0.620 928 906 036 742 024 297 51 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.620 928 906 036 742 024 297 51(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
0.620 928 906 036 742 024 297 51(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: 0.620 928 906 036 742 024 297 51.

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) 0.620 928 906 036 742 024 297 51 × 2 = 1 + 0.241 857 812 073 484 048 595 02;
  • 2) 0.241 857 812 073 484 048 595 02 × 2 = 0 + 0.483 715 624 146 968 097 190 04;
  • 3) 0.483 715 624 146 968 097 190 04 × 2 = 0 + 0.967 431 248 293 936 194 380 08;
  • 4) 0.967 431 248 293 936 194 380 08 × 2 = 1 + 0.934 862 496 587 872 388 760 16;
  • 5) 0.934 862 496 587 872 388 760 16 × 2 = 1 + 0.869 724 993 175 744 777 520 32;
  • 6) 0.869 724 993 175 744 777 520 32 × 2 = 1 + 0.739 449 986 351 489 555 040 64;
  • 7) 0.739 449 986 351 489 555 040 64 × 2 = 1 + 0.478 899 972 702 979 110 081 28;
  • 8) 0.478 899 972 702 979 110 081 28 × 2 = 0 + 0.957 799 945 405 958 220 162 56;
  • 9) 0.957 799 945 405 958 220 162 56 × 2 = 1 + 0.915 599 890 811 916 440 325 12;
  • 10) 0.915 599 890 811 916 440 325 12 × 2 = 1 + 0.831 199 781 623 832 880 650 24;
  • 11) 0.831 199 781 623 832 880 650 24 × 2 = 1 + 0.662 399 563 247 665 761 300 48;
  • 12) 0.662 399 563 247 665 761 300 48 × 2 = 1 + 0.324 799 126 495 331 522 600 96;
  • 13) 0.324 799 126 495 331 522 600 96 × 2 = 0 + 0.649 598 252 990 663 045 201 92;
  • 14) 0.649 598 252 990 663 045 201 92 × 2 = 1 + 0.299 196 505 981 326 090 403 84;
  • 15) 0.299 196 505 981 326 090 403 84 × 2 = 0 + 0.598 393 011 962 652 180 807 68;
  • 16) 0.598 393 011 962 652 180 807 68 × 2 = 1 + 0.196 786 023 925 304 361 615 36;
  • 17) 0.196 786 023 925 304 361 615 36 × 2 = 0 + 0.393 572 047 850 608 723 230 72;
  • 18) 0.393 572 047 850 608 723 230 72 × 2 = 0 + 0.787 144 095 701 217 446 461 44;
  • 19) 0.787 144 095 701 217 446 461 44 × 2 = 1 + 0.574 288 191 402 434 892 922 88;
  • 20) 0.574 288 191 402 434 892 922 88 × 2 = 1 + 0.148 576 382 804 869 785 845 76;
  • 21) 0.148 576 382 804 869 785 845 76 × 2 = 0 + 0.297 152 765 609 739 571 691 52;
  • 22) 0.297 152 765 609 739 571 691 52 × 2 = 0 + 0.594 305 531 219 479 143 383 04;
  • 23) 0.594 305 531 219 479 143 383 04 × 2 = 1 + 0.188 611 062 438 958 286 766 08;
  • 24) 0.188 611 062 438 958 286 766 08 × 2 = 0 + 0.377 222 124 877 916 573 532 16;
  • 25) 0.377 222 124 877 916 573 532 16 × 2 = 0 + 0.754 444 249 755 833 147 064 32;
  • 26) 0.754 444 249 755 833 147 064 32 × 2 = 1 + 0.508 888 499 511 666 294 128 64;
  • 27) 0.508 888 499 511 666 294 128 64 × 2 = 1 + 0.017 776 999 023 332 588 257 28;
  • 28) 0.017 776 999 023 332 588 257 28 × 2 = 0 + 0.035 553 998 046 665 176 514 56;
  • 29) 0.035 553 998 046 665 176 514 56 × 2 = 0 + 0.071 107 996 093 330 353 029 12;
  • 30) 0.071 107 996 093 330 353 029 12 × 2 = 0 + 0.142 215 992 186 660 706 058 24;
  • 31) 0.142 215 992 186 660 706 058 24 × 2 = 0 + 0.284 431 984 373 321 412 116 48;
  • 32) 0.284 431 984 373 321 412 116 48 × 2 = 0 + 0.568 863 968 746 642 824 232 96;
  • 33) 0.568 863 968 746 642 824 232 96 × 2 = 1 + 0.137 727 937 493 285 648 465 92;
  • 34) 0.137 727 937 493 285 648 465 92 × 2 = 0 + 0.275 455 874 986 571 296 931 84;
  • 35) 0.275 455 874 986 571 296 931 84 × 2 = 0 + 0.550 911 749 973 142 593 863 68;
  • 36) 0.550 911 749 973 142 593 863 68 × 2 = 1 + 0.101 823 499 946 285 187 727 36;
  • 37) 0.101 823 499 946 285 187 727 36 × 2 = 0 + 0.203 646 999 892 570 375 454 72;
  • 38) 0.203 646 999 892 570 375 454 72 × 2 = 0 + 0.407 293 999 785 140 750 909 44;
  • 39) 0.407 293 999 785 140 750 909 44 × 2 = 0 + 0.814 587 999 570 281 501 818 88;
  • 40) 0.814 587 999 570 281 501 818 88 × 2 = 1 + 0.629 175 999 140 563 003 637 76;
  • 41) 0.629 175 999 140 563 003 637 76 × 2 = 1 + 0.258 351 998 281 126 007 275 52;
  • 42) 0.258 351 998 281 126 007 275 52 × 2 = 0 + 0.516 703 996 562 252 014 551 04;
  • 43) 0.516 703 996 562 252 014 551 04 × 2 = 1 + 0.033 407 993 124 504 029 102 08;
  • 44) 0.033 407 993 124 504 029 102 08 × 2 = 0 + 0.066 815 986 249 008 058 204 16;
  • 45) 0.066 815 986 249 008 058 204 16 × 2 = 0 + 0.133 631 972 498 016 116 408 32;
  • 46) 0.133 631 972 498 016 116 408 32 × 2 = 0 + 0.267 263 944 996 032 232 816 64;
  • 47) 0.267 263 944 996 032 232 816 64 × 2 = 0 + 0.534 527 889 992 064 465 633 28;
  • 48) 0.534 527 889 992 064 465 633 28 × 2 = 1 + 0.069 055 779 984 128 931 266 56;
  • 49) 0.069 055 779 984 128 931 266 56 × 2 = 0 + 0.138 111 559 968 257 862 533 12;
  • 50) 0.138 111 559 968 257 862 533 12 × 2 = 0 + 0.276 223 119 936 515 725 066 24;
  • 51) 0.276 223 119 936 515 725 066 24 × 2 = 0 + 0.552 446 239 873 031 450 132 48;
  • 52) 0.552 446 239 873 031 450 132 48 × 2 = 1 + 0.104 892 479 746 062 900 264 96;
  • 53) 0.104 892 479 746 062 900 264 96 × 2 = 0 + 0.209 784 959 492 125 800 529 92;

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:


0.620 928 906 036 742 024 297 51(10) =

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

5. Positive number before normalization:

0.620 928 906 036 742 024 297 51(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:


0.620 928 906 036 742 024 297 51(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 0.620 928 906 036 742 024 297 51 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