64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 11.722 517 867 598 367 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 11.722 517 867 598 367(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: 11.
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;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.


11(10) =

1011(2)


3. Convert to binary (base 2) the fractional part: 0.722 517 867 598 367.

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) 0.722 517 867 598 367 × 2 = 1 + 0.445 035 735 196 734;
  • 2) 0.445 035 735 196 734 × 2 = 0 + 0.890 071 470 393 468;
  • 3) 0.890 071 470 393 468 × 2 = 1 + 0.780 142 940 786 936;
  • 4) 0.780 142 940 786 936 × 2 = 1 + 0.560 285 881 573 872;
  • 5) 0.560 285 881 573 872 × 2 = 1 + 0.120 571 763 147 744;
  • 6) 0.120 571 763 147 744 × 2 = 0 + 0.241 143 526 295 488;
  • 7) 0.241 143 526 295 488 × 2 = 0 + 0.482 287 052 590 976;
  • 8) 0.482 287 052 590 976 × 2 = 0 + 0.964 574 105 181 952;
  • 9) 0.964 574 105 181 952 × 2 = 1 + 0.929 148 210 363 904;
  • 10) 0.929 148 210 363 904 × 2 = 1 + 0.858 296 420 727 808;
  • 11) 0.858 296 420 727 808 × 2 = 1 + 0.716 592 841 455 616;
  • 12) 0.716 592 841 455 616 × 2 = 1 + 0.433 185 682 911 232;
  • 13) 0.433 185 682 911 232 × 2 = 0 + 0.866 371 365 822 464;
  • 14) 0.866 371 365 822 464 × 2 = 1 + 0.732 742 731 644 928;
  • 15) 0.732 742 731 644 928 × 2 = 1 + 0.465 485 463 289 856;
  • 16) 0.465 485 463 289 856 × 2 = 0 + 0.930 970 926 579 712;
  • 17) 0.930 970 926 579 712 × 2 = 1 + 0.861 941 853 159 424;
  • 18) 0.861 941 853 159 424 × 2 = 1 + 0.723 883 706 318 848;
  • 19) 0.723 883 706 318 848 × 2 = 1 + 0.447 767 412 637 696;
  • 20) 0.447 767 412 637 696 × 2 = 0 + 0.895 534 825 275 392;
  • 21) 0.895 534 825 275 392 × 2 = 1 + 0.791 069 650 550 784;
  • 22) 0.791 069 650 550 784 × 2 = 1 + 0.582 139 301 101 568;
  • 23) 0.582 139 301 101 568 × 2 = 1 + 0.164 278 602 203 136;
  • 24) 0.164 278 602 203 136 × 2 = 0 + 0.328 557 204 406 272;
  • 25) 0.328 557 204 406 272 × 2 = 0 + 0.657 114 408 812 544;
  • 26) 0.657 114 408 812 544 × 2 = 1 + 0.314 228 817 625 088;
  • 27) 0.314 228 817 625 088 × 2 = 0 + 0.628 457 635 250 176;
  • 28) 0.628 457 635 250 176 × 2 = 1 + 0.256 915 270 500 352;
  • 29) 0.256 915 270 500 352 × 2 = 0 + 0.513 830 541 000 704;
  • 30) 0.513 830 541 000 704 × 2 = 1 + 0.027 661 082 001 408;
  • 31) 0.027 661 082 001 408 × 2 = 0 + 0.055 322 164 002 816;
  • 32) 0.055 322 164 002 816 × 2 = 0 + 0.110 644 328 005 632;
  • 33) 0.110 644 328 005 632 × 2 = 0 + 0.221 288 656 011 264;
  • 34) 0.221 288 656 011 264 × 2 = 0 + 0.442 577 312 022 528;
  • 35) 0.442 577 312 022 528 × 2 = 0 + 0.885 154 624 045 056;
  • 36) 0.885 154 624 045 056 × 2 = 1 + 0.770 309 248 090 112;
  • 37) 0.770 309 248 090 112 × 2 = 1 + 0.540 618 496 180 224;
  • 38) 0.540 618 496 180 224 × 2 = 1 + 0.081 236 992 360 448;
  • 39) 0.081 236 992 360 448 × 2 = 0 + 0.162 473 984 720 896;
  • 40) 0.162 473 984 720 896 × 2 = 0 + 0.324 947 969 441 792;
  • 41) 0.324 947 969 441 792 × 2 = 0 + 0.649 895 938 883 584;
  • 42) 0.649 895 938 883 584 × 2 = 1 + 0.299 791 877 767 168;
  • 43) 0.299 791 877 767 168 × 2 = 0 + 0.599 583 755 534 336;
  • 44) 0.599 583 755 534 336 × 2 = 1 + 0.199 167 511 068 672;
  • 45) 0.199 167 511 068 672 × 2 = 0 + 0.398 335 022 137 344;
  • 46) 0.398 335 022 137 344 × 2 = 0 + 0.796 670 044 274 688;
  • 47) 0.796 670 044 274 688 × 2 = 1 + 0.593 340 088 549 376;
  • 48) 0.593 340 088 549 376 × 2 = 1 + 0.186 680 177 098 752;
  • 49) 0.186 680 177 098 752 × 2 = 0 + 0.373 360 354 197 504;
  • 50) 0.373 360 354 197 504 × 2 = 0 + 0.746 720 708 395 008;
  • 51) 0.746 720 708 395 008 × 2 = 1 + 0.493 441 416 790 016;
  • 52) 0.493 441 416 790 016 × 2 = 0 + 0.986 882 833 580 032;
  • 53) 0.986 882 833 580 032 × 2 = 1 + 0.973 765 667 160 064;

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:


0.722 517 867 598 367(10) =

0.1011 1000 1111 0110 1110 1110 0101 0100 0001 1100 0101 0011 0010 1(2)


5. Positive number before normalization:

11.722 517 867 598 367(10) =

1011.1011 1000 1111 0110 1110 1110 0101 0100 0001 1100 0101 0011 0010 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the left, so that only one non zero digit remains to the left of it:


11.722 517 867 598 367(10) =

1011.1011 1000 1111 0110 1110 1110 0101 0100 0001 1100 0101 0011 0010 1(2) =

1011.1011 1000 1111 0110 1110 1110 0101 0100 0001 1100 0101 0011 0010 1(2) × 20 =

1.0111 0111 0001 1110 1101 1101 1100 1010 1000 0011 1000 1010 0110 0101(2) × 23


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): 3

Mantissa (not normalized):
1.0111 0111 0001 1110 1101 1101 1100 1010 1000 0011 1000 1010 0110 0101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

3 + 2(11-1) - 1 =

(3 + 1 023)(10) =

1 026(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;
  • 1 026 ÷ 2 = 513 + 0;
  • 513 ÷ 2 = 256 + 1;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1026(10) =

100 0000 0010(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0111 0111 0001 1110 1101 1101 1100 1010 1000 0011 1000 1010 0110 0101 =

0111 0111 0001 1110 1101 1101 1100 1010 1000 0011 1000 1010 0110


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) =
100 0000 0010

Mantissa (52 bits) =
0111 0111 0001 1110 1101 1101 1100 1010 1000 0011 1000 1010 0110


The base ten decimal number 11.722 517 867 598 367 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 0010 - 0111 0111 0001 1110 1101 1101 1100 1010 1000 0011 1000 1010 0110

More operations with decimal numbers converted to 64 bit double precision IEEE 754 binary floating point representation:

» Number 11.722 517 867 598 366 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

» Number 11.722 517 867 598 368 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

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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