697 854 996 48 × 2 = 0 + 0.179 298 162 459 395 709 992 96;
  • 32) 0.179 298 162 459 395 709 992 96 × 2 = 0 + 0.358 596 324 918 791 419 985 92;
  • 33) 0.358 596 324 918 791 419 985 92 × 2 = 0 + 0.717 192 649 837 582 839 971 84;
  • 34) 0.717 192 649 837 582 839 971 84 × 2 = 1 + 0.434 385 299 675 165 679 943 68;
  • 35) 0.434 385 299 675 165 679 943 68 × 2 = 0 + 0.868 770 599 350 331 359 887 36;
  • 36) 0.868 770 599 350 331 359 887 36 × 2 = 1 + 0.737 541 198 700 662 719 774 72;
  • 37) 0.737 541 198 700 662 719 774 72 × 2 = 1 + 0.475 082 397 401 325 439 549 44;
  • 38) 0.475 082 397 401 325 439 549 44 × 2 = 0 + 0.950 164 794 802 650 879 098 88;
  • 39) 0.950 164 794 802 650 879 098 88 × 2 = 1 + 0.900 329 589 605 301 758 197 76;
  • 40) 0.900 329 589 605 301 758 197 76 × 2 = 1 + 0.800 659 179 210 603 516 395 52;
  • 41) 0.800 659 179 210 603 516 395 52 × 2 = 1 + 0.601 318 358 421 207 032 791 04;
  • 42) 0.601 318 358 421 207 032 791 04 × 2 = 1 + 0.202 636 716 842 414 065 582 08;
  • 43) 0.202 636 716 842 414 065 582 08 × 2 = 0 + 0.405 273 433 684 828 131 164 16;
  • 44) 0.405 273 433 684 828 131 164 16 × 2 = 0 + 0.810 546 867 369 656 262 328 32;
  • 45) 0.810 546 867 369 656 262 328 32 × 2 = 1 + 0.621 093 734 739 312 524 656 64;
  • 46) 0.621 093 734 739 312 524 656 64 × 2 = 1 + 0.242 187 469 478 625 049 313 28;
  • 47) 0.242 187 469 478 625 049 313 28 × 2 = 0 + 0.484 374 938 957 250 098 626 56;
  • 48) 0.484 374 938 957 250 098 626 56 × 2 = 0 + 0.968 749 877 914 500 197 253 12;
  • 49) 0.968 749 877 914 500 197 253 12 × 2 = 1 + 0.937 499 755 829 000 394 506 24;
  • 50) 0.937 499 755 829 000 394 506 24 × 2 = 1 + 0.874 999 511 658 000 789 012 48;
  • 51) 0.874 999 511 658 000 789 012 48 × 2 = 1 + 0.749 999 023 316 001 578 024 96;
  • 52) 0.749 999 023 316 001 578 024 96 × 2 = 1 + 0.499 998 046 632 003 156 049 92;
  • 53) 0.499 998 046 632 003 156 049 92 × 2 = 0 + 0.999 996 093 264 006 312 099 84;

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).


4. 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:


0.996 194 697 999 999 934 090 02(10) =

0.1111 1111 0000 0110 1001 1101 1010 0000 0101 1011 1100 1100 1111 0(2)

5. Positive number before normalization:

0.996 194 697 999 999 934 090 02(10) =

0.1111 1111 0000 0110 1001 1101 1010 0000 0101 1011 1100 1100 1111 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.996 194 697 999 999 934 090 02(10) =

0.1111 1111 0000 0110 1001 1101 1010 0000 0101 1011 1100 1100 1111 0(2) =

0.1111 1111 0000 0110 1001 1101 1010 0000 0101 1011 1100 1100 1111 0(2) × 20 =

1.1111 1110 0000 1101 0011 1011 0100 0000 1011 0111 1001 1001 1110(2) × 2-1


7. 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 0 (a positive number)

Exponent (unadjusted): -1

Mantissa (not normalized):
1.1111 1110 0000 1101 0011 1011 0100 0000 1011 0111 1001 1001 1110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-1 + 2(11-1) - 1 =

(-1 + 1 023)(10) =

1 022(10)


9. 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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 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;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1022(10) =

011 1111 1110(2)


11. 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. 1111 1110 0000 1101 0011 1011 0100 0000 1011 0111 1001 1001 1110 =

1111 1110 0000 1101 0011 1011 0100 0000 1011 0111 1001 1001 1110


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 1110

Mantissa (52 bits) =
1111 1110 0000 1101 0011 1011 0100 0000 1011 0111 1001 1001 1110


Decimal number 0.996 194 697 999 999 934 090 02 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1111 1110 0000 1101 0011 1011 0100 0000 1011 0111 1001 1001 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