0.805 245 165 974 627 154 089 01 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.805 245 165 974 627 154 089 01(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
0.805 245 165 974 627 154 089 01(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: 0.805 245 165 974 627 154 089 01.

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) 0.805 245 165 974 627 154 089 01 × 2 = 1 + 0.610 490 331 949 254 308 178 02;
  • 2) 0.610 490 331 949 254 308 178 02 × 2 = 1 + 0.220 980 663 898 508 616 356 04;
  • 3) 0.220 980 663 898 508 616 356 04 × 2 = 0 + 0.441 961 327 797 017 232 712 08;
  • 4) 0.441 961 327 797 017 232 712 08 × 2 = 0 + 0.883 922 655 594 034 465 424 16;
  • 5) 0.883 922 655 594 034 465 424 16 × 2 = 1 + 0.767 845 311 188 068 930 848 32;
  • 6) 0.767 845 311 188 068 930 848 32 × 2 = 1 + 0.535 690 622 376 137 861 696 64;
  • 7) 0.535 690 622 376 137 861 696 64 × 2 = 1 + 0.071 381 244 752 275 723 393 28;
  • 8) 0.071 381 244 752 275 723 393 28 × 2 = 0 + 0.142 762 489 504 551 446 786 56;
  • 9) 0.142 762 489 504 551 446 786 56 × 2 = 0 + 0.285 524 979 009 102 893 573 12;
  • 10) 0.285 524 979 009 102 893 573 12 × 2 = 0 + 0.571 049 958 018 205 787 146 24;
  • 11) 0.571 049 958 018 205 787 146 24 × 2 = 1 + 0.142 099 916 036 411 574 292 48;
  • 12) 0.142 099 916 036 411 574 292 48 × 2 = 0 + 0.284 199 832 072 823 148 584 96;
  • 13) 0.284 199 832 072 823 148 584 96 × 2 = 0 + 0.568 399 664 145 646 297 169 92;
  • 14) 0.568 399 664 145 646 297 169 92 × 2 = 1 + 0.136 799 328 291 292 594 339 84;
  • 15) 0.136 799 328 291 292 594 339 84 × 2 = 0 + 0.273 598 656 582 585 188 679 68;
  • 16) 0.273 598 656 582 585 188 679 68 × 2 = 0 + 0.547 197 313 165 170 377 359 36;
  • 17) 0.547 197 313 165 170 377 359 36 × 2 = 1 + 0.094 394 626 330 340 754 718 72;
  • 18) 0.094 394 626 330 340 754 718 72 × 2 = 0 + 0.188 789 252 660 681 509 437 44;
  • 19) 0.188 789 252 660 681 509 437 44 × 2 = 0 + 0.377 578 505 321 363 018 874 88;
  • 20) 0.377 578 505 321 363 018 874 88 × 2 = 0 + 0.755 157 010 642 726 037 749 76;
  • 21) 0.755 157 010 642 726 037 749 76 × 2 = 1 + 0.510 314 021 285 452 075 499 52;
  • 22) 0.510 314 021 285 452 075 499 52 × 2 = 1 + 0.020 628 042 570 904 150 999 04;
  • 23) 0.020 628 042 570 904 150 999 04 × 2 = 0 + 0.041 256 085 141 808 301 998 08;
  • 24) 0.041 256 085 141 808 301 998 08 × 2 = 0 + 0.082 512 170 283 616 603 996 16;
  • 25) 0.082 512 170 283 616 603 996 16 × 2 = 0 + 0.165 024 340 567 233 207 992 32;
  • 26) 0.165 024 340 567 233 207 992 32 × 2 = 0 + 0.330 048 681 134 466 415 984 64;
  • 27) 0.330 048 681 134 466 415 984 64 × 2 = 0 + 0.660 097 362 268 932 831 969 28;
  • 28) 0.660 097 362 268 932 831 969 28 × 2 = 1 + 0.320 194 724 537 865 663 938 56;
  • 29) 0.320 194 724 537 865 663 938 56 × 2 = 0 + 0.640 389 449 075 731 327 877 12;
  • 30) 0.640 389 449 075 731 327 877 12 × 2 = 1 + 0.280 778 898 151 462 655 754 24;
  • 31) 0.280 778 898 151 462 655 754 24 × 2 = 0 + 0.561 557 796 302 925 311 508 48;
  • 32) 0.561 557 796 302 925 311 508 48 × 2 = 1 + 0.123 115 592 605 850 623 016 96;
  • 33) 0.123 115 592 605 850 623 016 96 × 2 = 0 + 0.246 231 185 211 701 246 033 92;
  • 34) 0.246 231 185 211 701 246 033 92 × 2 = 0 + 0.492 462 370 423 402 492 067 84;
  • 35) 0.492 462 370 423 402 492 067 84 × 2 = 0 + 0.984 924 740 846 804 984 135 68;
  • 36) 0.984 924 740 846 804 984 135 68 × 2 = 1 + 0.969 849 481 693 609 968 271 36;
  • 37) 0.969 849 481 693 609 968 271 36 × 2 = 1 + 0.939 698 963 387 219 936 542 72;
  • 38) 0.939 698 963 387 219 936 542 72 × 2 = 1 + 0.879 397 926 774 439 873 085 44;
  • 39) 0.879 397 926 774 439 873 085 44 × 2 = 1 + 0.758 795 853 548 879 746 170 88;
  • 40) 0.758 795 853 548 879 746 170 88 × 2 = 1 + 0.517 591 707 097 759 492 341 76;
  • 41) 0.517 591 707 097 759 492 341 76 × 2 = 1 + 0.035 183 414 195 518 984 683 52;
  • 42) 0.035 183 414 195 518 984 683 52 × 2 = 0 + 0.070 366 828 391 037 969 367 04;
  • 43) 0.070 366 828 391 037 969 367 04 × 2 = 0 + 0.140 733 656 782 075 938 734 08;
  • 44) 0.140 733 656 782 075 938 734 08 × 2 = 0 + 0.281 467 313 564 151 877 468 16;
  • 45) 0.281 467 313 564 151 877 468 16 × 2 = 0 + 0.562 934 627 128 303 754 936 32;
  • 46) 0.562 934 627 128 303 754 936 32 × 2 = 1 + 0.125 869 254 256 607 509 872 64;
  • 47) 0.125 869 254 256 607 509 872 64 × 2 = 0 + 0.251 738 508 513 215 019 745 28;
  • 48) 0.251 738 508 513 215 019 745 28 × 2 = 0 + 0.503 477 017 026 430 039 490 56;
  • 49) 0.503 477 017 026 430 039 490 56 × 2 = 1 + 0.006 954 034 052 860 078 981 12;
  • 50) 0.006 954 034 052 860 078 981 12 × 2 = 0 + 0.013 908 068 105 720 157 962 24;
  • 51) 0.013 908 068 105 720 157 962 24 × 2 = 0 + 0.027 816 136 211 440 315 924 48;
  • 52) 0.027 816 136 211 440 315 924 48 × 2 = 0 + 0.055 632 272 422 880 631 848 96;
  • 53) 0.055 632 272 422 880 631 848 96 × 2 = 0 + 0.111 264 544 845 761 263 697 92;

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:


0.805 245 165 974 627 154 089 01(10) =

0.1100 1110 0010 0100 1000 1100 0001 0101 0001 1111 1000 0100 1000 0(2)

5. Positive number before normalization:

0.805 245 165 974 627 154 089 01(10) =

0.1100 1110 0010 0100 1000 1100 0001 0101 0001 1111 1000 0100 1000 0(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:


0.805 245 165 974 627 154 089 01(10) =

0.1100 1110 0010 0100 1000 1100 0001 0101 0001 1111 1000 0100 1000 0(2) =

0.1100 1110 0010 0100 1000 1100 0001 0101 0001 1111 1000 0100 1000 0(2) × 20 =

1.1001 1100 0100 1001 0001 1000 0010 1010 0011 1111 0000 1001 0000(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):
1.1001 1100 0100 1001 0001 1000 0010 1010 0011 1111 0000 1001 0000


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. 1001 1100 0100 1001 0001 1000 0010 1010 0011 1111 0000 1001 0000 =

1001 1100 0100 1001 0001 1000 0010 1010 0011 1111 0000 1001 0000


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) =
1001 1100 0100 1001 0001 1000 0010 1010 0011 1111 0000 1001 0000


Decimal number 0.805 245 165 974 627 154 089 01 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1001 1100 0100 1001 0001 1000 0010 1010 0011 1111 0000 1001 0000

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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