0.771 105 412 703 972 01 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.771 105 412 703 972 01(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 01(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 01.

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 01 × 2 = 1 + 0.542 210 825 407 944 02;
  • 2) 0.542 210 825 407 944 02 × 2 = 1 + 0.084 421 650 815 888 04;
  • 3) 0.084 421 650 815 888 04 × 2 = 0 + 0.168 843 301 631 776 08;
  • 4) 0.168 843 301 631 776 08 × 2 = 0 + 0.337 686 603 263 552 16;
  • 5) 0.337 686 603 263 552 16 × 2 = 0 + 0.675 373 206 527 104 32;
  • 6) 0.675 373 206 527 104 32 × 2 = 1 + 0.350 746 413 054 208 64;
  • 7) 0.350 746 413 054 208 64 × 2 = 0 + 0.701 492 826 108 417 28;
  • 8) 0.701 492 826 108 417 28 × 2 = 1 + 0.402 985 652 216 834 56;
  • 9) 0.402 985 652 216 834 56 × 2 = 0 + 0.805 971 304 433 669 12;
  • 10) 0.805 971 304 433 669 12 × 2 = 1 + 0.611 942 608 867 338 24;
  • 11) 0.611 942 608 867 338 24 × 2 = 1 + 0.223 885 217 734 676 48;
  • 12) 0.223 885 217 734 676 48 × 2 = 0 + 0.447 770 435 469 352 96;
  • 13) 0.447 770 435 469 352 96 × 2 = 0 + 0.895 540 870 938 705 92;
  • 14) 0.895 540 870 938 705 92 × 2 = 1 + 0.791 081 741 877 411 84;
  • 15) 0.791 081 741 877 411 84 × 2 = 1 + 0.582 163 483 754 823 68;
  • 16) 0.582 163 483 754 823 68 × 2 = 1 + 0.164 326 967 509 647 36;
  • 17) 0.164 326 967 509 647 36 × 2 = 0 + 0.328 653 935 019 294 72;
  • 18) 0.328 653 935 019 294 72 × 2 = 0 + 0.657 307 870 038 589 44;
  • 19) 0.657 307 870 038 589 44 × 2 = 1 + 0.314 615 740 077 178 88;
  • 20) 0.314 615 740 077 178 88 × 2 = 0 + 0.629 231 480 154 357 76;
  • 21) 0.629 231 480 154 357 76 × 2 = 1 + 0.258 462 960 308 715 52;
  • 22) 0.258 462 960 308 715 52 × 2 = 0 + 0.516 925 920 617 431 04;
  • 23) 0.516 925 920 617 431 04 × 2 = 1 + 0.033 851 841 234 862 08;
  • 24) 0.033 851 841 234 862 08 × 2 = 0 + 0.067 703 682 469 724 16;
  • 25) 0.067 703 682 469 724 16 × 2 = 0 + 0.135 407 364 939 448 32;
  • 26) 0.135 407 364 939 448 32 × 2 = 0 + 0.270 814 729 878 896 64;
  • 27) 0.270 814 729 878 896 64 × 2 = 0 + 0.541 629 459 757 793 28;
  • 28) 0.541 629 459 757 793 28 × 2 = 1 + 0.083 258 919 515 586 56;
  • 29) 0.083 258 919 515 586 56 × 2 = 0 + 0.166 517 839 031 173 12;
  • 30) 0.166 517 839 031 173 12 × 2 = 0 + 0.333 035 678 062 346 24;
  • 31) 0.333 035 678 062 346 24 × 2 = 0 + 0.666 071 356 124 692 48;
  • 32) 0.666 071 356 124 692 48 × 2 = 1 + 0.332 142 712 249 384 96;
  • 33) 0.332 142 712 249 384 96 × 2 = 0 + 0.664 285 424 498 769 92;
  • 34) 0.664 285 424 498 769 92 × 2 = 1 + 0.328 570 848 997 539 84;
  • 35) 0.328 570 848 997 539 84 × 2 = 0 + 0.657 141 697 995 079 68;
  • 36) 0.657 141 697 995 079 68 × 2 = 1 + 0.314 283 395 990 159 36;
  • 37) 0.314 283 395 990 159 36 × 2 = 0 + 0.628 566 791 980 318 72;
  • 38) 0.628 566 791 980 318 72 × 2 = 1 + 0.257 133 583 960 637 44;
  • 39) 0.257 133 583 960 637 44 × 2 = 0 + 0.514 267 167 921 274 88;
  • 40) 0.514 267 167 921 274 88 × 2 = 1 + 0.028 534 335 842 549 76;
  • 41) 0.028 534 335 842 549 76 × 2 = 0 + 0.057 068 671 685 099 52;
  • 42) 0.057 068 671 685 099 52 × 2 = 0 + 0.114 137 343 370 199 04;
  • 43) 0.114 137 343 370 199 04 × 2 = 0 + 0.228 274 686 740 398 08;
  • 44) 0.228 274 686 740 398 08 × 2 = 0 + 0.456 549 373 480 796 16;
  • 45) 0.456 549 373 480 796 16 × 2 = 0 + 0.913 098 746 961 592 32;
  • 46) 0.913 098 746 961 592 32 × 2 = 1 + 0.826 197 493 923 184 64;
  • 47) 0.826 197 493 923 184 64 × 2 = 1 + 0.652 394 987 846 369 28;
  • 48) 0.652 394 987 846 369 28 × 2 = 1 + 0.304 789 975 692 738 56;
  • 49) 0.304 789 975 692 738 56 × 2 = 0 + 0.609 579 951 385 477 12;
  • 50) 0.609 579 951 385 477 12 × 2 = 1 + 0.219 159 902 770 954 24;
  • 51) 0.219 159 902 770 954 24 × 2 = 0 + 0.438 319 805 541 908 48;
  • 52) 0.438 319 805 541 908 48 × 2 = 0 + 0.876 639 611 083 816 96;
  • 53) 0.876 639 611 083 816 96 × 2 = 1 + 0.753 279 222 167 633 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:


0.771 105 412 703 972 01(10) =

0.1100 0101 0110 0111 0010 1010 0001 0001 0101 0101 0000 0111 0100 1(2)

5. Positive number before normalization:

0.771 105 412 703 972 01(10) =

0.1100 0101 0110 0111 0010 1010 0001 0001 0101 0101 0000 0111 0100 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 01(10) =

0.1100 0101 0110 0111 0010 1010 0001 0001 0101 0101 0000 0111 0100 1(2) =

0.1100 0101 0110 0111 0010 1010 0001 0001 0101 0101 0000 0111 0100 1(2) × 20 =

1.1000 1010 1100 1110 0101 0100 0010 0010 1010 1010 0000 1110 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):
1.1000 1010 1100 1110 0101 0100 0010 0010 1010 1010 0000 1110 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. 1000 1010 1100 1110 0101 0100 0010 0010 1010 1010 0000 1110 1001 =

1000 1010 1100 1110 0101 0100 0010 0010 1010 1010 0000 1110 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) =
1000 1010 1100 1110 0101 0100 0010 0010 1010 1010 0000 1110 1001


Decimal number 0.771 105 412 703 972 01 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 1001

» Convert decimal 0.771 105 412 703 972 64 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