64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 0.000 000 127 591 3 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 0.000 000 127 591 3(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: 0.000 000 127 591 3.

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) 0.000 000 127 591 3 × 2 = 0 + 0.000 000 255 182 6;
  • 2) 0.000 000 255 182 6 × 2 = 0 + 0.000 000 510 365 2;
  • 3) 0.000 000 510 365 2 × 2 = 0 + 0.000 001 020 730 4;
  • 4) 0.000 001 020 730 4 × 2 = 0 + 0.000 002 041 460 8;
  • 5) 0.000 002 041 460 8 × 2 = 0 + 0.000 004 082 921 6;
  • 6) 0.000 004 082 921 6 × 2 = 0 + 0.000 008 165 843 2;
  • 7) 0.000 008 165 843 2 × 2 = 0 + 0.000 016 331 686 4;
  • 8) 0.000 016 331 686 4 × 2 = 0 + 0.000 032 663 372 8;
  • 9) 0.000 032 663 372 8 × 2 = 0 + 0.000 065 326 745 6;
  • 10) 0.000 065 326 745 6 × 2 = 0 + 0.000 130 653 491 2;
  • 11) 0.000 130 653 491 2 × 2 = 0 + 0.000 261 306 982 4;
  • 12) 0.000 261 306 982 4 × 2 = 0 + 0.000 522 613 964 8;
  • 13) 0.000 522 613 964 8 × 2 = 0 + 0.001 045 227 929 6;
  • 14) 0.001 045 227 929 6 × 2 = 0 + 0.002 090 455 859 2;
  • 15) 0.002 090 455 859 2 × 2 = 0 + 0.004 180 911 718 4;
  • 16) 0.004 180 911 718 4 × 2 = 0 + 0.008 361 823 436 8;
  • 17) 0.008 361 823 436 8 × 2 = 0 + 0.016 723 646 873 6;
  • 18) 0.016 723 646 873 6 × 2 = 0 + 0.033 447 293 747 2;
  • 19) 0.033 447 293 747 2 × 2 = 0 + 0.066 894 587 494 4;
  • 20) 0.066 894 587 494 4 × 2 = 0 + 0.133 789 174 988 8;
  • 21) 0.133 789 174 988 8 × 2 = 0 + 0.267 578 349 977 6;
  • 22) 0.267 578 349 977 6 × 2 = 0 + 0.535 156 699 955 2;
  • 23) 0.535 156 699 955 2 × 2 = 1 + 0.070 313 399 910 4;
  • 24) 0.070 313 399 910 4 × 2 = 0 + 0.140 626 799 820 8;
  • 25) 0.140 626 799 820 8 × 2 = 0 + 0.281 253 599 641 6;
  • 26) 0.281 253 599 641 6 × 2 = 0 + 0.562 507 199 283 2;
  • 27) 0.562 507 199 283 2 × 2 = 1 + 0.125 014 398 566 4;
  • 28) 0.125 014 398 566 4 × 2 = 0 + 0.250 028 797 132 8;
  • 29) 0.250 028 797 132 8 × 2 = 0 + 0.500 057 594 265 6;
  • 30) 0.500 057 594 265 6 × 2 = 1 + 0.000 115 188 531 2;
  • 31) 0.000 115 188 531 2 × 2 = 0 + 0.000 230 377 062 4;
  • 32) 0.000 230 377 062 4 × 2 = 0 + 0.000 460 754 124 8;
  • 33) 0.000 460 754 124 8 × 2 = 0 + 0.000 921 508 249 6;
  • 34) 0.000 921 508 249 6 × 2 = 0 + 0.001 843 016 499 2;
  • 35) 0.001 843 016 499 2 × 2 = 0 + 0.003 686 032 998 4;
  • 36) 0.003 686 032 998 4 × 2 = 0 + 0.007 372 065 996 8;
  • 37) 0.007 372 065 996 8 × 2 = 0 + 0.014 744 131 993 6;
  • 38) 0.014 744 131 993 6 × 2 = 0 + 0.029 488 263 987 2;
  • 39) 0.029 488 263 987 2 × 2 = 0 + 0.058 976 527 974 4;
  • 40) 0.058 976 527 974 4 × 2 = 0 + 0.117 953 055 948 8;
  • 41) 0.117 953 055 948 8 × 2 = 0 + 0.235 906 111 897 6;
  • 42) 0.235 906 111 897 6 × 2 = 0 + 0.471 812 223 795 2;
  • 43) 0.471 812 223 795 2 × 2 = 0 + 0.943 624 447 590 4;
  • 44) 0.943 624 447 590 4 × 2 = 1 + 0.887 248 895 180 8;
  • 45) 0.887 248 895 180 8 × 2 = 1 + 0.774 497 790 361 6;
  • 46) 0.774 497 790 361 6 × 2 = 1 + 0.548 995 580 723 2;
  • 47) 0.548 995 580 723 2 × 2 = 1 + 0.097 991 161 446 4;
  • 48) 0.097 991 161 446 4 × 2 = 0 + 0.195 982 322 892 8;
  • 49) 0.195 982 322 892 8 × 2 = 0 + 0.391 964 645 785 6;
  • 50) 0.391 964 645 785 6 × 2 = 0 + 0.783 929 291 571 2;
  • 51) 0.783 929 291 571 2 × 2 = 1 + 0.567 858 583 142 4;
  • 52) 0.567 858 583 142 4 × 2 = 1 + 0.135 717 166 284 8;
  • 53) 0.135 717 166 284 8 × 2 = 0 + 0.271 434 332 569 6;
  • 54) 0.271 434 332 569 6 × 2 = 0 + 0.542 868 665 139 2;
  • 55) 0.542 868 665 139 2 × 2 = 1 + 0.085 737 330 278 4;
  • 56) 0.085 737 330 278 4 × 2 = 0 + 0.171 474 660 556 8;
  • 57) 0.171 474 660 556 8 × 2 = 0 + 0.342 949 321 113 6;
  • 58) 0.342 949 321 113 6 × 2 = 0 + 0.685 898 642 227 2;
  • 59) 0.685 898 642 227 2 × 2 = 1 + 0.371 797 284 454 4;
  • 60) 0.371 797 284 454 4 × 2 = 0 + 0.743 594 568 908 8;
  • 61) 0.743 594 568 908 8 × 2 = 1 + 0.487 189 137 817 6;
  • 62) 0.487 189 137 817 6 × 2 = 0 + 0.974 378 275 635 2;
  • 63) 0.974 378 275 635 2 × 2 = 1 + 0.948 756 551 270 4;
  • 64) 0.948 756 551 270 4 × 2 = 1 + 0.897 513 102 540 8;
  • 65) 0.897 513 102 540 8 × 2 = 1 + 0.795 026 205 081 6;
  • 66) 0.795 026 205 081 6 × 2 = 1 + 0.590 052 410 163 2;
  • 67) 0.590 052 410 163 2 × 2 = 1 + 0.180 104 820 326 4;
  • 68) 0.180 104 820 326 4 × 2 = 0 + 0.360 209 640 652 8;
  • 69) 0.360 209 640 652 8 × 2 = 0 + 0.720 419 281 305 6;
  • 70) 0.720 419 281 305 6 × 2 = 1 + 0.440 838 562 611 2;
  • 71) 0.440 838 562 611 2 × 2 = 0 + 0.881 677 125 222 4;
  • 72) 0.881 677 125 222 4 × 2 = 1 + 0.763 354 250 444 8;
  • 73) 0.763 354 250 444 8 × 2 = 1 + 0.526 708 500 889 6;
  • 74) 0.526 708 500 889 6 × 2 = 1 + 0.053 417 001 779 2;
  • 75) 0.053 417 001 779 2 × 2 = 0 + 0.106 834 003 558 4;

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:


0.000 000 127 591 3(10) =

0.0000 0000 0000 0000 0000 0010 0010 0100 0000 0000 0001 1110 0011 0010 0010 1011 1110 0101 110(2)


5. Positive number before normalization:

0.000 000 127 591 3(10) =

0.0000 0000 0000 0000 0000 0010 0010 0100 0000 0000 0001 1110 0011 0010 0010 1011 1110 0101 110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 23 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 127 591 3(10) =

0.0000 0000 0000 0000 0000 0010 0010 0100 0000 0000 0001 1110 0011 0010 0010 1011 1110 0101 110(2) =

0.0000 0000 0000 0000 0000 0010 0010 0100 0000 0000 0001 1110 0011 0010 0010 1011 1110 0101 110(2) × 20 =

1.0001 0010 0000 0000 0000 1111 0001 1001 0001 0101 1111 0010 1110(2) × 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): -23

Mantissa (not normalized):
1.0001 0010 0000 0000 0000 1111 0001 1001 0001 0101 1111 0010 1110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-23 + 2(11-1) - 1 =

(-23 + 1 023)(10) =

1 000(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;
  • 1 000 ÷ 2 = 500 + 0;
  • 500 ÷ 2 = 250 + 0;
  • 250 ÷ 2 = 125 + 0;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

1000(10) =

011 1110 1000(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. 0001 0010 0000 0000 0000 1111 0001 1001 0001 0101 1111 0010 1110 =

0001 0010 0000 0000 0000 1111 0001 1001 0001 0101 1111 0010 1110


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 1110 1000

Mantissa (52 bits) =
0001 0010 0000 0000 0000 1111 0001 1001 0001 0101 1111 0010 1110


The base ten decimal number 0.000 000 127 591 3 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1110 1000 - 0001 0010 0000 0000 0000 1111 0001 1001 0001 0101 1111 0010 1110

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

» Number 0.000 000 127 591 2 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