0.620 928 906 036 742 024 296 62 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.620 928 906 036 742 024 296 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
0.620 928 906 036 742 024 296 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: 0.620 928 906 036 742 024 296 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) 0.620 928 906 036 742 024 296 62 × 2 = 1 + 0.241 857 812 073 484 048 593 24;
  • 2) 0.241 857 812 073 484 048 593 24 × 2 = 0 + 0.483 715 624 146 968 097 186 48;
  • 3) 0.483 715 624 146 968 097 186 48 × 2 = 0 + 0.967 431 248 293 936 194 372 96;
  • 4) 0.967 431 248 293 936 194 372 96 × 2 = 1 + 0.934 862 496 587 872 388 745 92;
  • 5) 0.934 862 496 587 872 388 745 92 × 2 = 1 + 0.869 724 993 175 744 777 491 84;
  • 6) 0.869 724 993 175 744 777 491 84 × 2 = 1 + 0.739 449 986 351 489 554 983 68;
  • 7) 0.739 449 986 351 489 554 983 68 × 2 = 1 + 0.478 899 972 702 979 109 967 36;
  • 8) 0.478 899 972 702 979 109 967 36 × 2 = 0 + 0.957 799 945 405 958 219 934 72;
  • 9) 0.957 799 945 405 958 219 934 72 × 2 = 1 + 0.915 599 890 811 916 439 869 44;
  • 10) 0.915 599 890 811 916 439 869 44 × 2 = 1 + 0.831 199 781 623 832 879 738 88;
  • 11) 0.831 199 781 623 832 879 738 88 × 2 = 1 + 0.662 399 563 247 665 759 477 76;
  • 12) 0.662 399 563 247 665 759 477 76 × 2 = 1 + 0.324 799 126 495 331 518 955 52;
  • 13) 0.324 799 126 495 331 518 955 52 × 2 = 0 + 0.649 598 252 990 663 037 911 04;
  • 14) 0.649 598 252 990 663 037 911 04 × 2 = 1 + 0.299 196 505 981 326 075 822 08;
  • 15) 0.299 196 505 981 326 075 822 08 × 2 = 0 + 0.598 393 011 962 652 151 644 16;
  • 16) 0.598 393 011 962 652 151 644 16 × 2 = 1 + 0.196 786 023 925 304 303 288 32;
  • 17) 0.196 786 023 925 304 303 288 32 × 2 = 0 + 0.393 572 047 850 608 606 576 64;
  • 18) 0.393 572 047 850 608 606 576 64 × 2 = 0 + 0.787 144 095 701 217 213 153 28;
  • 19) 0.787 144 095 701 217 213 153 28 × 2 = 1 + 0.574 288 191 402 434 426 306 56;
  • 20) 0.574 288 191 402 434 426 306 56 × 2 = 1 + 0.148 576 382 804 868 852 613 12;
  • 21) 0.148 576 382 804 868 852 613 12 × 2 = 0 + 0.297 152 765 609 737 705 226 24;
  • 22) 0.297 152 765 609 737 705 226 24 × 2 = 0 + 0.594 305 531 219 475 410 452 48;
  • 23) 0.594 305 531 219 475 410 452 48 × 2 = 1 + 0.188 611 062 438 950 820 904 96;
  • 24) 0.188 611 062 438 950 820 904 96 × 2 = 0 + 0.377 222 124 877 901 641 809 92;
  • 25) 0.377 222 124 877 901 641 809 92 × 2 = 0 + 0.754 444 249 755 803 283 619 84;
  • 26) 0.754 444 249 755 803 283 619 84 × 2 = 1 + 0.508 888 499 511 606 567 239 68;
  • 27) 0.508 888 499 511 606 567 239 68 × 2 = 1 + 0.017 776 999 023 213 134 479 36;
  • 28) 0.017 776 999 023 213 134 479 36 × 2 = 0 + 0.035 553 998 046 426 268 958 72;
  • 29) 0.035 553 998 046 426 268 958 72 × 2 = 0 + 0.071 107 996 092 852 537 917 44;
  • 30) 0.071 107 996 092 852 537 917 44 × 2 = 0 + 0.142 215 992 185 705 075 834 88;
  • 31) 0.142 215 992 185 705 075 834 88 × 2 = 0 + 0.284 431 984 371 410 151 669 76;
  • 32) 0.284 431 984 371 410 151 669 76 × 2 = 0 + 0.568 863 968 742 820 303 339 52;
  • 33) 0.568 863 968 742 820 303 339 52 × 2 = 1 + 0.137 727 937 485 640 606 679 04;
  • 34) 0.137 727 937 485 640 606 679 04 × 2 = 0 + 0.275 455 874 971 281 213 358 08;
  • 35) 0.275 455 874 971 281 213 358 08 × 2 = 0 + 0.550 911 749 942 562 426 716 16;
  • 36) 0.550 911 749 942 562 426 716 16 × 2 = 1 + 0.101 823 499 885 124 853 432 32;
  • 37) 0.101 823 499 885 124 853 432 32 × 2 = 0 + 0.203 646 999 770 249 706 864 64;
  • 38) 0.203 646 999 770 249 706 864 64 × 2 = 0 + 0.407 293 999 540 499 413 729 28;
  • 39) 0.407 293 999 540 499 413 729 28 × 2 = 0 + 0.814 587 999 080 998 827 458 56;
  • 40) 0.814 587 999 080 998 827 458 56 × 2 = 1 + 0.629 175 998 161 997 654 917 12;
  • 41) 0.629 175 998 161 997 654 917 12 × 2 = 1 + 0.258 351 996 323 995 309 834 24;
  • 42) 0.258 351 996 323 995 309 834 24 × 2 = 0 + 0.516 703 992 647 990 619 668 48;
  • 43) 0.516 703 992 647 990 619 668 48 × 2 = 1 + 0.033 407 985 295 981 239 336 96;
  • 44) 0.033 407 985 295 981 239 336 96 × 2 = 0 + 0.066 815 970 591 962 478 673 92;
  • 45) 0.066 815 970 591 962 478 673 92 × 2 = 0 + 0.133 631 941 183 924 957 347 84;
  • 46) 0.133 631 941 183 924 957 347 84 × 2 = 0 + 0.267 263 882 367 849 914 695 68;
  • 47) 0.267 263 882 367 849 914 695 68 × 2 = 0 + 0.534 527 764 735 699 829 391 36;
  • 48) 0.534 527 764 735 699 829 391 36 × 2 = 1 + 0.069 055 529 471 399 658 782 72;
  • 49) 0.069 055 529 471 399 658 782 72 × 2 = 0 + 0.138 111 058 942 799 317 565 44;
  • 50) 0.138 111 058 942 799 317 565 44 × 2 = 0 + 0.276 222 117 885 598 635 130 88;
  • 51) 0.276 222 117 885 598 635 130 88 × 2 = 0 + 0.552 444 235 771 197 270 261 76;
  • 52) 0.552 444 235 771 197 270 261 76 × 2 = 1 + 0.104 888 471 542 394 540 523 52;
  • 53) 0.104 888 471 542 394 540 523 52 × 2 = 0 + 0.209 776 943 084 789 081 047 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:


0.620 928 906 036 742 024 296 62(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 62(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 62(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 62 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