0.620 928 906 036 742 024 296 838 734 377 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.620 928 906 036 742 024 296 838 734 377 3(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
0.620 928 906 036 742 024 296 838 734 377 3(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: 0.620 928 906 036 742 024 296 838 734 377 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) 0.620 928 906 036 742 024 296 838 734 377 3 × 2 = 1 + 0.241 857 812 073 484 048 593 677 468 754 6;
  • 2) 0.241 857 812 073 484 048 593 677 468 754 6 × 2 = 0 + 0.483 715 624 146 968 097 187 354 937 509 2;
  • 3) 0.483 715 624 146 968 097 187 354 937 509 2 × 2 = 0 + 0.967 431 248 293 936 194 374 709 875 018 4;
  • 4) 0.967 431 248 293 936 194 374 709 875 018 4 × 2 = 1 + 0.934 862 496 587 872 388 749 419 750 036 8;
  • 5) 0.934 862 496 587 872 388 749 419 750 036 8 × 2 = 1 + 0.869 724 993 175 744 777 498 839 500 073 6;
  • 6) 0.869 724 993 175 744 777 498 839 500 073 6 × 2 = 1 + 0.739 449 986 351 489 554 997 679 000 147 2;
  • 7) 0.739 449 986 351 489 554 997 679 000 147 2 × 2 = 1 + 0.478 899 972 702 979 109 995 358 000 294 4;
  • 8) 0.478 899 972 702 979 109 995 358 000 294 4 × 2 = 0 + 0.957 799 945 405 958 219 990 716 000 588 8;
  • 9) 0.957 799 945 405 958 219 990 716 000 588 8 × 2 = 1 + 0.915 599 890 811 916 439 981 432 001 177 6;
  • 10) 0.915 599 890 811 916 439 981 432 001 177 6 × 2 = 1 + 0.831 199 781 623 832 879 962 864 002 355 2;
  • 11) 0.831 199 781 623 832 879 962 864 002 355 2 × 2 = 1 + 0.662 399 563 247 665 759 925 728 004 710 4;
  • 12) 0.662 399 563 247 665 759 925 728 004 710 4 × 2 = 1 + 0.324 799 126 495 331 519 851 456 009 420 8;
  • 13) 0.324 799 126 495 331 519 851 456 009 420 8 × 2 = 0 + 0.649 598 252 990 663 039 702 912 018 841 6;
  • 14) 0.649 598 252 990 663 039 702 912 018 841 6 × 2 = 1 + 0.299 196 505 981 326 079 405 824 037 683 2;
  • 15) 0.299 196 505 981 326 079 405 824 037 683 2 × 2 = 0 + 0.598 393 011 962 652 158 811 648 075 366 4;
  • 16) 0.598 393 011 962 652 158 811 648 075 366 4 × 2 = 1 + 0.196 786 023 925 304 317 623 296 150 732 8;
  • 17) 0.196 786 023 925 304 317 623 296 150 732 8 × 2 = 0 + 0.393 572 047 850 608 635 246 592 301 465 6;
  • 18) 0.393 572 047 850 608 635 246 592 301 465 6 × 2 = 0 + 0.787 144 095 701 217 270 493 184 602 931 2;
  • 19) 0.787 144 095 701 217 270 493 184 602 931 2 × 2 = 1 + 0.574 288 191 402 434 540 986 369 205 862 4;
  • 20) 0.574 288 191 402 434 540 986 369 205 862 4 × 2 = 1 + 0.148 576 382 804 869 081 972 738 411 724 8;
  • 21) 0.148 576 382 804 869 081 972 738 411 724 8 × 2 = 0 + 0.297 152 765 609 738 163 945 476 823 449 6;
  • 22) 0.297 152 765 609 738 163 945 476 823 449 6 × 2 = 0 + 0.594 305 531 219 476 327 890 953 646 899 2;
  • 23) 0.594 305 531 219 476 327 890 953 646 899 2 × 2 = 1 + 0.188 611 062 438 952 655 781 907 293 798 4;
  • 24) 0.188 611 062 438 952 655 781 907 293 798 4 × 2 = 0 + 0.377 222 124 877 905 311 563 814 587 596 8;
  • 25) 0.377 222 124 877 905 311 563 814 587 596 8 × 2 = 0 + 0.754 444 249 755 810 623 127 629 175 193 6;
  • 26) 0.754 444 249 755 810 623 127 629 175 193 6 × 2 = 1 + 0.508 888 499 511 621 246 255 258 350 387 2;
  • 27) 0.508 888 499 511 621 246 255 258 350 387 2 × 2 = 1 + 0.017 776 999 023 242 492 510 516 700 774 4;
  • 28) 0.017 776 999 023 242 492 510 516 700 774 4 × 2 = 0 + 0.035 553 998 046 484 985 021 033 401 548 8;
  • 29) 0.035 553 998 046 484 985 021 033 401 548 8 × 2 = 0 + 0.071 107 996 092 969 970 042 066 803 097 6;
  • 30) 0.071 107 996 092 969 970 042 066 803 097 6 × 2 = 0 + 0.142 215 992 185 939 940 084 133 606 195 2;
  • 31) 0.142 215 992 185 939 940 084 133 606 195 2 × 2 = 0 + 0.284 431 984 371 879 880 168 267 212 390 4;
  • 32) 0.284 431 984 371 879 880 168 267 212 390 4 × 2 = 0 + 0.568 863 968 743 759 760 336 534 424 780 8;
  • 33) 0.568 863 968 743 759 760 336 534 424 780 8 × 2 = 1 + 0.137 727 937 487 519 520 673 068 849 561 6;
  • 34) 0.137 727 937 487 519 520 673 068 849 561 6 × 2 = 0 + 0.275 455 874 975 039 041 346 137 699 123 2;
  • 35) 0.275 455 874 975 039 041 346 137 699 123 2 × 2 = 0 + 0.550 911 749 950 078 082 692 275 398 246 4;
  • 36) 0.550 911 749 950 078 082 692 275 398 246 4 × 2 = 1 + 0.101 823 499 900 156 165 384 550 796 492 8;
  • 37) 0.101 823 499 900 156 165 384 550 796 492 8 × 2 = 0 + 0.203 646 999 800 312 330 769 101 592 985 6;
  • 38) 0.203 646 999 800 312 330 769 101 592 985 6 × 2 = 0 + 0.407 293 999 600 624 661 538 203 185 971 2;
  • 39) 0.407 293 999 600 624 661 538 203 185 971 2 × 2 = 0 + 0.814 587 999 201 249 323 076 406 371 942 4;
  • 40) 0.814 587 999 201 249 323 076 406 371 942 4 × 2 = 1 + 0.629 175 998 402 498 646 152 812 743 884 8;
  • 41) 0.629 175 998 402 498 646 152 812 743 884 8 × 2 = 1 + 0.258 351 996 804 997 292 305 625 487 769 6;
  • 42) 0.258 351 996 804 997 292 305 625 487 769 6 × 2 = 0 + 0.516 703 993 609 994 584 611 250 975 539 2;
  • 43) 0.516 703 993 609 994 584 611 250 975 539 2 × 2 = 1 + 0.033 407 987 219 989 169 222 501 951 078 4;
  • 44) 0.033 407 987 219 989 169 222 501 951 078 4 × 2 = 0 + 0.066 815 974 439 978 338 445 003 902 156 8;
  • 45) 0.066 815 974 439 978 338 445 003 902 156 8 × 2 = 0 + 0.133 631 948 879 956 676 890 007 804 313 6;
  • 46) 0.133 631 948 879 956 676 890 007 804 313 6 × 2 = 0 + 0.267 263 897 759 913 353 780 015 608 627 2;
  • 47) 0.267 263 897 759 913 353 780 015 608 627 2 × 2 = 0 + 0.534 527 795 519 826 707 560 031 217 254 4;
  • 48) 0.534 527 795 519 826 707 560 031 217 254 4 × 2 = 1 + 0.069 055 591 039 653 415 120 062 434 508 8;
  • 49) 0.069 055 591 039 653 415 120 062 434 508 8 × 2 = 0 + 0.138 111 182 079 306 830 240 124 869 017 6;
  • 50) 0.138 111 182 079 306 830 240 124 869 017 6 × 2 = 0 + 0.276 222 364 158 613 660 480 249 738 035 2;
  • 51) 0.276 222 364 158 613 660 480 249 738 035 2 × 2 = 0 + 0.552 444 728 317 227 320 960 499 476 070 4;
  • 52) 0.552 444 728 317 227 320 960 499 476 070 4 × 2 = 1 + 0.104 889 456 634 454 641 920 998 952 140 8;
  • 53) 0.104 889 456 634 454 641 920 998 952 140 8 × 2 = 0 + 0.209 778 913 268 909 283 841 997 904 281 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 - 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:


0.620 928 906 036 742 024 296 838 734 377 3(10) =

0.1001 1110 1111 0101 0011 0010 0110 0000 1001 0001 1010 0001 0001 0(2)

5. Positive number before normalization:

0.620 928 906 036 742 024 296 838 734 377 3(10) =

0.1001 1110 1111 0101 0011 0010 0110 0000 1001 0001 1010 0001 0001 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:


0.620 928 906 036 742 024 296 838 734 377 3(10) =

0.1001 1110 1111 0101 0011 0010 0110 0000 1001 0001 1010 0001 0001 0(2) =

0.1001 1110 1111 0101 0011 0010 0110 0000 1001 0001 1010 0001 0001 0(2) × 20 =

1.0011 1101 1110 1010 0110 0100 1100 0001 0010 0011 0100 0010 0010(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):
1.0011 1101 1110 1010 0110 0100 1100 0001 0010 0011 0100 0010 0010


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. 0011 1101 1110 1010 0110 0100 1100 0001 0010 0011 0100 0010 0010 =

0011 1101 1110 1010 0110 0100 1100 0001 0010 0011 0100 0010 0010


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) =
0011 1101 1110 1010 0110 0100 1100 0001 0010 0011 0100 0010 0010


Decimal number 0.620 928 906 036 742 024 296 838 734 377 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0011 1101 1110 1010 0110 0100 1100 0001 0010 0011 0100 0010 0010

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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