64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 999 999.999 999 999 4 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 999 999.999 999 999 4(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: 999 999.
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;
  • 999 999 ÷ 2 = 499 999 + 1;
  • 499 999 ÷ 2 = 249 999 + 1;
  • 249 999 ÷ 2 = 124 999 + 1;
  • 124 999 ÷ 2 = 62 499 + 1;
  • 62 499 ÷ 2 = 31 249 + 1;
  • 31 249 ÷ 2 = 15 624 + 1;
  • 15 624 ÷ 2 = 7 812 + 0;
  • 7 812 ÷ 2 = 3 906 + 0;
  • 3 906 ÷ 2 = 1 953 + 0;
  • 1 953 ÷ 2 = 976 + 1;
  • 976 ÷ 2 = 488 + 0;
  • 488 ÷ 2 = 244 + 0;
  • 244 ÷ 2 = 122 + 0;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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.


999 999(10) =


1111 0100 0010 0011 1111(2)


3. Convert to binary (base 2) the fractional part: 0.999 999 999 4.

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 4 × 2 = 1 + 0.999 999 998 8;
  • 2) 0.999 999 998 8 × 2 = 1 + 0.999 999 997 6;
  • 3) 0.999 999 997 6 × 2 = 1 + 0.999 999 995 2;
  • 4) 0.999 999 995 2 × 2 = 1 + 0.999 999 990 4;
  • 5) 0.999 999 990 4 × 2 = 1 + 0.999 999 980 8;
  • 6) 0.999 999 980 8 × 2 = 1 + 0.999 999 961 6;
  • 7) 0.999 999 961 6 × 2 = 1 + 0.999 999 923 2;
  • 8) 0.999 999 923 2 × 2 = 1 + 0.999 999 846 4;
  • 9) 0.999 999 846 4 × 2 = 1 + 0.999 999 692 8;
  • 10) 0.999 999 692 8 × 2 = 1 + 0.999 999 385 6;
  • 11) 0.999 999 385 6 × 2 = 1 + 0.999 998 771 2;
  • 12) 0.999 998 771 2 × 2 = 1 + 0.999 997 542 4;
  • 13) 0.999 997 542 4 × 2 = 1 + 0.999 995 084 8;
  • 14) 0.999 995 084 8 × 2 = 1 + 0.999 990 169 6;
  • 15) 0.999 990 169 6 × 2 = 1 + 0.999 980 339 2;
  • 16) 0.999 980 339 2 × 2 = 1 + 0.999 960 678 4;
  • 17) 0.999 960 678 4 × 2 = 1 + 0.999 921 356 8;
  • 18) 0.999 921 356 8 × 2 = 1 + 0.999 842 713 6;
  • 19) 0.999 842 713 6 × 2 = 1 + 0.999 685 427 2;
  • 20) 0.999 685 427 2 × 2 = 1 + 0.999 370 854 4;
  • 21) 0.999 370 854 4 × 2 = 1 + 0.998 741 708 8;
  • 22) 0.998 741 708 8 × 2 = 1 + 0.997 483 417 6;
  • 23) 0.997 483 417 6 × 2 = 1 + 0.994 966 835 2;
  • 24) 0.994 966 835 2 × 2 = 1 + 0.989 933 670 4;
  • 25) 0.989 933 670 4 × 2 = 1 + 0.979 867 340 8;
  • 26) 0.979 867 340 8 × 2 = 1 + 0.959 734 681 6;
  • 27) 0.959 734 681 6 × 2 = 1 + 0.919 469 363 2;
  • 28) 0.919 469 363 2 × 2 = 1 + 0.838 938 726 4;
  • 29) 0.838 938 726 4 × 2 = 1 + 0.677 877 452 8;
  • 30) 0.677 877 452 8 × 2 = 1 + 0.355 754 905 6;
  • 31) 0.355 754 905 6 × 2 = 0 + 0.711 509 811 2;
  • 32) 0.711 509 811 2 × 2 = 1 + 0.423 019 622 4;
  • 33) 0.423 019 622 4 × 2 = 0 + 0.846 039 244 8;
  • 34) 0.846 039 244 8 × 2 = 1 + 0.692 078 489 6;
  • 35) 0.692 078 489 6 × 2 = 1 + 0.384 156 979 2;
  • 36) 0.384 156 979 2 × 2 = 0 + 0.768 313 958 4;
  • 37) 0.768 313 958 4 × 2 = 1 + 0.536 627 916 8;
  • 38) 0.536 627 916 8 × 2 = 1 + 0.073 255 833 6;
  • 39) 0.073 255 833 6 × 2 = 0 + 0.146 511 667 2;
  • 40) 0.146 511 667 2 × 2 = 0 + 0.293 023 334 4;
  • 41) 0.293 023 334 4 × 2 = 0 + 0.586 046 668 8;
  • 42) 0.586 046 668 8 × 2 = 1 + 0.172 093 337 6;
  • 43) 0.172 093 337 6 × 2 = 0 + 0.344 186 675 2;
  • 44) 0.344 186 675 2 × 2 = 0 + 0.688 373 350 4;
  • 45) 0.688 373 350 4 × 2 = 1 + 0.376 746 700 8;
  • 46) 0.376 746 700 8 × 2 = 0 + 0.753 493 401 6;
  • 47) 0.753 493 401 6 × 2 = 1 + 0.506 986 803 2;
  • 48) 0.506 986 803 2 × 2 = 1 + 0.013 973 606 4;
  • 49) 0.013 973 606 4 × 2 = 0 + 0.027 947 212 8;
  • 50) 0.027 947 212 8 × 2 = 0 + 0.055 894 425 6;
  • 51) 0.055 894 425 6 × 2 = 0 + 0.111 788 851 2;
  • 52) 0.111 788 851 2 × 2 = 0 + 0.223 577 702 4;
  • 53) 0.223 577 702 4 × 2 = 0 + 0.447 155 404 8;

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 4(10) =


0.1111 1111 1111 1111 1111 1111 1111 1101 0110 1100 0100 1011 0000 0(2)


5. Positive number before normalization:

999 999.999 999 999 4(10) =


1111 0100 0010 0011 1111.1111 1111 1111 1111 1111 1111 1111 1101 0110 1100 0100 1011 0000 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 19 positions to the left, so that only one non zero digit remains to the left of it:


999 999.999 999 999 4(10) =


1111 0100 0010 0011 1111.1111 1111 1111 1111 1111 1111 1111 1101 0110 1100 0100 1011 0000 0(2) =


1111 0100 0010 0011 1111.1111 1111 1111 1111 1111 1111 1111 1101 0110 1100 0100 1011 0000 0(2) × 20 =


1.1110 1000 0100 0111 1111 1111 1111 1111 1111 1111 1111 1111 1010 1101 1000 1001 0110 0000(2) × 219


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): 19


Mantissa (not normalized):
1.1110 1000 0100 0111 1111 1111 1111 1111 1111 1111 1111 1111 1010 1101 1000 1001 0110 0000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


19 + 2(11-1) - 1 =


(19 + 1 023)(10) =


1 042(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;
  • 1 042 ÷ 2 = 521 + 0;
  • 521 ÷ 2 = 260 + 1;
  • 260 ÷ 2 = 130 + 0;
  • 130 ÷ 2 = 65 + 0;
  • 65 ÷ 2 = 32 + 1;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =


1042(10) =


100 0001 0010(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. 1110 1000 0100 0111 1111 1111 1111 1111 1111 1111 1111 1111 1010 1101 1000 1001 0110 0000 =


1110 1000 0100 0111 1111 1111 1111 1111 1111 1111 1111 1111 1010


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) =
100 0001 0010


Mantissa (52 bits) =
1110 1000 0100 0111 1111 1111 1111 1111 1111 1111 1111 1111 1010


The base ten decimal number 999 999.999 999 999 4 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0001 0010 - 1110 1000 0100 0111 1111 1111 1111 1111 1111 1111 1111 1111 1010

The latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

Number 19.034 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Jul 13 13:05 UTC (GMT)
Number 36 046 815 018 089 597 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Jul 13 13:05 UTC (GMT)
Number 1 019 033 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Jul 13 13:05 UTC (GMT)
Number -1.014 006 126 349 342 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Jul 13 13:05 UTC (GMT)
Number 1 054 867 405 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Jul 13 13:05 UTC (GMT)
Number 23 127.9 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Jul 13 13:05 UTC (GMT)
Number 78 816 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Jul 13 13:04 UTC (GMT)
Number 1 111 117 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Jul 13 13:04 UTC (GMT)
Number 524 403 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Jul 13 13:04 UTC (GMT)
Number 5.126 365 606 866 167 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Jul 13 13:04 UTC (GMT)
All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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