0.555 555 555 555 555 580 227 178 325 002 78 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.555 555 555 555 555 580 227 178 325 002 78(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
0.555 555 555 555 555 580 227 178 325 002 78(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: 0.555 555 555 555 555 580 227 178 325 002 78.

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) 0.555 555 555 555 555 580 227 178 325 002 78 × 2 = 1 + 0.111 111 111 111 111 160 454 356 650 005 56;
  • 2) 0.111 111 111 111 111 160 454 356 650 005 56 × 2 = 0 + 0.222 222 222 222 222 320 908 713 300 011 12;
  • 3) 0.222 222 222 222 222 320 908 713 300 011 12 × 2 = 0 + 0.444 444 444 444 444 641 817 426 600 022 24;
  • 4) 0.444 444 444 444 444 641 817 426 600 022 24 × 2 = 0 + 0.888 888 888 888 889 283 634 853 200 044 48;
  • 5) 0.888 888 888 888 889 283 634 853 200 044 48 × 2 = 1 + 0.777 777 777 777 778 567 269 706 400 088 96;
  • 6) 0.777 777 777 777 778 567 269 706 400 088 96 × 2 = 1 + 0.555 555 555 555 557 134 539 412 800 177 92;
  • 7) 0.555 555 555 555 557 134 539 412 800 177 92 × 2 = 1 + 0.111 111 111 111 114 269 078 825 600 355 84;
  • 8) 0.111 111 111 111 114 269 078 825 600 355 84 × 2 = 0 + 0.222 222 222 222 228 538 157 651 200 711 68;
  • 9) 0.222 222 222 222 228 538 157 651 200 711 68 × 2 = 0 + 0.444 444 444 444 457 076 315 302 401 423 36;
  • 10) 0.444 444 444 444 457 076 315 302 401 423 36 × 2 = 0 + 0.888 888 888 888 914 152 630 604 802 846 72;
  • 11) 0.888 888 888 888 914 152 630 604 802 846 72 × 2 = 1 + 0.777 777 777 777 828 305 261 209 605 693 44;
  • 12) 0.777 777 777 777 828 305 261 209 605 693 44 × 2 = 1 + 0.555 555 555 555 656 610 522 419 211 386 88;
  • 13) 0.555 555 555 555 656 610 522 419 211 386 88 × 2 = 1 + 0.111 111 111 111 313 221 044 838 422 773 76;
  • 14) 0.111 111 111 111 313 221 044 838 422 773 76 × 2 = 0 + 0.222 222 222 222 626 442 089 676 845 547 52;
  • 15) 0.222 222 222 222 626 442 089 676 845 547 52 × 2 = 0 + 0.444 444 444 445 252 884 179 353 691 095 04;
  • 16) 0.444 444 444 445 252 884 179 353 691 095 04 × 2 = 0 + 0.888 888 888 890 505 768 358 707 382 190 08;
  • 17) 0.888 888 888 890 505 768 358 707 382 190 08 × 2 = 1 + 0.777 777 777 781 011 536 717 414 764 380 16;
  • 18) 0.777 777 777 781 011 536 717 414 764 380 16 × 2 = 1 + 0.555 555 555 562 023 073 434 829 528 760 32;
  • 19) 0.555 555 555 562 023 073 434 829 528 760 32 × 2 = 1 + 0.111 111 111 124 046 146 869 659 057 520 64;
  • 20) 0.111 111 111 124 046 146 869 659 057 520 64 × 2 = 0 + 0.222 222 222 248 092 293 739 318 115 041 28;
  • 21) 0.222 222 222 248 092 293 739 318 115 041 28 × 2 = 0 + 0.444 444 444 496 184 587 478 636 230 082 56;
  • 22) 0.444 444 444 496 184 587 478 636 230 082 56 × 2 = 0 + 0.888 888 888 992 369 174 957 272 460 165 12;
  • 23) 0.888 888 888 992 369 174 957 272 460 165 12 × 2 = 1 + 0.777 777 777 984 738 349 914 544 920 330 24;
  • 24) 0.777 777 777 984 738 349 914 544 920 330 24 × 2 = 1 + 0.555 555 555 969 476 699 829 089 840 660 48;
  • 25) 0.555 555 555 969 476 699 829 089 840 660 48 × 2 = 1 + 0.111 111 111 938 953 399 658 179 681 320 96;
  • 26) 0.111 111 111 938 953 399 658 179 681 320 96 × 2 = 0 + 0.222 222 223 877 906 799 316 359 362 641 92;
  • 27) 0.222 222 223 877 906 799 316 359 362 641 92 × 2 = 0 + 0.444 444 447 755 813 598 632 718 725 283 84;
  • 28) 0.444 444 447 755 813 598 632 718 725 283 84 × 2 = 0 + 0.888 888 895 511 627 197 265 437 450 567 68;
  • 29) 0.888 888 895 511 627 197 265 437 450 567 68 × 2 = 1 + 0.777 777 791 023 254 394 530 874 901 135 36;
  • 30) 0.777 777 791 023 254 394 530 874 901 135 36 × 2 = 1 + 0.555 555 582 046 508 789 061 749 802 270 72;
  • 31) 0.555 555 582 046 508 789 061 749 802 270 72 × 2 = 1 + 0.111 111 164 093 017 578 123 499 604 541 44;
  • 32) 0.111 111 164 093 017 578 123 499 604 541 44 × 2 = 0 + 0.222 222 328 186 035 156 246 999 209 082 88;
  • 33) 0.222 222 328 186 035 156 246 999 209 082 88 × 2 = 0 + 0.444 444 656 372 070 312 493 998 418 165 76;
  • 34) 0.444 444 656 372 070 312 493 998 418 165 76 × 2 = 0 + 0.888 889 312 744 140 624 987 996 836 331 52;
  • 35) 0.888 889 312 744 140 624 987 996 836 331 52 × 2 = 1 + 0.777 778 625 488 281 249 975 993 672 663 04;
  • 36) 0.777 778 625 488 281 249 975 993 672 663 04 × 2 = 1 + 0.555 557 250 976 562 499 951 987 345 326 08;
  • 37) 0.555 557 250 976 562 499 951 987 345 326 08 × 2 = 1 + 0.111 114 501 953 124 999 903 974 690 652 16;
  • 38) 0.111 114 501 953 124 999 903 974 690 652 16 × 2 = 0 + 0.222 229 003 906 249 999 807 949 381 304 32;
  • 39) 0.222 229 003 906 249 999 807 949 381 304 32 × 2 = 0 + 0.444 458 007 812 499 999 615 898 762 608 64;
  • 40) 0.444 458 007 812 499 999 615 898 762 608 64 × 2 = 0 + 0.888 916 015 624 999 999 231 797 525 217 28;
  • 41) 0.888 916 015 624 999 999 231 797 525 217 28 × 2 = 1 + 0.777 832 031 249 999 998 463 595 050 434 56;
  • 42) 0.777 832 031 249 999 998 463 595 050 434 56 × 2 = 1 + 0.555 664 062 499 999 996 927 190 100 869 12;
  • 43) 0.555 664 062 499 999 996 927 190 100 869 12 × 2 = 1 + 0.111 328 124 999 999 993 854 380 201 738 24;
  • 44) 0.111 328 124 999 999 993 854 380 201 738 24 × 2 = 0 + 0.222 656 249 999 999 987 708 760 403 476 48;
  • 45) 0.222 656 249 999 999 987 708 760 403 476 48 × 2 = 0 + 0.445 312 499 999 999 975 417 520 806 952 96;
  • 46) 0.445 312 499 999 999 975 417 520 806 952 96 × 2 = 0 + 0.890 624 999 999 999 950 835 041 613 905 92;
  • 47) 0.890 624 999 999 999 950 835 041 613 905 92 × 2 = 1 + 0.781 249 999 999 999 901 670 083 227 811 84;
  • 48) 0.781 249 999 999 999 901 670 083 227 811 84 × 2 = 1 + 0.562 499 999 999 999 803 340 166 455 623 68;
  • 49) 0.562 499 999 999 999 803 340 166 455 623 68 × 2 = 1 + 0.124 999 999 999 999 606 680 332 911 247 36;
  • 50) 0.124 999 999 999 999 606 680 332 911 247 36 × 2 = 0 + 0.249 999 999 999 999 213 360 665 822 494 72;
  • 51) 0.249 999 999 999 999 213 360 665 822 494 72 × 2 = 0 + 0.499 999 999 999 998 426 721 331 644 989 44;
  • 52) 0.499 999 999 999 998 426 721 331 644 989 44 × 2 = 0 + 0.999 999 999 999 996 853 442 663 289 978 88;
  • 53) 0.999 999 999 999 996 853 442 663 289 978 88 × 2 = 1 + 0.999 999 999 999 993 706 885 326 579 957 76;

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:


0.555 555 555 555 555 580 227 178 325 002 78(10) =

0.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1(2)

5. Positive number before normalization:

0.555 555 555 555 555 580 227 178 325 002 78(10) =

0.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1(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:


0.555 555 555 555 555 580 227 178 325 002 78(10) =

0.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1(2) =

0.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1(2) × 20 =

1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001(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):
1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001


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. 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 =

0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 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 1110

Mantissa (52 bits) =
0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001


Decimal number 0.555 555 555 555 555 580 227 178 325 002 78 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001

» Convert decimal 0.555 555 555 555 555 580 227 178 325 003 71 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