0.957 603 280 698 573 66 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.957 603 280 698 573 66(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
0.957 603 280 698 573 66(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: 0.957 603 280 698 573 66.

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) 0.957 603 280 698 573 66 × 2 = 1 + 0.915 206 561 397 147 32;
  • 2) 0.915 206 561 397 147 32 × 2 = 1 + 0.830 413 122 794 294 64;
  • 3) 0.830 413 122 794 294 64 × 2 = 1 + 0.660 826 245 588 589 28;
  • 4) 0.660 826 245 588 589 28 × 2 = 1 + 0.321 652 491 177 178 56;
  • 5) 0.321 652 491 177 178 56 × 2 = 0 + 0.643 304 982 354 357 12;
  • 6) 0.643 304 982 354 357 12 × 2 = 1 + 0.286 609 964 708 714 24;
  • 7) 0.286 609 964 708 714 24 × 2 = 0 + 0.573 219 929 417 428 48;
  • 8) 0.573 219 929 417 428 48 × 2 = 1 + 0.146 439 858 834 856 96;
  • 9) 0.146 439 858 834 856 96 × 2 = 0 + 0.292 879 717 669 713 92;
  • 10) 0.292 879 717 669 713 92 × 2 = 0 + 0.585 759 435 339 427 84;
  • 11) 0.585 759 435 339 427 84 × 2 = 1 + 0.171 518 870 678 855 68;
  • 12) 0.171 518 870 678 855 68 × 2 = 0 + 0.343 037 741 357 711 36;
  • 13) 0.343 037 741 357 711 36 × 2 = 0 + 0.686 075 482 715 422 72;
  • 14) 0.686 075 482 715 422 72 × 2 = 1 + 0.372 150 965 430 845 44;
  • 15) 0.372 150 965 430 845 44 × 2 = 0 + 0.744 301 930 861 690 88;
  • 16) 0.744 301 930 861 690 88 × 2 = 1 + 0.488 603 861 723 381 76;
  • 17) 0.488 603 861 723 381 76 × 2 = 0 + 0.977 207 723 446 763 52;
  • 18) 0.977 207 723 446 763 52 × 2 = 1 + 0.954 415 446 893 527 04;
  • 19) 0.954 415 446 893 527 04 × 2 = 1 + 0.908 830 893 787 054 08;
  • 20) 0.908 830 893 787 054 08 × 2 = 1 + 0.817 661 787 574 108 16;
  • 21) 0.817 661 787 574 108 16 × 2 = 1 + 0.635 323 575 148 216 32;
  • 22) 0.635 323 575 148 216 32 × 2 = 1 + 0.270 647 150 296 432 64;
  • 23) 0.270 647 150 296 432 64 × 2 = 0 + 0.541 294 300 592 865 28;
  • 24) 0.541 294 300 592 865 28 × 2 = 1 + 0.082 588 601 185 730 56;
  • 25) 0.082 588 601 185 730 56 × 2 = 0 + 0.165 177 202 371 461 12;
  • 26) 0.165 177 202 371 461 12 × 2 = 0 + 0.330 354 404 742 922 24;
  • 27) 0.330 354 404 742 922 24 × 2 = 0 + 0.660 708 809 485 844 48;
  • 28) 0.660 708 809 485 844 48 × 2 = 1 + 0.321 417 618 971 688 96;
  • 29) 0.321 417 618 971 688 96 × 2 = 0 + 0.642 835 237 943 377 92;
  • 30) 0.642 835 237 943 377 92 × 2 = 1 + 0.285 670 475 886 755 84;
  • 31) 0.285 670 475 886 755 84 × 2 = 0 + 0.571 340 951 773 511 68;
  • 32) 0.571 340 951 773 511 68 × 2 = 1 + 0.142 681 903 547 023 36;
  • 33) 0.142 681 903 547 023 36 × 2 = 0 + 0.285 363 807 094 046 72;
  • 34) 0.285 363 807 094 046 72 × 2 = 0 + 0.570 727 614 188 093 44;
  • 35) 0.570 727 614 188 093 44 × 2 = 1 + 0.141 455 228 376 186 88;
  • 36) 0.141 455 228 376 186 88 × 2 = 0 + 0.282 910 456 752 373 76;
  • 37) 0.282 910 456 752 373 76 × 2 = 0 + 0.565 820 913 504 747 52;
  • 38) 0.565 820 913 504 747 52 × 2 = 1 + 0.131 641 827 009 495 04;
  • 39) 0.131 641 827 009 495 04 × 2 = 0 + 0.263 283 654 018 990 08;
  • 40) 0.263 283 654 018 990 08 × 2 = 0 + 0.526 567 308 037 980 16;
  • 41) 0.526 567 308 037 980 16 × 2 = 1 + 0.053 134 616 075 960 32;
  • 42) 0.053 134 616 075 960 32 × 2 = 0 + 0.106 269 232 151 920 64;
  • 43) 0.106 269 232 151 920 64 × 2 = 0 + 0.212 538 464 303 841 28;
  • 44) 0.212 538 464 303 841 28 × 2 = 0 + 0.425 076 928 607 682 56;
  • 45) 0.425 076 928 607 682 56 × 2 = 0 + 0.850 153 857 215 365 12;
  • 46) 0.850 153 857 215 365 12 × 2 = 1 + 0.700 307 714 430 730 24;
  • 47) 0.700 307 714 430 730 24 × 2 = 1 + 0.400 615 428 861 460 48;
  • 48) 0.400 615 428 861 460 48 × 2 = 0 + 0.801 230 857 722 920 96;
  • 49) 0.801 230 857 722 920 96 × 2 = 1 + 0.602 461 715 445 841 92;
  • 50) 0.602 461 715 445 841 92 × 2 = 1 + 0.204 923 430 891 683 84;
  • 51) 0.204 923 430 891 683 84 × 2 = 0 + 0.409 846 861 783 367 68;
  • 52) 0.409 846 861 783 367 68 × 2 = 0 + 0.819 693 723 566 735 36;
  • 53) 0.819 693 723 566 735 36 × 2 = 1 + 0.639 387 447 133 470 72;

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:


0.957 603 280 698 573 66(10) =

0.1111 0101 0010 0101 0111 1101 0001 0101 0010 0100 1000 0110 1100 1(2)

5. Positive number before normalization:

0.957 603 280 698 573 66(10) =

0.1111 0101 0010 0101 0111 1101 0001 0101 0010 0100 1000 0110 1100 1(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:


0.957 603 280 698 573 66(10) =

0.1111 0101 0010 0101 0111 1101 0001 0101 0010 0100 1000 0110 1100 1(2) =

0.1111 0101 0010 0101 0111 1101 0001 0101 0010 0100 1000 0110 1100 1(2) × 20 =

1.1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1101 1001(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):
1.1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1101 1001


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. 1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1101 1001 =

1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1101 1001


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) =
1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1101 1001


Decimal number 0.957 603 280 698 573 66 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1101 1001

» Convert decimal 0.957 603 280 698 574 55 to 64 bit double precision IEEE 754 binary floating point representation standard

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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