0.771 105 412 703 972 64 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.771 105 412 703 972 64(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
0.771 105 412 703 972 64(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: 0.771 105 412 703 972 64.

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) 0.771 105 412 703 972 64 × 2 = 1 + 0.542 210 825 407 945 28;
  • 2) 0.542 210 825 407 945 28 × 2 = 1 + 0.084 421 650 815 890 56;
  • 3) 0.084 421 650 815 890 56 × 2 = 0 + 0.168 843 301 631 781 12;
  • 4) 0.168 843 301 631 781 12 × 2 = 0 + 0.337 686 603 263 562 24;
  • 5) 0.337 686 603 263 562 24 × 2 = 0 + 0.675 373 206 527 124 48;
  • 6) 0.675 373 206 527 124 48 × 2 = 1 + 0.350 746 413 054 248 96;
  • 7) 0.350 746 413 054 248 96 × 2 = 0 + 0.701 492 826 108 497 92;
  • 8) 0.701 492 826 108 497 92 × 2 = 1 + 0.402 985 652 216 995 84;
  • 9) 0.402 985 652 216 995 84 × 2 = 0 + 0.805 971 304 433 991 68;
  • 10) 0.805 971 304 433 991 68 × 2 = 1 + 0.611 942 608 867 983 36;
  • 11) 0.611 942 608 867 983 36 × 2 = 1 + 0.223 885 217 735 966 72;
  • 12) 0.223 885 217 735 966 72 × 2 = 0 + 0.447 770 435 471 933 44;
  • 13) 0.447 770 435 471 933 44 × 2 = 0 + 0.895 540 870 943 866 88;
  • 14) 0.895 540 870 943 866 88 × 2 = 1 + 0.791 081 741 887 733 76;
  • 15) 0.791 081 741 887 733 76 × 2 = 1 + 0.582 163 483 775 467 52;
  • 16) 0.582 163 483 775 467 52 × 2 = 1 + 0.164 326 967 550 935 04;
  • 17) 0.164 326 967 550 935 04 × 2 = 0 + 0.328 653 935 101 870 08;
  • 18) 0.328 653 935 101 870 08 × 2 = 0 + 0.657 307 870 203 740 16;
  • 19) 0.657 307 870 203 740 16 × 2 = 1 + 0.314 615 740 407 480 32;
  • 20) 0.314 615 740 407 480 32 × 2 = 0 + 0.629 231 480 814 960 64;
  • 21) 0.629 231 480 814 960 64 × 2 = 1 + 0.258 462 961 629 921 28;
  • 22) 0.258 462 961 629 921 28 × 2 = 0 + 0.516 925 923 259 842 56;
  • 23) 0.516 925 923 259 842 56 × 2 = 1 + 0.033 851 846 519 685 12;
  • 24) 0.033 851 846 519 685 12 × 2 = 0 + 0.067 703 693 039 370 24;
  • 25) 0.067 703 693 039 370 24 × 2 = 0 + 0.135 407 386 078 740 48;
  • 26) 0.135 407 386 078 740 48 × 2 = 0 + 0.270 814 772 157 480 96;
  • 27) 0.270 814 772 157 480 96 × 2 = 0 + 0.541 629 544 314 961 92;
  • 28) 0.541 629 544 314 961 92 × 2 = 1 + 0.083 259 088 629 923 84;
  • 29) 0.083 259 088 629 923 84 × 2 = 0 + 0.166 518 177 259 847 68;
  • 30) 0.166 518 177 259 847 68 × 2 = 0 + 0.333 036 354 519 695 36;
  • 31) 0.333 036 354 519 695 36 × 2 = 0 + 0.666 072 709 039 390 72;
  • 32) 0.666 072 709 039 390 72 × 2 = 1 + 0.332 145 418 078 781 44;
  • 33) 0.332 145 418 078 781 44 × 2 = 0 + 0.664 290 836 157 562 88;
  • 34) 0.664 290 836 157 562 88 × 2 = 1 + 0.328 581 672 315 125 76;
  • 35) 0.328 581 672 315 125 76 × 2 = 0 + 0.657 163 344 630 251 52;
  • 36) 0.657 163 344 630 251 52 × 2 = 1 + 0.314 326 689 260 503 04;
  • 37) 0.314 326 689 260 503 04 × 2 = 0 + 0.628 653 378 521 006 08;
  • 38) 0.628 653 378 521 006 08 × 2 = 1 + 0.257 306 757 042 012 16;
  • 39) 0.257 306 757 042 012 16 × 2 = 0 + 0.514 613 514 084 024 32;
  • 40) 0.514 613 514 084 024 32 × 2 = 1 + 0.029 227 028 168 048 64;
  • 41) 0.029 227 028 168 048 64 × 2 = 0 + 0.058 454 056 336 097 28;
  • 42) 0.058 454 056 336 097 28 × 2 = 0 + 0.116 908 112 672 194 56;
  • 43) 0.116 908 112 672 194 56 × 2 = 0 + 0.233 816 225 344 389 12;
  • 44) 0.233 816 225 344 389 12 × 2 = 0 + 0.467 632 450 688 778 24;
  • 45) 0.467 632 450 688 778 24 × 2 = 0 + 0.935 264 901 377 556 48;
  • 46) 0.935 264 901 377 556 48 × 2 = 1 + 0.870 529 802 755 112 96;
  • 47) 0.870 529 802 755 112 96 × 2 = 1 + 0.741 059 605 510 225 92;
  • 48) 0.741 059 605 510 225 92 × 2 = 1 + 0.482 119 211 020 451 84;
  • 49) 0.482 119 211 020 451 84 × 2 = 0 + 0.964 238 422 040 903 68;
  • 50) 0.964 238 422 040 903 68 × 2 = 1 + 0.928 476 844 081 807 36;
  • 51) 0.928 476 844 081 807 36 × 2 = 1 + 0.856 953 688 163 614 72;
  • 52) 0.856 953 688 163 614 72 × 2 = 1 + 0.713 907 376 327 229 44;
  • 53) 0.713 907 376 327 229 44 × 2 = 1 + 0.427 814 752 654 458 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:


0.771 105 412 703 972 64(10) =

0.1100 0101 0110 0111 0010 1010 0001 0001 0101 0101 0000 0111 0111 1(2)

5. Positive number before normalization:

0.771 105 412 703 972 64(10) =

0.1100 0101 0110 0111 0010 1010 0001 0001 0101 0101 0000 0111 0111 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:


0.771 105 412 703 972 64(10) =

0.1100 0101 0110 0111 0010 1010 0001 0001 0101 0101 0000 0111 0111 1(2) =

0.1100 0101 0110 0111 0010 1010 0001 0001 0101 0101 0000 0111 0111 1(2) × 20 =

1.1000 1010 1100 1110 0101 0100 0010 0010 1010 1010 0000 1110 1111(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):
1.1000 1010 1100 1110 0101 0100 0010 0010 1010 1010 0000 1110 1111


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. 1000 1010 1100 1110 0101 0100 0010 0010 1010 1010 0000 1110 1111 =

1000 1010 1100 1110 0101 0100 0010 0010 1010 1010 0000 1110 1111


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) =
1000 1010 1100 1110 0101 0100 0010 0010 1010 1010 0000 1110 1111


Decimal number 0.771 105 412 703 972 64 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1000 1010 1100 1110 0101 0100 0010 0010 1010 1010 0000 1110 1111

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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