0.805 245 165 974 627 154 089 62 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.805 245 165 974 627 154 089 62(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 62(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 62.

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 62 × 2 = 1 + 0.610 490 331 949 254 308 179 24;
  • 2) 0.610 490 331 949 254 308 179 24 × 2 = 1 + 0.220 980 663 898 508 616 358 48;
  • 3) 0.220 980 663 898 508 616 358 48 × 2 = 0 + 0.441 961 327 797 017 232 716 96;
  • 4) 0.441 961 327 797 017 232 716 96 × 2 = 0 + 0.883 922 655 594 034 465 433 92;
  • 5) 0.883 922 655 594 034 465 433 92 × 2 = 1 + 0.767 845 311 188 068 930 867 84;
  • 6) 0.767 845 311 188 068 930 867 84 × 2 = 1 + 0.535 690 622 376 137 861 735 68;
  • 7) 0.535 690 622 376 137 861 735 68 × 2 = 1 + 0.071 381 244 752 275 723 471 36;
  • 8) 0.071 381 244 752 275 723 471 36 × 2 = 0 + 0.142 762 489 504 551 446 942 72;
  • 9) 0.142 762 489 504 551 446 942 72 × 2 = 0 + 0.285 524 979 009 102 893 885 44;
  • 10) 0.285 524 979 009 102 893 885 44 × 2 = 0 + 0.571 049 958 018 205 787 770 88;
  • 11) 0.571 049 958 018 205 787 770 88 × 2 = 1 + 0.142 099 916 036 411 575 541 76;
  • 12) 0.142 099 916 036 411 575 541 76 × 2 = 0 + 0.284 199 832 072 823 151 083 52;
  • 13) 0.284 199 832 072 823 151 083 52 × 2 = 0 + 0.568 399 664 145 646 302 167 04;
  • 14) 0.568 399 664 145 646 302 167 04 × 2 = 1 + 0.136 799 328 291 292 604 334 08;
  • 15) 0.136 799 328 291 292 604 334 08 × 2 = 0 + 0.273 598 656 582 585 208 668 16;
  • 16) 0.273 598 656 582 585 208 668 16 × 2 = 0 + 0.547 197 313 165 170 417 336 32;
  • 17) 0.547 197 313 165 170 417 336 32 × 2 = 1 + 0.094 394 626 330 340 834 672 64;
  • 18) 0.094 394 626 330 340 834 672 64 × 2 = 0 + 0.188 789 252 660 681 669 345 28;
  • 19) 0.188 789 252 660 681 669 345 28 × 2 = 0 + 0.377 578 505 321 363 338 690 56;
  • 20) 0.377 578 505 321 363 338 690 56 × 2 = 0 + 0.755 157 010 642 726 677 381 12;
  • 21) 0.755 157 010 642 726 677 381 12 × 2 = 1 + 0.510 314 021 285 453 354 762 24;
  • 22) 0.510 314 021 285 453 354 762 24 × 2 = 1 + 0.020 628 042 570 906 709 524 48;
  • 23) 0.020 628 042 570 906 709 524 48 × 2 = 0 + 0.041 256 085 141 813 419 048 96;
  • 24) 0.041 256 085 141 813 419 048 96 × 2 = 0 + 0.082 512 170 283 626 838 097 92;
  • 25) 0.082 512 170 283 626 838 097 92 × 2 = 0 + 0.165 024 340 567 253 676 195 84;
  • 26) 0.165 024 340 567 253 676 195 84 × 2 = 0 + 0.330 048 681 134 507 352 391 68;
  • 27) 0.330 048 681 134 507 352 391 68 × 2 = 0 + 0.660 097 362 269 014 704 783 36;
  • 28) 0.660 097 362 269 014 704 783 36 × 2 = 1 + 0.320 194 724 538 029 409 566 72;
  • 29) 0.320 194 724 538 029 409 566 72 × 2 = 0 + 0.640 389 449 076 058 819 133 44;
  • 30) 0.640 389 449 076 058 819 133 44 × 2 = 1 + 0.280 778 898 152 117 638 266 88;
  • 31) 0.280 778 898 152 117 638 266 88 × 2 = 0 + 0.561 557 796 304 235 276 533 76;
  • 32) 0.561 557 796 304 235 276 533 76 × 2 = 1 + 0.123 115 592 608 470 553 067 52;
  • 33) 0.123 115 592 608 470 553 067 52 × 2 = 0 + 0.246 231 185 216 941 106 135 04;
  • 34) 0.246 231 185 216 941 106 135 04 × 2 = 0 + 0.492 462 370 433 882 212 270 08;
  • 35) 0.492 462 370 433 882 212 270 08 × 2 = 0 + 0.984 924 740 867 764 424 540 16;
  • 36) 0.984 924 740 867 764 424 540 16 × 2 = 1 + 0.969 849 481 735 528 849 080 32;
  • 37) 0.969 849 481 735 528 849 080 32 × 2 = 1 + 0.939 698 963 471 057 698 160 64;
  • 38) 0.939 698 963 471 057 698 160 64 × 2 = 1 + 0.879 397 926 942 115 396 321 28;
  • 39) 0.879 397 926 942 115 396 321 28 × 2 = 1 + 0.758 795 853 884 230 792 642 56;
  • 40) 0.758 795 853 884 230 792 642 56 × 2 = 1 + 0.517 591 707 768 461 585 285 12;
  • 41) 0.517 591 707 768 461 585 285 12 × 2 = 1 + 0.035 183 415 536 923 170 570 24;
  • 42) 0.035 183 415 536 923 170 570 24 × 2 = 0 + 0.070 366 831 073 846 341 140 48;
  • 43) 0.070 366 831 073 846 341 140 48 × 2 = 0 + 0.140 733 662 147 692 682 280 96;
  • 44) 0.140 733 662 147 692 682 280 96 × 2 = 0 + 0.281 467 324 295 385 364 561 92;
  • 45) 0.281 467 324 295 385 364 561 92 × 2 = 0 + 0.562 934 648 590 770 729 123 84;
  • 46) 0.562 934 648 590 770 729 123 84 × 2 = 1 + 0.125 869 297 181 541 458 247 68;
  • 47) 0.125 869 297 181 541 458 247 68 × 2 = 0 + 0.251 738 594 363 082 916 495 36;
  • 48) 0.251 738 594 363 082 916 495 36 × 2 = 0 + 0.503 477 188 726 165 832 990 72;
  • 49) 0.503 477 188 726 165 832 990 72 × 2 = 1 + 0.006 954 377 452 331 665 981 44;
  • 50) 0.006 954 377 452 331 665 981 44 × 2 = 0 + 0.013 908 754 904 663 331 962 88;
  • 51) 0.013 908 754 904 663 331 962 88 × 2 = 0 + 0.027 817 509 809 326 663 925 76;
  • 52) 0.027 817 509 809 326 663 925 76 × 2 = 0 + 0.055 635 019 618 653 327 851 52;
  • 53) 0.055 635 019 618 653 327 851 52 × 2 = 0 + 0.111 270 039 237 306 655 703 04;

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 62(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 62(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 62(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 62 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