64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 0.785 398 163 346 387 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 0.785 398 163 346 387(10) converted and written in 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: 0.785 398 163 346 387.

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) 0.785 398 163 346 387 × 2 = 1 + 0.570 796 326 692 774;
  • 2) 0.570 796 326 692 774 × 2 = 1 + 0.141 592 653 385 548;
  • 3) 0.141 592 653 385 548 × 2 = 0 + 0.283 185 306 771 096;
  • 4) 0.283 185 306 771 096 × 2 = 0 + 0.566 370 613 542 192;
  • 5) 0.566 370 613 542 192 × 2 = 1 + 0.132 741 227 084 384;
  • 6) 0.132 741 227 084 384 × 2 = 0 + 0.265 482 454 168 768;
  • 7) 0.265 482 454 168 768 × 2 = 0 + 0.530 964 908 337 536;
  • 8) 0.530 964 908 337 536 × 2 = 1 + 0.061 929 816 675 072;
  • 9) 0.061 929 816 675 072 × 2 = 0 + 0.123 859 633 350 144;
  • 10) 0.123 859 633 350 144 × 2 = 0 + 0.247 719 266 700 288;
  • 11) 0.247 719 266 700 288 × 2 = 0 + 0.495 438 533 400 576;
  • 12) 0.495 438 533 400 576 × 2 = 0 + 0.990 877 066 801 152;
  • 13) 0.990 877 066 801 152 × 2 = 1 + 0.981 754 133 602 304;
  • 14) 0.981 754 133 602 304 × 2 = 1 + 0.963 508 267 204 608;
  • 15) 0.963 508 267 204 608 × 2 = 1 + 0.927 016 534 409 216;
  • 16) 0.927 016 534 409 216 × 2 = 1 + 0.854 033 068 818 432;
  • 17) 0.854 033 068 818 432 × 2 = 1 + 0.708 066 137 636 864;
  • 18) 0.708 066 137 636 864 × 2 = 1 + 0.416 132 275 273 728;
  • 19) 0.416 132 275 273 728 × 2 = 0 + 0.832 264 550 547 456;
  • 20) 0.832 264 550 547 456 × 2 = 1 + 0.664 529 101 094 912;
  • 21) 0.664 529 101 094 912 × 2 = 1 + 0.329 058 202 189 824;
  • 22) 0.329 058 202 189 824 × 2 = 0 + 0.658 116 404 379 648;
  • 23) 0.658 116 404 379 648 × 2 = 1 + 0.316 232 808 759 296;
  • 24) 0.316 232 808 759 296 × 2 = 0 + 0.632 465 617 518 592;
  • 25) 0.632 465 617 518 592 × 2 = 1 + 0.264 931 235 037 184;
  • 26) 0.264 931 235 037 184 × 2 = 0 + 0.529 862 470 074 368;
  • 27) 0.529 862 470 074 368 × 2 = 1 + 0.059 724 940 148 736;
  • 28) 0.059 724 940 148 736 × 2 = 0 + 0.119 449 880 297 472;
  • 29) 0.119 449 880 297 472 × 2 = 0 + 0.238 899 760 594 944;
  • 30) 0.238 899 760 594 944 × 2 = 0 + 0.477 799 521 189 888;
  • 31) 0.477 799 521 189 888 × 2 = 0 + 0.955 599 042 379 776;
  • 32) 0.955 599 042 379 776 × 2 = 1 + 0.911 198 084 759 552;
  • 33) 0.911 198 084 759 552 × 2 = 1 + 0.822 396 169 519 104;
  • 34) 0.822 396 169 519 104 × 2 = 1 + 0.644 792 339 038 208;
  • 35) 0.644 792 339 038 208 × 2 = 1 + 0.289 584 678 076 416;
  • 36) 0.289 584 678 076 416 × 2 = 0 + 0.579 169 356 152 832;
  • 37) 0.579 169 356 152 832 × 2 = 1 + 0.158 338 712 305 664;
  • 38) 0.158 338 712 305 664 × 2 = 0 + 0.316 677 424 611 328;
  • 39) 0.316 677 424 611 328 × 2 = 0 + 0.633 354 849 222 656;
  • 40) 0.633 354 849 222 656 × 2 = 1 + 0.266 709 698 445 312;
  • 41) 0.266 709 698 445 312 × 2 = 0 + 0.533 419 396 890 624;
  • 42) 0.533 419 396 890 624 × 2 = 1 + 0.066 838 793 781 248;
  • 43) 0.066 838 793 781 248 × 2 = 0 + 0.133 677 587 562 496;
  • 44) 0.133 677 587 562 496 × 2 = 0 + 0.267 355 175 124 992;
  • 45) 0.267 355 175 124 992 × 2 = 0 + 0.534 710 350 249 984;
  • 46) 0.534 710 350 249 984 × 2 = 1 + 0.069 420 700 499 968;
  • 47) 0.069 420 700 499 968 × 2 = 0 + 0.138 841 400 999 936;
  • 48) 0.138 841 400 999 936 × 2 = 0 + 0.277 682 801 999 872;
  • 49) 0.277 682 801 999 872 × 2 = 0 + 0.555 365 603 999 744;
  • 50) 0.555 365 603 999 744 × 2 = 1 + 0.110 731 207 999 488;
  • 51) 0.110 731 207 999 488 × 2 = 0 + 0.221 462 415 998 976;
  • 52) 0.221 462 415 998 976 × 2 = 0 + 0.442 924 831 997 952;
  • 53) 0.442 924 831 997 952 × 2 = 0 + 0.885 849 663 995 904;

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...)


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:


0.785 398 163 346 387(10) =


0.1100 1001 0000 1111 1101 1010 1010 0001 1110 1001 0100 0100 0100 0(2)


5. Positive number before normalization:

0.785 398 163 346 387(10) =


0.1100 1001 0000 1111 1101 1010 1010 0001 1110 1001 0100 0100 0100 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:


0.785 398 163 346 387(10) =


0.1100 1001 0000 1111 1101 1010 1010 0001 1110 1001 0100 0100 0100 0(2) =


0.1100 1001 0000 1111 1101 1010 1010 0001 1110 1001 0100 0100 0100 0(2) × 20 =


1.1001 0010 0001 1111 1011 0101 0100 0011 1101 0010 1000 1000 1000(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):
1.1001 0010 0001 1111 1011 0101 0100 0011 1101 0010 1000 1000 1000


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. 1001 0010 0001 1111 1011 0101 0100 0011 1101 0010 1000 1000 1000 =


1001 0010 0001 1111 1011 0101 0100 0011 1101 0010 1000 1000 1000


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) =
1001 0010 0001 1111 1011 0101 0100 0011 1101 0010 1000 1000 1000


The base ten decimal number 0.785 398 163 346 387 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 1110 - 1001 0010 0001 1111 1011 0101 0100 0011 1101 0010 1000 1000 1000

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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