64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 2.666 666 666 666 666 666 666 666 666 1 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 2.666 666 666 666 666 666 666 666 666 1(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: 2.
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;
  • 2 ÷ 2 = 1 + 0;
  • 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.


2(10) =

10(2)


3. Convert to binary (base 2) the fractional part: 0.666 666 666 666 666 666 666 666 666 1.

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) 0.666 666 666 666 666 666 666 666 666 1 × 2 = 1 + 0.333 333 333 333 333 333 333 333 332 2;
  • 2) 0.333 333 333 333 333 333 333 333 332 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 664 4;
  • 3) 0.666 666 666 666 666 666 666 666 664 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 328 8;
  • 4) 0.333 333 333 333 333 333 333 333 328 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 657 6;
  • 5) 0.666 666 666 666 666 666 666 666 657 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 315 2;
  • 6) 0.333 333 333 333 333 333 333 333 315 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 630 4;
  • 7) 0.666 666 666 666 666 666 666 666 630 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 260 8;
  • 8) 0.333 333 333 333 333 333 333 333 260 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 521 6;
  • 9) 0.666 666 666 666 666 666 666 666 521 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 043 2;
  • 10) 0.333 333 333 333 333 333 333 333 043 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 086 4;
  • 11) 0.666 666 666 666 666 666 666 666 086 4 × 2 = 1 + 0.333 333 333 333 333 333 333 332 172 8;
  • 12) 0.333 333 333 333 333 333 333 332 172 8 × 2 = 0 + 0.666 666 666 666 666 666 666 664 345 6;
  • 13) 0.666 666 666 666 666 666 666 664 345 6 × 2 = 1 + 0.333 333 333 333 333 333 333 328 691 2;
  • 14) 0.333 333 333 333 333 333 333 328 691 2 × 2 = 0 + 0.666 666 666 666 666 666 666 657 382 4;
  • 15) 0.666 666 666 666 666 666 666 657 382 4 × 2 = 1 + 0.333 333 333 333 333 333 333 314 764 8;
  • 16) 0.333 333 333 333 333 333 333 314 764 8 × 2 = 0 + 0.666 666 666 666 666 666 666 629 529 6;
  • 17) 0.666 666 666 666 666 666 666 629 529 6 × 2 = 1 + 0.333 333 333 333 333 333 333 259 059 2;
  • 18) 0.333 333 333 333 333 333 333 259 059 2 × 2 = 0 + 0.666 666 666 666 666 666 666 518 118 4;
  • 19) 0.666 666 666 666 666 666 666 518 118 4 × 2 = 1 + 0.333 333 333 333 333 333 333 036 236 8;
  • 20) 0.333 333 333 333 333 333 333 036 236 8 × 2 = 0 + 0.666 666 666 666 666 666 666 072 473 6;
  • 21) 0.666 666 666 666 666 666 666 072 473 6 × 2 = 1 + 0.333 333 333 333 333 333 332 144 947 2;
  • 22) 0.333 333 333 333 333 333 332 144 947 2 × 2 = 0 + 0.666 666 666 666 666 666 664 289 894 4;
  • 23) 0.666 666 666 666 666 666 664 289 894 4 × 2 = 1 + 0.333 333 333 333 333 333 328 579 788 8;
  • 24) 0.333 333 333 333 333 333 328 579 788 8 × 2 = 0 + 0.666 666 666 666 666 666 657 159 577 6;
  • 25) 0.666 666 666 666 666 666 657 159 577 6 × 2 = 1 + 0.333 333 333 333 333 333 314 319 155 2;
  • 26) 0.333 333 333 333 333 333 314 319 155 2 × 2 = 0 + 0.666 666 666 666 666 666 628 638 310 4;
  • 27) 0.666 666 666 666 666 666 628 638 310 4 × 2 = 1 + 0.333 333 333 333 333 333 257 276 620 8;
  • 28) 0.333 333 333 333 333 333 257 276 620 8 × 2 = 0 + 0.666 666 666 666 666 666 514 553 241 6;
  • 29) 0.666 666 666 666 666 666 514 553 241 6 × 2 = 1 + 0.333 333 333 333 333 333 029 106 483 2;
  • 30) 0.333 333 333 333 333 333 029 106 483 2 × 2 = 0 + 0.666 666 666 666 666 666 058 212 966 4;
  • 31) 0.666 666 666 666 666 666 058 212 966 4 × 2 = 1 + 0.333 333 333 333 333 332 116 425 932 8;
  • 32) 0.333 333 333 333 333 332 116 425 932 8 × 2 = 0 + 0.666 666 666 666 666 664 232 851 865 6;
  • 33) 0.666 666 666 666 666 664 232 851 865 6 × 2 = 1 + 0.333 333 333 333 333 328 465 703 731 2;
  • 34) 0.333 333 333 333 333 328 465 703 731 2 × 2 = 0 + 0.666 666 666 666 666 656 931 407 462 4;
  • 35) 0.666 666 666 666 666 656 931 407 462 4 × 2 = 1 + 0.333 333 333 333 333 313 862 814 924 8;
  • 36) 0.333 333 333 333 333 313 862 814 924 8 × 2 = 0 + 0.666 666 666 666 666 627 725 629 849 6;
  • 37) 0.666 666 666 666 666 627 725 629 849 6 × 2 = 1 + 0.333 333 333 333 333 255 451 259 699 2;
  • 38) 0.333 333 333 333 333 255 451 259 699 2 × 2 = 0 + 0.666 666 666 666 666 510 902 519 398 4;
  • 39) 0.666 666 666 666 666 510 902 519 398 4 × 2 = 1 + 0.333 333 333 333 333 021 805 038 796 8;
  • 40) 0.333 333 333 333 333 021 805 038 796 8 × 2 = 0 + 0.666 666 666 666 666 043 610 077 593 6;
  • 41) 0.666 666 666 666 666 043 610 077 593 6 × 2 = 1 + 0.333 333 333 333 332 087 220 155 187 2;
  • 42) 0.333 333 333 333 332 087 220 155 187 2 × 2 = 0 + 0.666 666 666 666 664 174 440 310 374 4;
  • 43) 0.666 666 666 666 664 174 440 310 374 4 × 2 = 1 + 0.333 333 333 333 328 348 880 620 748 8;
  • 44) 0.333 333 333 333 328 348 880 620 748 8 × 2 = 0 + 0.666 666 666 666 656 697 761 241 497 6;
  • 45) 0.666 666 666 666 656 697 761 241 497 6 × 2 = 1 + 0.333 333 333 333 313 395 522 482 995 2;
  • 46) 0.333 333 333 333 313 395 522 482 995 2 × 2 = 0 + 0.666 666 666 666 626 791 044 965 990 4;
  • 47) 0.666 666 666 666 626 791 044 965 990 4 × 2 = 1 + 0.333 333 333 333 253 582 089 931 980 8;
  • 48) 0.333 333 333 333 253 582 089 931 980 8 × 2 = 0 + 0.666 666 666 666 507 164 179 863 961 6;
  • 49) 0.666 666 666 666 507 164 179 863 961 6 × 2 = 1 + 0.333 333 333 333 014 328 359 727 923 2;
  • 50) 0.333 333 333 333 014 328 359 727 923 2 × 2 = 0 + 0.666 666 666 666 028 656 719 455 846 4;
  • 51) 0.666 666 666 666 028 656 719 455 846 4 × 2 = 1 + 0.333 333 333 332 057 313 438 911 692 8;
  • 52) 0.333 333 333 332 057 313 438 911 692 8 × 2 = 0 + 0.666 666 666 664 114 626 877 823 385 6;
  • 53) 0.666 666 666 664 114 626 877 823 385 6 × 2 = 1 + 0.333 333 333 328 229 253 755 646 771 2;

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:


0.666 666 666 666 666 666 666 666 666 1(10) =

0.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1(2)


5. Positive number before normalization:

2.666 666 666 666 666 666 666 666 666 1(10) =

10.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1(2)

6. Normalize the binary representation of the number.

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


2.666 666 666 666 666 666 666 666 666 1(10) =

10.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1(2) =

10.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1(2) × 20 =

1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 01(2) × 21


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):
1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 01


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) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 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) =

1024(10) =

100 0000 0000(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. 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 01 =

0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101


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 0000 0000

Mantissa (52 bits) =
0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101


The base ten decimal number 2.666 666 666 666 666 666 666 666 666 1 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 0000 - 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101

More operations with decimal numbers converted to 64 bit double precision IEEE 754 binary floating point representation:

» Number 2.666 666 666 666 666 666 666 666 666 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

» Number 2.666 666 666 666 666 666 666 666 666 2 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

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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