0.805 245 165 974 627 154 089 44 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.805 245 165 974 627 154 089 44(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 44(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 44.

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 44 × 2 = 1 + 0.610 490 331 949 254 308 178 88;
  • 2) 0.610 490 331 949 254 308 178 88 × 2 = 1 + 0.220 980 663 898 508 616 357 76;
  • 3) 0.220 980 663 898 508 616 357 76 × 2 = 0 + 0.441 961 327 797 017 232 715 52;
  • 4) 0.441 961 327 797 017 232 715 52 × 2 = 0 + 0.883 922 655 594 034 465 431 04;
  • 5) 0.883 922 655 594 034 465 431 04 × 2 = 1 + 0.767 845 311 188 068 930 862 08;
  • 6) 0.767 845 311 188 068 930 862 08 × 2 = 1 + 0.535 690 622 376 137 861 724 16;
  • 7) 0.535 690 622 376 137 861 724 16 × 2 = 1 + 0.071 381 244 752 275 723 448 32;
  • 8) 0.071 381 244 752 275 723 448 32 × 2 = 0 + 0.142 762 489 504 551 446 896 64;
  • 9) 0.142 762 489 504 551 446 896 64 × 2 = 0 + 0.285 524 979 009 102 893 793 28;
  • 10) 0.285 524 979 009 102 893 793 28 × 2 = 0 + 0.571 049 958 018 205 787 586 56;
  • 11) 0.571 049 958 018 205 787 586 56 × 2 = 1 + 0.142 099 916 036 411 575 173 12;
  • 12) 0.142 099 916 036 411 575 173 12 × 2 = 0 + 0.284 199 832 072 823 150 346 24;
  • 13) 0.284 199 832 072 823 150 346 24 × 2 = 0 + 0.568 399 664 145 646 300 692 48;
  • 14) 0.568 399 664 145 646 300 692 48 × 2 = 1 + 0.136 799 328 291 292 601 384 96;
  • 15) 0.136 799 328 291 292 601 384 96 × 2 = 0 + 0.273 598 656 582 585 202 769 92;
  • 16) 0.273 598 656 582 585 202 769 92 × 2 = 0 + 0.547 197 313 165 170 405 539 84;
  • 17) 0.547 197 313 165 170 405 539 84 × 2 = 1 + 0.094 394 626 330 340 811 079 68;
  • 18) 0.094 394 626 330 340 811 079 68 × 2 = 0 + 0.188 789 252 660 681 622 159 36;
  • 19) 0.188 789 252 660 681 622 159 36 × 2 = 0 + 0.377 578 505 321 363 244 318 72;
  • 20) 0.377 578 505 321 363 244 318 72 × 2 = 0 + 0.755 157 010 642 726 488 637 44;
  • 21) 0.755 157 010 642 726 488 637 44 × 2 = 1 + 0.510 314 021 285 452 977 274 88;
  • 22) 0.510 314 021 285 452 977 274 88 × 2 = 1 + 0.020 628 042 570 905 954 549 76;
  • 23) 0.020 628 042 570 905 954 549 76 × 2 = 0 + 0.041 256 085 141 811 909 099 52;
  • 24) 0.041 256 085 141 811 909 099 52 × 2 = 0 + 0.082 512 170 283 623 818 199 04;
  • 25) 0.082 512 170 283 623 818 199 04 × 2 = 0 + 0.165 024 340 567 247 636 398 08;
  • 26) 0.165 024 340 567 247 636 398 08 × 2 = 0 + 0.330 048 681 134 495 272 796 16;
  • 27) 0.330 048 681 134 495 272 796 16 × 2 = 0 + 0.660 097 362 268 990 545 592 32;
  • 28) 0.660 097 362 268 990 545 592 32 × 2 = 1 + 0.320 194 724 537 981 091 184 64;
  • 29) 0.320 194 724 537 981 091 184 64 × 2 = 0 + 0.640 389 449 075 962 182 369 28;
  • 30) 0.640 389 449 075 962 182 369 28 × 2 = 1 + 0.280 778 898 151 924 364 738 56;
  • 31) 0.280 778 898 151 924 364 738 56 × 2 = 0 + 0.561 557 796 303 848 729 477 12;
  • 32) 0.561 557 796 303 848 729 477 12 × 2 = 1 + 0.123 115 592 607 697 458 954 24;
  • 33) 0.123 115 592 607 697 458 954 24 × 2 = 0 + 0.246 231 185 215 394 917 908 48;
  • 34) 0.246 231 185 215 394 917 908 48 × 2 = 0 + 0.492 462 370 430 789 835 816 96;
  • 35) 0.492 462 370 430 789 835 816 96 × 2 = 0 + 0.984 924 740 861 579 671 633 92;
  • 36) 0.984 924 740 861 579 671 633 92 × 2 = 1 + 0.969 849 481 723 159 343 267 84;
  • 37) 0.969 849 481 723 159 343 267 84 × 2 = 1 + 0.939 698 963 446 318 686 535 68;
  • 38) 0.939 698 963 446 318 686 535 68 × 2 = 1 + 0.879 397 926 892 637 373 071 36;
  • 39) 0.879 397 926 892 637 373 071 36 × 2 = 1 + 0.758 795 853 785 274 746 142 72;
  • 40) 0.758 795 853 785 274 746 142 72 × 2 = 1 + 0.517 591 707 570 549 492 285 44;
  • 41) 0.517 591 707 570 549 492 285 44 × 2 = 1 + 0.035 183 415 141 098 984 570 88;
  • 42) 0.035 183 415 141 098 984 570 88 × 2 = 0 + 0.070 366 830 282 197 969 141 76;
  • 43) 0.070 366 830 282 197 969 141 76 × 2 = 0 + 0.140 733 660 564 395 938 283 52;
  • 44) 0.140 733 660 564 395 938 283 52 × 2 = 0 + 0.281 467 321 128 791 876 567 04;
  • 45) 0.281 467 321 128 791 876 567 04 × 2 = 0 + 0.562 934 642 257 583 753 134 08;
  • 46) 0.562 934 642 257 583 753 134 08 × 2 = 1 + 0.125 869 284 515 167 506 268 16;
  • 47) 0.125 869 284 515 167 506 268 16 × 2 = 0 + 0.251 738 569 030 335 012 536 32;
  • 48) 0.251 738 569 030 335 012 536 32 × 2 = 0 + 0.503 477 138 060 670 025 072 64;
  • 49) 0.503 477 138 060 670 025 072 64 × 2 = 1 + 0.006 954 276 121 340 050 145 28;
  • 50) 0.006 954 276 121 340 050 145 28 × 2 = 0 + 0.013 908 552 242 680 100 290 56;
  • 51) 0.013 908 552 242 680 100 290 56 × 2 = 0 + 0.027 817 104 485 360 200 581 12;
  • 52) 0.027 817 104 485 360 200 581 12 × 2 = 0 + 0.055 634 208 970 720 401 162 24;
  • 53) 0.055 634 208 970 720 401 162 24 × 2 = 0 + 0.111 268 417 941 440 802 324 48;

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 44(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 44(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 44(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 44 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

» Convert decimal 0.805 245 165 974 627 154 089 01 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