64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 123 456 789.123 456 779 8 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 123 456 789.123 456 779 8(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: 123 456 789.
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;
  • 123 456 789 ÷ 2 = 61 728 394 + 1;
  • 61 728 394 ÷ 2 = 30 864 197 + 0;
  • 30 864 197 ÷ 2 = 15 432 098 + 1;
  • 15 432 098 ÷ 2 = 7 716 049 + 0;
  • 7 716 049 ÷ 2 = 3 858 024 + 1;
  • 3 858 024 ÷ 2 = 1 929 012 + 0;
  • 1 929 012 ÷ 2 = 964 506 + 0;
  • 964 506 ÷ 2 = 482 253 + 0;
  • 482 253 ÷ 2 = 241 126 + 1;
  • 241 126 ÷ 2 = 120 563 + 0;
  • 120 563 ÷ 2 = 60 281 + 1;
  • 60 281 ÷ 2 = 30 140 + 1;
  • 30 140 ÷ 2 = 15 070 + 0;
  • 15 070 ÷ 2 = 7 535 + 0;
  • 7 535 ÷ 2 = 3 767 + 1;
  • 3 767 ÷ 2 = 1 883 + 1;
  • 1 883 ÷ 2 = 941 + 1;
  • 941 ÷ 2 = 470 + 1;
  • 470 ÷ 2 = 235 + 0;
  • 235 ÷ 2 = 117 + 1;
  • 117 ÷ 2 = 58 + 1;
  • 58 ÷ 2 = 29 + 0;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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.


123 456 789(10) =


111 0101 1011 1100 1101 0001 0101(2)


3. Convert to binary (base 2) the fractional part: 0.123 456 779 8.

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) 0.123 456 779 8 × 2 = 0 + 0.246 913 559 6;
  • 2) 0.246 913 559 6 × 2 = 0 + 0.493 827 119 2;
  • 3) 0.493 827 119 2 × 2 = 0 + 0.987 654 238 4;
  • 4) 0.987 654 238 4 × 2 = 1 + 0.975 308 476 8;
  • 5) 0.975 308 476 8 × 2 = 1 + 0.950 616 953 6;
  • 6) 0.950 616 953 6 × 2 = 1 + 0.901 233 907 2;
  • 7) 0.901 233 907 2 × 2 = 1 + 0.802 467 814 4;
  • 8) 0.802 467 814 4 × 2 = 1 + 0.604 935 628 8;
  • 9) 0.604 935 628 8 × 2 = 1 + 0.209 871 257 6;
  • 10) 0.209 871 257 6 × 2 = 0 + 0.419 742 515 2;
  • 11) 0.419 742 515 2 × 2 = 0 + 0.839 485 030 4;
  • 12) 0.839 485 030 4 × 2 = 1 + 0.678 970 060 8;
  • 13) 0.678 970 060 8 × 2 = 1 + 0.357 940 121 6;
  • 14) 0.357 940 121 6 × 2 = 0 + 0.715 880 243 2;
  • 15) 0.715 880 243 2 × 2 = 1 + 0.431 760 486 4;
  • 16) 0.431 760 486 4 × 2 = 0 + 0.863 520 972 8;
  • 17) 0.863 520 972 8 × 2 = 1 + 0.727 041 945 6;
  • 18) 0.727 041 945 6 × 2 = 1 + 0.454 083 891 2;
  • 19) 0.454 083 891 2 × 2 = 0 + 0.908 167 782 4;
  • 20) 0.908 167 782 4 × 2 = 1 + 0.816 335 564 8;
  • 21) 0.816 335 564 8 × 2 = 1 + 0.632 671 129 6;
  • 22) 0.632 671 129 6 × 2 = 1 + 0.265 342 259 2;
  • 23) 0.265 342 259 2 × 2 = 0 + 0.530 684 518 4;
  • 24) 0.530 684 518 4 × 2 = 1 + 0.061 369 036 8;
  • 25) 0.061 369 036 8 × 2 = 0 + 0.122 738 073 6;
  • 26) 0.122 738 073 6 × 2 = 0 + 0.245 476 147 2;
  • 27) 0.245 476 147 2 × 2 = 0 + 0.490 952 294 4;
  • 28) 0.490 952 294 4 × 2 = 0 + 0.981 904 588 8;
  • 29) 0.981 904 588 8 × 2 = 1 + 0.963 809 177 6;
  • 30) 0.963 809 177 6 × 2 = 1 + 0.927 618 355 2;
  • 31) 0.927 618 355 2 × 2 = 1 + 0.855 236 710 4;
  • 32) 0.855 236 710 4 × 2 = 1 + 0.710 473 420 8;
  • 33) 0.710 473 420 8 × 2 = 1 + 0.420 946 841 6;
  • 34) 0.420 946 841 6 × 2 = 0 + 0.841 893 683 2;
  • 35) 0.841 893 683 2 × 2 = 1 + 0.683 787 366 4;
  • 36) 0.683 787 366 4 × 2 = 1 + 0.367 574 732 8;
  • 37) 0.367 574 732 8 × 2 = 0 + 0.735 149 465 6;
  • 38) 0.735 149 465 6 × 2 = 1 + 0.470 298 931 2;
  • 39) 0.470 298 931 2 × 2 = 0 + 0.940 597 862 4;
  • 40) 0.940 597 862 4 × 2 = 1 + 0.881 195 724 8;
  • 41) 0.881 195 724 8 × 2 = 1 + 0.762 391 449 6;
  • 42) 0.762 391 449 6 × 2 = 1 + 0.524 782 899 2;
  • 43) 0.524 782 899 2 × 2 = 1 + 0.049 565 798 4;
  • 44) 0.049 565 798 4 × 2 = 0 + 0.099 131 596 8;
  • 45) 0.099 131 596 8 × 2 = 0 + 0.198 263 193 6;
  • 46) 0.198 263 193 6 × 2 = 0 + 0.396 526 387 2;
  • 47) 0.396 526 387 2 × 2 = 0 + 0.793 052 774 4;
  • 48) 0.793 052 774 4 × 2 = 1 + 0.586 105 548 8;
  • 49) 0.586 105 548 8 × 2 = 1 + 0.172 211 097 6;
  • 50) 0.172 211 097 6 × 2 = 0 + 0.344 422 195 2;
  • 51) 0.344 422 195 2 × 2 = 0 + 0.688 844 390 4;
  • 52) 0.688 844 390 4 × 2 = 1 + 0.377 688 780 8;
  • 53) 0.377 688 780 8 × 2 = 0 + 0.755 377 561 6;

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:


0.123 456 779 8(10) =


0.0001 1111 1001 1010 1101 1101 0000 1111 1011 0101 1110 0001 1001 0(2)


5. Positive number before normalization:

123 456 789.123 456 779 8(10) =


111 0101 1011 1100 1101 0001 0101.0001 1111 1001 1010 1101 1101 0000 1111 1011 0101 1110 0001 1001 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the left, so that only one non zero digit remains to the left of it:


123 456 789.123 456 779 8(10) =


111 0101 1011 1100 1101 0001 0101.0001 1111 1001 1010 1101 1101 0000 1111 1011 0101 1110 0001 1001 0(2) =


111 0101 1011 1100 1101 0001 0101.0001 1111 1001 1010 1101 1101 0000 1111 1011 0101 1110 0001 1001 0(2) × 20 =


1.1101 0110 1111 0011 0100 0101 0100 0111 1110 0110 1011 0111 0100 0011 1110 1101 0111 1000 0110 010(2) × 226


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): 26


Mantissa (not normalized):
1.1101 0110 1111 0011 0100 0101 0100 0111 1110 0110 1011 0111 0100 0011 1110 1101 0111 1000 0110 010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


26 + 2(11-1) - 1 =


(26 + 1 023)(10) =


1 049(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;
  • 1 049 ÷ 2 = 524 + 1;
  • 524 ÷ 2 = 262 + 0;
  • 262 ÷ 2 = 131 + 0;
  • 131 ÷ 2 = 65 + 1;
  • 65 ÷ 2 = 32 + 1;
  • 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) =


1049(10) =


100 0001 1001(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. 1101 0110 1111 0011 0100 0101 0100 0111 1110 0110 1011 0111 0100 001 1111 0110 1011 1100 0011 0010 =


1101 0110 1111 0011 0100 0101 0100 0111 1110 0110 1011 0111 0100


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 0001 1001


Mantissa (52 bits) =
1101 0110 1111 0011 0100 0101 0100 0111 1110 0110 1011 0111 0100


The base ten decimal number 123 456 789.123 456 779 8 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0001 1001 - 1101 0110 1111 0011 0100 0101 0100 0111 1110 0110 1011 0111 0100

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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