0.620 928 906 036 742 024 295 89 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.620 928 906 036 742 024 295 89(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
0.620 928 906 036 742 024 295 89(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: 0.620 928 906 036 742 024 295 89.

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) 0.620 928 906 036 742 024 295 89 × 2 = 1 + 0.241 857 812 073 484 048 591 78;
  • 2) 0.241 857 812 073 484 048 591 78 × 2 = 0 + 0.483 715 624 146 968 097 183 56;
  • 3) 0.483 715 624 146 968 097 183 56 × 2 = 0 + 0.967 431 248 293 936 194 367 12;
  • 4) 0.967 431 248 293 936 194 367 12 × 2 = 1 + 0.934 862 496 587 872 388 734 24;
  • 5) 0.934 862 496 587 872 388 734 24 × 2 = 1 + 0.869 724 993 175 744 777 468 48;
  • 6) 0.869 724 993 175 744 777 468 48 × 2 = 1 + 0.739 449 986 351 489 554 936 96;
  • 7) 0.739 449 986 351 489 554 936 96 × 2 = 1 + 0.478 899 972 702 979 109 873 92;
  • 8) 0.478 899 972 702 979 109 873 92 × 2 = 0 + 0.957 799 945 405 958 219 747 84;
  • 9) 0.957 799 945 405 958 219 747 84 × 2 = 1 + 0.915 599 890 811 916 439 495 68;
  • 10) 0.915 599 890 811 916 439 495 68 × 2 = 1 + 0.831 199 781 623 832 878 991 36;
  • 11) 0.831 199 781 623 832 878 991 36 × 2 = 1 + 0.662 399 563 247 665 757 982 72;
  • 12) 0.662 399 563 247 665 757 982 72 × 2 = 1 + 0.324 799 126 495 331 515 965 44;
  • 13) 0.324 799 126 495 331 515 965 44 × 2 = 0 + 0.649 598 252 990 663 031 930 88;
  • 14) 0.649 598 252 990 663 031 930 88 × 2 = 1 + 0.299 196 505 981 326 063 861 76;
  • 15) 0.299 196 505 981 326 063 861 76 × 2 = 0 + 0.598 393 011 962 652 127 723 52;
  • 16) 0.598 393 011 962 652 127 723 52 × 2 = 1 + 0.196 786 023 925 304 255 447 04;
  • 17) 0.196 786 023 925 304 255 447 04 × 2 = 0 + 0.393 572 047 850 608 510 894 08;
  • 18) 0.393 572 047 850 608 510 894 08 × 2 = 0 + 0.787 144 095 701 217 021 788 16;
  • 19) 0.787 144 095 701 217 021 788 16 × 2 = 1 + 0.574 288 191 402 434 043 576 32;
  • 20) 0.574 288 191 402 434 043 576 32 × 2 = 1 + 0.148 576 382 804 868 087 152 64;
  • 21) 0.148 576 382 804 868 087 152 64 × 2 = 0 + 0.297 152 765 609 736 174 305 28;
  • 22) 0.297 152 765 609 736 174 305 28 × 2 = 0 + 0.594 305 531 219 472 348 610 56;
  • 23) 0.594 305 531 219 472 348 610 56 × 2 = 1 + 0.188 611 062 438 944 697 221 12;
  • 24) 0.188 611 062 438 944 697 221 12 × 2 = 0 + 0.377 222 124 877 889 394 442 24;
  • 25) 0.377 222 124 877 889 394 442 24 × 2 = 0 + 0.754 444 249 755 778 788 884 48;
  • 26) 0.754 444 249 755 778 788 884 48 × 2 = 1 + 0.508 888 499 511 557 577 768 96;
  • 27) 0.508 888 499 511 557 577 768 96 × 2 = 1 + 0.017 776 999 023 115 155 537 92;
  • 28) 0.017 776 999 023 115 155 537 92 × 2 = 0 + 0.035 553 998 046 230 311 075 84;
  • 29) 0.035 553 998 046 230 311 075 84 × 2 = 0 + 0.071 107 996 092 460 622 151 68;
  • 30) 0.071 107 996 092 460 622 151 68 × 2 = 0 + 0.142 215 992 184 921 244 303 36;
  • 31) 0.142 215 992 184 921 244 303 36 × 2 = 0 + 0.284 431 984 369 842 488 606 72;
  • 32) 0.284 431 984 369 842 488 606 72 × 2 = 0 + 0.568 863 968 739 684 977 213 44;
  • 33) 0.568 863 968 739 684 977 213 44 × 2 = 1 + 0.137 727 937 479 369 954 426 88;
  • 34) 0.137 727 937 479 369 954 426 88 × 2 = 0 + 0.275 455 874 958 739 908 853 76;
  • 35) 0.275 455 874 958 739 908 853 76 × 2 = 0 + 0.550 911 749 917 479 817 707 52;
  • 36) 0.550 911 749 917 479 817 707 52 × 2 = 1 + 0.101 823 499 834 959 635 415 04;
  • 37) 0.101 823 499 834 959 635 415 04 × 2 = 0 + 0.203 646 999 669 919 270 830 08;
  • 38) 0.203 646 999 669 919 270 830 08 × 2 = 0 + 0.407 293 999 339 838 541 660 16;
  • 39) 0.407 293 999 339 838 541 660 16 × 2 = 0 + 0.814 587 998 679 677 083 320 32;
  • 40) 0.814 587 998 679 677 083 320 32 × 2 = 1 + 0.629 175 997 359 354 166 640 64;
  • 41) 0.629 175 997 359 354 166 640 64 × 2 = 1 + 0.258 351 994 718 708 333 281 28;
  • 42) 0.258 351 994 718 708 333 281 28 × 2 = 0 + 0.516 703 989 437 416 666 562 56;
  • 43) 0.516 703 989 437 416 666 562 56 × 2 = 1 + 0.033 407 978 874 833 333 125 12;
  • 44) 0.033 407 978 874 833 333 125 12 × 2 = 0 + 0.066 815 957 749 666 666 250 24;
  • 45) 0.066 815 957 749 666 666 250 24 × 2 = 0 + 0.133 631 915 499 333 332 500 48;
  • 46) 0.133 631 915 499 333 332 500 48 × 2 = 0 + 0.267 263 830 998 666 665 000 96;
  • 47) 0.267 263 830 998 666 665 000 96 × 2 = 0 + 0.534 527 661 997 333 330 001 92;
  • 48) 0.534 527 661 997 333 330 001 92 × 2 = 1 + 0.069 055 323 994 666 660 003 84;
  • 49) 0.069 055 323 994 666 660 003 84 × 2 = 0 + 0.138 110 647 989 333 320 007 68;
  • 50) 0.138 110 647 989 333 320 007 68 × 2 = 0 + 0.276 221 295 978 666 640 015 36;
  • 51) 0.276 221 295 978 666 640 015 36 × 2 = 0 + 0.552 442 591 957 333 280 030 72;
  • 52) 0.552 442 591 957 333 280 030 72 × 2 = 1 + 0.104 885 183 914 666 560 061 44;
  • 53) 0.104 885 183 914 666 560 061 44 × 2 = 0 + 0.209 770 367 829 333 120 122 88;

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:


0.620 928 906 036 742 024 295 89(10) =

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

5. Positive number before normalization:

0.620 928 906 036 742 024 295 89(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:


0.620 928 906 036 742 024 295 89(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 0.620 928 906 036 742 024 295 89 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