64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 1.999 999 999 999 999 777 957 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 1.999 999 999 999 999 777 957(10) converted and written in 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: 1.
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;
  • 1 ÷ 2 = 0 + 1;

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.


1(10) =


1(2)


3. Convert to binary (base 2) the fractional part: 0.999 999 999 999 999 777 957.

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) 0.999 999 999 999 999 777 957 × 2 = 1 + 0.999 999 999 999 999 555 914;
  • 2) 0.999 999 999 999 999 555 914 × 2 = 1 + 0.999 999 999 999 999 111 828;
  • 3) 0.999 999 999 999 999 111 828 × 2 = 1 + 0.999 999 999 999 998 223 656;
  • 4) 0.999 999 999 999 998 223 656 × 2 = 1 + 0.999 999 999 999 996 447 312;
  • 5) 0.999 999 999 999 996 447 312 × 2 = 1 + 0.999 999 999 999 992 894 624;
  • 6) 0.999 999 999 999 992 894 624 × 2 = 1 + 0.999 999 999 999 985 789 248;
  • 7) 0.999 999 999 999 985 789 248 × 2 = 1 + 0.999 999 999 999 971 578 496;
  • 8) 0.999 999 999 999 971 578 496 × 2 = 1 + 0.999 999 999 999 943 156 992;
  • 9) 0.999 999 999 999 943 156 992 × 2 = 1 + 0.999 999 999 999 886 313 984;
  • 10) 0.999 999 999 999 886 313 984 × 2 = 1 + 0.999 999 999 999 772 627 968;
  • 11) 0.999 999 999 999 772 627 968 × 2 = 1 + 0.999 999 999 999 545 255 936;
  • 12) 0.999 999 999 999 545 255 936 × 2 = 1 + 0.999 999 999 999 090 511 872;
  • 13) 0.999 999 999 999 090 511 872 × 2 = 1 + 0.999 999 999 998 181 023 744;
  • 14) 0.999 999 999 998 181 023 744 × 2 = 1 + 0.999 999 999 996 362 047 488;
  • 15) 0.999 999 999 996 362 047 488 × 2 = 1 + 0.999 999 999 992 724 094 976;
  • 16) 0.999 999 999 992 724 094 976 × 2 = 1 + 0.999 999 999 985 448 189 952;
  • 17) 0.999 999 999 985 448 189 952 × 2 = 1 + 0.999 999 999 970 896 379 904;
  • 18) 0.999 999 999 970 896 379 904 × 2 = 1 + 0.999 999 999 941 792 759 808;
  • 19) 0.999 999 999 941 792 759 808 × 2 = 1 + 0.999 999 999 883 585 519 616;
  • 20) 0.999 999 999 883 585 519 616 × 2 = 1 + 0.999 999 999 767 171 039 232;
  • 21) 0.999 999 999 767 171 039 232 × 2 = 1 + 0.999 999 999 534 342 078 464;
  • 22) 0.999 999 999 534 342 078 464 × 2 = 1 + 0.999 999 999 068 684 156 928;
  • 23) 0.999 999 999 068 684 156 928 × 2 = 1 + 0.999 999 998 137 368 313 856;
  • 24) 0.999 999 998 137 368 313 856 × 2 = 1 + 0.999 999 996 274 736 627 712;
  • 25) 0.999 999 996 274 736 627 712 × 2 = 1 + 0.999 999 992 549 473 255 424;
  • 26) 0.999 999 992 549 473 255 424 × 2 = 1 + 0.999 999 985 098 946 510 848;
  • 27) 0.999 999 985 098 946 510 848 × 2 = 1 + 0.999 999 970 197 893 021 696;
  • 28) 0.999 999 970 197 893 021 696 × 2 = 1 + 0.999 999 940 395 786 043 392;
  • 29) 0.999 999 940 395 786 043 392 × 2 = 1 + 0.999 999 880 791 572 086 784;
  • 30) 0.999 999 880 791 572 086 784 × 2 = 1 + 0.999 999 761 583 144 173 568;
  • 31) 0.999 999 761 583 144 173 568 × 2 = 1 + 0.999 999 523 166 288 347 136;
  • 32) 0.999 999 523 166 288 347 136 × 2 = 1 + 0.999 999 046 332 576 694 272;
  • 33) 0.999 999 046 332 576 694 272 × 2 = 1 + 0.999 998 092 665 153 388 544;
  • 34) 0.999 998 092 665 153 388 544 × 2 = 1 + 0.999 996 185 330 306 777 088;
  • 35) 0.999 996 185 330 306 777 088 × 2 = 1 + 0.999 992 370 660 613 554 176;
  • 36) 0.999 992 370 660 613 554 176 × 2 = 1 + 0.999 984 741 321 227 108 352;
  • 37) 0.999 984 741 321 227 108 352 × 2 = 1 + 0.999 969 482 642 454 216 704;
  • 38) 0.999 969 482 642 454 216 704 × 2 = 1 + 0.999 938 965 284 908 433 408;
  • 39) 0.999 938 965 284 908 433 408 × 2 = 1 + 0.999 877 930 569 816 866 816;
  • 40) 0.999 877 930 569 816 866 816 × 2 = 1 + 0.999 755 861 139 633 733 632;
  • 41) 0.999 755 861 139 633 733 632 × 2 = 1 + 0.999 511 722 279 267 467 264;
  • 42) 0.999 511 722 279 267 467 264 × 2 = 1 + 0.999 023 444 558 534 934 528;
  • 43) 0.999 023 444 558 534 934 528 × 2 = 1 + 0.998 046 889 117 069 869 056;
  • 44) 0.998 046 889 117 069 869 056 × 2 = 1 + 0.996 093 778 234 139 738 112;
  • 45) 0.996 093 778 234 139 738 112 × 2 = 1 + 0.992 187 556 468 279 476 224;
  • 46) 0.992 187 556 468 279 476 224 × 2 = 1 + 0.984 375 112 936 558 952 448;
  • 47) 0.984 375 112 936 558 952 448 × 2 = 1 + 0.968 750 225 873 117 904 896;
  • 48) 0.968 750 225 873 117 904 896 × 2 = 1 + 0.937 500 451 746 235 809 792;
  • 49) 0.937 500 451 746 235 809 792 × 2 = 1 + 0.875 000 903 492 471 619 584;
  • 50) 0.875 000 903 492 471 619 584 × 2 = 1 + 0.750 001 806 984 943 239 168;
  • 51) 0.750 001 806 984 943 239 168 × 2 = 1 + 0.500 003 613 969 886 478 336;
  • 52) 0.500 003 613 969 886 478 336 × 2 = 1 + 0.000 007 227 939 772 956 672;
  • 53) 0.000 007 227 939 772 956 672 × 2 = 0 + 0.000 014 455 879 545 913 344;

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...)


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:


0.999 999 999 999 999 777 957(10) =


0.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0(2)


5. Positive number before normalization:

1.999 999 999 999 999 777 957(10) =


1.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.999 999 999 999 999 777 957(10) =


1.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0(2) =


1.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0(2) × 20


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): 0


Mantissa (not normalized):
1.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


0 + 2(11-1) - 1 =


(0 + 1 023)(10) =


1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 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) =


1023(10) =


011 1111 1111(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0 =


1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


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 1111


Mantissa (52 bits) =
1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


The base ten decimal number 1.999 999 999 999 999 777 957 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 1111 - 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111

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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