0.066 666 666 666 666 666 666 666 666 666 666 667 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.066 666 666 666 666 666 666 666 666 666 666 667 3(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
0.066 666 666 666 666 666 666 666 666 666 666 667 3(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: 0.066 666 666 666 666 666 666 666 666 666 666 667 3.

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) 0.066 666 666 666 666 666 666 666 666 666 666 667 3 × 2 = 0 + 0.133 333 333 333 333 333 333 333 333 333 333 334 6;
  • 2) 0.133 333 333 333 333 333 333 333 333 333 333 334 6 × 2 = 0 + 0.266 666 666 666 666 666 666 666 666 666 666 669 2;
  • 3) 0.266 666 666 666 666 666 666 666 666 666 666 669 2 × 2 = 0 + 0.533 333 333 333 333 333 333 333 333 333 333 338 4;
  • 4) 0.533 333 333 333 333 333 333 333 333 333 333 338 4 × 2 = 1 + 0.066 666 666 666 666 666 666 666 666 666 666 676 8;
  • 5) 0.066 666 666 666 666 666 666 666 666 666 666 676 8 × 2 = 0 + 0.133 333 333 333 333 333 333 333 333 333 333 353 6;
  • 6) 0.133 333 333 333 333 333 333 333 333 333 333 353 6 × 2 = 0 + 0.266 666 666 666 666 666 666 666 666 666 666 707 2;
  • 7) 0.266 666 666 666 666 666 666 666 666 666 666 707 2 × 2 = 0 + 0.533 333 333 333 333 333 333 333 333 333 333 414 4;
  • 8) 0.533 333 333 333 333 333 333 333 333 333 333 414 4 × 2 = 1 + 0.066 666 666 666 666 666 666 666 666 666 666 828 8;
  • 9) 0.066 666 666 666 666 666 666 666 666 666 666 828 8 × 2 = 0 + 0.133 333 333 333 333 333 333 333 333 333 333 657 6;
  • 10) 0.133 333 333 333 333 333 333 333 333 333 333 657 6 × 2 = 0 + 0.266 666 666 666 666 666 666 666 666 666 667 315 2;
  • 11) 0.266 666 666 666 666 666 666 666 666 666 667 315 2 × 2 = 0 + 0.533 333 333 333 333 333 333 333 333 333 334 630 4;
  • 12) 0.533 333 333 333 333 333 333 333 333 333 334 630 4 × 2 = 1 + 0.066 666 666 666 666 666 666 666 666 666 669 260 8;
  • 13) 0.066 666 666 666 666 666 666 666 666 666 669 260 8 × 2 = 0 + 0.133 333 333 333 333 333 333 333 333 333 338 521 6;
  • 14) 0.133 333 333 333 333 333 333 333 333 333 338 521 6 × 2 = 0 + 0.266 666 666 666 666 666 666 666 666 666 677 043 2;
  • 15) 0.266 666 666 666 666 666 666 666 666 666 677 043 2 × 2 = 0 + 0.533 333 333 333 333 333 333 333 333 333 354 086 4;
  • 16) 0.533 333 333 333 333 333 333 333 333 333 354 086 4 × 2 = 1 + 0.066 666 666 666 666 666 666 666 666 666 708 172 8;
  • 17) 0.066 666 666 666 666 666 666 666 666 666 708 172 8 × 2 = 0 + 0.133 333 333 333 333 333 333 333 333 333 416 345 6;
  • 18) 0.133 333 333 333 333 333 333 333 333 333 416 345 6 × 2 = 0 + 0.266 666 666 666 666 666 666 666 666 666 832 691 2;
  • 19) 0.266 666 666 666 666 666 666 666 666 666 832 691 2 × 2 = 0 + 0.533 333 333 333 333 333 333 333 333 333 665 382 4;
  • 20) 0.533 333 333 333 333 333 333 333 333 333 665 382 4 × 2 = 1 + 0.066 666 666 666 666 666 666 666 666 667 330 764 8;
  • 21) 0.066 666 666 666 666 666 666 666 666 667 330 764 8 × 2 = 0 + 0.133 333 333 333 333 333 333 333 333 334 661 529 6;
  • 22) 0.133 333 333 333 333 333 333 333 333 334 661 529 6 × 2 = 0 + 0.266 666 666 666 666 666 666 666 666 669 323 059 2;
  • 23) 0.266 666 666 666 666 666 666 666 666 669 323 059 2 × 2 = 0 + 0.533 333 333 333 333 333 333 333 333 338 646 118 4;
  • 24) 0.533 333 333 333 333 333 333 333 333 338 646 118 4 × 2 = 1 + 0.066 666 666 666 666 666 666 666 666 677 292 236 8;
  • 25) 0.066 666 666 666 666 666 666 666 666 677 292 236 8 × 2 = 0 + 0.133 333 333 333 333 333 333 333 333 354 584 473 6;
  • 26) 0.133 333 333 333 333 333 333 333 333 354 584 473 6 × 2 = 0 + 0.266 666 666 666 666 666 666 666 666 709 168 947 2;
  • 27) 0.266 666 666 666 666 666 666 666 666 709 168 947 2 × 2 = 0 + 0.533 333 333 333 333 333 333 333 333 418 337 894 4;
  • 28) 0.533 333 333 333 333 333 333 333 333 418 337 894 4 × 2 = 1 + 0.066 666 666 666 666 666 666 666 666 836 675 788 8;
  • 29) 0.066 666 666 666 666 666 666 666 666 836 675 788 8 × 2 = 0 + 0.133 333 333 333 333 333 333 333 333 673 351 577 6;
  • 30) 0.133 333 333 333 333 333 333 333 333 673 351 577 6 × 2 = 0 + 0.266 666 666 666 666 666 666 666 667 346 703 155 2;
  • 31) 0.266 666 666 666 666 666 666 666 667 346 703 155 2 × 2 = 0 + 0.533 333 333 333 333 333 333 333 334 693 406 310 4;
  • 32) 0.533 333 333 333 333 333 333 333 334 693 406 310 4 × 2 = 1 + 0.066 666 666 666 666 666 666 666 669 386 812 620 8;
  • 33) 0.066 666 666 666 666 666 666 666 669 386 812 620 8 × 2 = 0 + 0.133 333 333 333 333 333 333 333 338 773 625 241 6;
  • 34) 0.133 333 333 333 333 333 333 333 338 773 625 241 6 × 2 = 0 + 0.266 666 666 666 666 666 666 666 677 547 250 483 2;
  • 35) 0.266 666 666 666 666 666 666 666 677 547 250 483 2 × 2 = 0 + 0.533 333 333 333 333 333 333 333 355 094 500 966 4;
  • 36) 0.533 333 333 333 333 333 333 333 355 094 500 966 4 × 2 = 1 + 0.066 666 666 666 666 666 666 666 710 189 001 932 8;
  • 37) 0.066 666 666 666 666 666 666 666 710 189 001 932 8 × 2 = 0 + 0.133 333 333 333 333 333 333 333 420 378 003 865 6;
  • 38) 0.133 333 333 333 333 333 333 333 420 378 003 865 6 × 2 = 0 + 0.266 666 666 666 666 666 666 666 840 756 007 731 2;
  • 39) 0.266 666 666 666 666 666 666 666 840 756 007 731 2 × 2 = 0 + 0.533 333 333 333 333 333 333 333 681 512 015 462 4;
  • 40) 0.533 333 333 333 333 333 333 333 681 512 015 462 4 × 2 = 1 + 0.066 666 666 666 666 666 666 667 363 024 030 924 8;
  • 41) 0.066 666 666 666 666 666 666 667 363 024 030 924 8 × 2 = 0 + 0.133 333 333 333 333 333 333 334 726 048 061 849 6;
  • 42) 0.133 333 333 333 333 333 333 334 726 048 061 849 6 × 2 = 0 + 0.266 666 666 666 666 666 666 669 452 096 123 699 2;
  • 43) 0.266 666 666 666 666 666 666 669 452 096 123 699 2 × 2 = 0 + 0.533 333 333 333 333 333 333 338 904 192 247 398 4;
  • 44) 0.533 333 333 333 333 333 333 338 904 192 247 398 4 × 2 = 1 + 0.066 666 666 666 666 666 666 677 808 384 494 796 8;
  • 45) 0.066 666 666 666 666 666 666 677 808 384 494 796 8 × 2 = 0 + 0.133 333 333 333 333 333 333 355 616 768 989 593 6;
  • 46) 0.133 333 333 333 333 333 333 355 616 768 989 593 6 × 2 = 0 + 0.266 666 666 666 666 666 666 711 233 537 979 187 2;
  • 47) 0.266 666 666 666 666 666 666 711 233 537 979 187 2 × 2 = 0 + 0.533 333 333 333 333 333 333 422 467 075 958 374 4;
  • 48) 0.533 333 333 333 333 333 333 422 467 075 958 374 4 × 2 = 1 + 0.066 666 666 666 666 666 666 844 934 151 916 748 8;
  • 49) 0.066 666 666 666 666 666 666 844 934 151 916 748 8 × 2 = 0 + 0.133 333 333 333 333 333 333 689 868 303 833 497 6;
  • 50) 0.133 333 333 333 333 333 333 689 868 303 833 497 6 × 2 = 0 + 0.266 666 666 666 666 666 667 379 736 607 666 995 2;
  • 51) 0.266 666 666 666 666 666 667 379 736 607 666 995 2 × 2 = 0 + 0.533 333 333 333 333 333 334 759 473 215 333 990 4;
  • 52) 0.533 333 333 333 333 333 334 759 473 215 333 990 4 × 2 = 1 + 0.066 666 666 666 666 666 669 518 946 430 667 980 8;
  • 53) 0.066 666 666 666 666 666 669 518 946 430 667 980 8 × 2 = 0 + 0.133 333 333 333 333 333 339 037 892 861 335 961 6;
  • 54) 0.133 333 333 333 333 333 339 037 892 861 335 961 6 × 2 = 0 + 0.266 666 666 666 666 666 678 075 785 722 671 923 2;
  • 55) 0.266 666 666 666 666 666 678 075 785 722 671 923 2 × 2 = 0 + 0.533 333 333 333 333 333 356 151 571 445 343 846 4;
  • 56) 0.533 333 333 333 333 333 356 151 571 445 343 846 4 × 2 = 1 + 0.066 666 666 666 666 666 712 303 142 890 687 692 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 - 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:


0.066 666 666 666 666 666 666 666 666 666 666 667 3(10) =

0.0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001(2)

5. Positive number before normalization:

0.066 666 666 666 666 666 666 666 666 666 666 667 3(10) =

0.0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.066 666 666 666 666 666 666 666 666 666 666 667 3(10) =

0.0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001(2) =

0.0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001(2) × 20 =

1.0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001(2) × 2-4


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): -4

Mantissa (not normalized):
1.0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 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;

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) =

1019(10) =

011 1111 1011(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. 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 =

0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001


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 1011

Mantissa (52 bits) =
0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001


Decimal number 0.066 666 666 666 666 666 666 666 666 666 666 667 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1011 - 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001

» Convert decimal 0.066 666 666 666 666 666 666 666 666 666 666 670 6 to 64 bit double precision IEEE 754 binary floating point representation standard

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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