64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 0.805 245 165 974 627 154 089 760 3 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 0.805 245 165 974 627 154 089 760 3(10) converted and written in 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: 0.805 245 165 974 627 154 089 760 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) 0.805 245 165 974 627 154 089 760 3 × 2 = 1 + 0.610 490 331 949 254 308 179 520 6;
  • 2) 0.610 490 331 949 254 308 179 520 6 × 2 = 1 + 0.220 980 663 898 508 616 359 041 2;
  • 3) 0.220 980 663 898 508 616 359 041 2 × 2 = 0 + 0.441 961 327 797 017 232 718 082 4;
  • 4) 0.441 961 327 797 017 232 718 082 4 × 2 = 0 + 0.883 922 655 594 034 465 436 164 8;
  • 5) 0.883 922 655 594 034 465 436 164 8 × 2 = 1 + 0.767 845 311 188 068 930 872 329 6;
  • 6) 0.767 845 311 188 068 930 872 329 6 × 2 = 1 + 0.535 690 622 376 137 861 744 659 2;
  • 7) 0.535 690 622 376 137 861 744 659 2 × 2 = 1 + 0.071 381 244 752 275 723 489 318 4;
  • 8) 0.071 381 244 752 275 723 489 318 4 × 2 = 0 + 0.142 762 489 504 551 446 978 636 8;
  • 9) 0.142 762 489 504 551 446 978 636 8 × 2 = 0 + 0.285 524 979 009 102 893 957 273 6;
  • 10) 0.285 524 979 009 102 893 957 273 6 × 2 = 0 + 0.571 049 958 018 205 787 914 547 2;
  • 11) 0.571 049 958 018 205 787 914 547 2 × 2 = 1 + 0.142 099 916 036 411 575 829 094 4;
  • 12) 0.142 099 916 036 411 575 829 094 4 × 2 = 0 + 0.284 199 832 072 823 151 658 188 8;
  • 13) 0.284 199 832 072 823 151 658 188 8 × 2 = 0 + 0.568 399 664 145 646 303 316 377 6;
  • 14) 0.568 399 664 145 646 303 316 377 6 × 2 = 1 + 0.136 799 328 291 292 606 632 755 2;
  • 15) 0.136 799 328 291 292 606 632 755 2 × 2 = 0 + 0.273 598 656 582 585 213 265 510 4;
  • 16) 0.273 598 656 582 585 213 265 510 4 × 2 = 0 + 0.547 197 313 165 170 426 531 020 8;
  • 17) 0.547 197 313 165 170 426 531 020 8 × 2 = 1 + 0.094 394 626 330 340 853 062 041 6;
  • 18) 0.094 394 626 330 340 853 062 041 6 × 2 = 0 + 0.188 789 252 660 681 706 124 083 2;
  • 19) 0.188 789 252 660 681 706 124 083 2 × 2 = 0 + 0.377 578 505 321 363 412 248 166 4;
  • 20) 0.377 578 505 321 363 412 248 166 4 × 2 = 0 + 0.755 157 010 642 726 824 496 332 8;
  • 21) 0.755 157 010 642 726 824 496 332 8 × 2 = 1 + 0.510 314 021 285 453 648 992 665 6;
  • 22) 0.510 314 021 285 453 648 992 665 6 × 2 = 1 + 0.020 628 042 570 907 297 985 331 2;
  • 23) 0.020 628 042 570 907 297 985 331 2 × 2 = 0 + 0.041 256 085 141 814 595 970 662 4;
  • 24) 0.041 256 085 141 814 595 970 662 4 × 2 = 0 + 0.082 512 170 283 629 191 941 324 8;
  • 25) 0.082 512 170 283 629 191 941 324 8 × 2 = 0 + 0.165 024 340 567 258 383 882 649 6;
  • 26) 0.165 024 340 567 258 383 882 649 6 × 2 = 0 + 0.330 048 681 134 516 767 765 299 2;
  • 27) 0.330 048 681 134 516 767 765 299 2 × 2 = 0 + 0.660 097 362 269 033 535 530 598 4;
  • 28) 0.660 097 362 269 033 535 530 598 4 × 2 = 1 + 0.320 194 724 538 067 071 061 196 8;
  • 29) 0.320 194 724 538 067 071 061 196 8 × 2 = 0 + 0.640 389 449 076 134 142 122 393 6;
  • 30) 0.640 389 449 076 134 142 122 393 6 × 2 = 1 + 0.280 778 898 152 268 284 244 787 2;
  • 31) 0.280 778 898 152 268 284 244 787 2 × 2 = 0 + 0.561 557 796 304 536 568 489 574 4;
  • 32) 0.561 557 796 304 536 568 489 574 4 × 2 = 1 + 0.123 115 592 609 073 136 979 148 8;
  • 33) 0.123 115 592 609 073 136 979 148 8 × 2 = 0 + 0.246 231 185 218 146 273 958 297 6;
  • 34) 0.246 231 185 218 146 273 958 297 6 × 2 = 0 + 0.492 462 370 436 292 547 916 595 2;
  • 35) 0.492 462 370 436 292 547 916 595 2 × 2 = 0 + 0.984 924 740 872 585 095 833 190 4;
  • 36) 0.984 924 740 872 585 095 833 190 4 × 2 = 1 + 0.969 849 481 745 170 191 666 380 8;
  • 37) 0.969 849 481 745 170 191 666 380 8 × 2 = 1 + 0.939 698 963 490 340 383 332 761 6;
  • 38) 0.939 698 963 490 340 383 332 761 6 × 2 = 1 + 0.879 397 926 980 680 766 665 523 2;
  • 39) 0.879 397 926 980 680 766 665 523 2 × 2 = 1 + 0.758 795 853 961 361 533 331 046 4;
  • 40) 0.758 795 853 961 361 533 331 046 4 × 2 = 1 + 0.517 591 707 922 723 066 662 092 8;
  • 41) 0.517 591 707 922 723 066 662 092 8 × 2 = 1 + 0.035 183 415 845 446 133 324 185 6;
  • 42) 0.035 183 415 845 446 133 324 185 6 × 2 = 0 + 0.070 366 831 690 892 266 648 371 2;
  • 43) 0.070 366 831 690 892 266 648 371 2 × 2 = 0 + 0.140 733 663 381 784 533 296 742 4;
  • 44) 0.140 733 663 381 784 533 296 742 4 × 2 = 0 + 0.281 467 326 763 569 066 593 484 8;
  • 45) 0.281 467 326 763 569 066 593 484 8 × 2 = 0 + 0.562 934 653 527 138 133 186 969 6;
  • 46) 0.562 934 653 527 138 133 186 969 6 × 2 = 1 + 0.125 869 307 054 276 266 373 939 2;
  • 47) 0.125 869 307 054 276 266 373 939 2 × 2 = 0 + 0.251 738 614 108 552 532 747 878 4;
  • 48) 0.251 738 614 108 552 532 747 878 4 × 2 = 0 + 0.503 477 228 217 105 065 495 756 8;
  • 49) 0.503 477 228 217 105 065 495 756 8 × 2 = 1 + 0.006 954 456 434 210 130 991 513 6;
  • 50) 0.006 954 456 434 210 130 991 513 6 × 2 = 0 + 0.013 908 912 868 420 261 983 027 2;
  • 51) 0.013 908 912 868 420 261 983 027 2 × 2 = 0 + 0.027 817 825 736 840 523 966 054 4;
  • 52) 0.027 817 825 736 840 523 966 054 4 × 2 = 0 + 0.055 635 651 473 681 047 932 108 8;
  • 53) 0.055 635 651 473 681 047 932 108 8 × 2 = 0 + 0.111 271 302 947 362 095 864 217 6;

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...)


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:


0.805 245 165 974 627 154 089 760 3(10) =

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


5. Positive number before normalization:

0.805 245 165 974 627 154 089 760 3(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:


0.805 245 165 974 627 154 089 760 3(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


The base ten decimal number 0.805 245 165 974 627 154 089 760 3 converted and written in 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

More operations with decimal numbers converted to 64 bit double precision IEEE 754 binary floating point representation:

» Number 0.805 245 165 974 627 154 089 760 2 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

» Number 0.805 245 165 974 627 154 089 760 4 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

The latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

Number 110 000 010 001 010 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 013 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Apr 18 12:24 UTC (GMT)
Number 123 791 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Apr 18 12:24 UTC (GMT)
Number 4 294 171 488 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Apr 18 12:24 UTC (GMT)
Number 345.267 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Apr 18 12:24 UTC (GMT)
Number 0.000 000 000 000 000 013 248 735 989 328 012 498 123 389 7 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Apr 18 12:24 UTC (GMT)
Number 1 152 921 508 901 814 266 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Apr 18 12:24 UTC (GMT)
Number 9 838 263 505 978 427 527 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Apr 18 12:24 UTC (GMT)
Number 6 177 218 022 601 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Apr 18 12:24 UTC (GMT)
Number 7 061 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Apr 18 12:24 UTC (GMT)
Number 62 156 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Apr 18 12:24 UTC (GMT)
All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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