Decimal to 64 Bit IEEE 754 Binary: Convert Number 2.560 879 608 4 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 2.560 879 608 4(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 2.
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;
  • 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.

2(10) =


10(2)


3. Convert to binary (base 2) the fractional part: 0.560 879 608 4.

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) 0.560 879 608 4 × 2 = 1 + 0.121 759 216 8;
  • 2) 0.121 759 216 8 × 2 = 0 + 0.243 518 433 6;
  • 3) 0.243 518 433 6 × 2 = 0 + 0.487 036 867 2;
  • 4) 0.487 036 867 2 × 2 = 0 + 0.974 073 734 4;
  • 5) 0.974 073 734 4 × 2 = 1 + 0.948 147 468 8;
  • 6) 0.948 147 468 8 × 2 = 1 + 0.896 294 937 6;
  • 7) 0.896 294 937 6 × 2 = 1 + 0.792 589 875 2;
  • 8) 0.792 589 875 2 × 2 = 1 + 0.585 179 750 4;
  • 9) 0.585 179 750 4 × 2 = 1 + 0.170 359 500 8;
  • 10) 0.170 359 500 8 × 2 = 0 + 0.340 719 001 6;
  • 11) 0.340 719 001 6 × 2 = 0 + 0.681 438 003 2;
  • 12) 0.681 438 003 2 × 2 = 1 + 0.362 876 006 4;
  • 13) 0.362 876 006 4 × 2 = 0 + 0.725 752 012 8;
  • 14) 0.725 752 012 8 × 2 = 1 + 0.451 504 025 6;
  • 15) 0.451 504 025 6 × 2 = 0 + 0.903 008 051 2;
  • 16) 0.903 008 051 2 × 2 = 1 + 0.806 016 102 4;
  • 17) 0.806 016 102 4 × 2 = 1 + 0.612 032 204 8;
  • 18) 0.612 032 204 8 × 2 = 1 + 0.224 064 409 6;
  • 19) 0.224 064 409 6 × 2 = 0 + 0.448 128 819 2;
  • 20) 0.448 128 819 2 × 2 = 0 + 0.896 257 638 4;
  • 21) 0.896 257 638 4 × 2 = 1 + 0.792 515 276 8;
  • 22) 0.792 515 276 8 × 2 = 1 + 0.585 030 553 6;
  • 23) 0.585 030 553 6 × 2 = 1 + 0.170 061 107 2;
  • 24) 0.170 061 107 2 × 2 = 0 + 0.340 122 214 4;
  • 25) 0.340 122 214 4 × 2 = 0 + 0.680 244 428 8;
  • 26) 0.680 244 428 8 × 2 = 1 + 0.360 488 857 6;
  • 27) 0.360 488 857 6 × 2 = 0 + 0.720 977 715 2;
  • 28) 0.720 977 715 2 × 2 = 1 + 0.441 955 430 4;
  • 29) 0.441 955 430 4 × 2 = 0 + 0.883 910 860 8;
  • 30) 0.883 910 860 8 × 2 = 1 + 0.767 821 721 6;
  • 31) 0.767 821 721 6 × 2 = 1 + 0.535 643 443 2;
  • 32) 0.535 643 443 2 × 2 = 1 + 0.071 286 886 4;
  • 33) 0.071 286 886 4 × 2 = 0 + 0.142 573 772 8;
  • 34) 0.142 573 772 8 × 2 = 0 + 0.285 147 545 6;
  • 35) 0.285 147 545 6 × 2 = 0 + 0.570 295 091 2;
  • 36) 0.570 295 091 2 × 2 = 1 + 0.140 590 182 4;
  • 37) 0.140 590 182 4 × 2 = 0 + 0.281 180 364 8;
  • 38) 0.281 180 364 8 × 2 = 0 + 0.562 360 729 6;
  • 39) 0.562 360 729 6 × 2 = 1 + 0.124 721 459 2;
  • 40) 0.124 721 459 2 × 2 = 0 + 0.249 442 918 4;
  • 41) 0.249 442 918 4 × 2 = 0 + 0.498 885 836 8;
  • 42) 0.498 885 836 8 × 2 = 0 + 0.997 771 673 6;
  • 43) 0.997 771 673 6 × 2 = 1 + 0.995 543 347 2;
  • 44) 0.995 543 347 2 × 2 = 1 + 0.991 086 694 4;
  • 45) 0.991 086 694 4 × 2 = 1 + 0.982 173 388 8;
  • 46) 0.982 173 388 8 × 2 = 1 + 0.964 346 777 6;
  • 47) 0.964 346 777 6 × 2 = 1 + 0.928 693 555 2;
  • 48) 0.928 693 555 2 × 2 = 1 + 0.857 387 110 4;
  • 49) 0.857 387 110 4 × 2 = 1 + 0.714 774 220 8;
  • 50) 0.714 774 220 8 × 2 = 1 + 0.429 548 441 6;
  • 51) 0.429 548 441 6 × 2 = 0 + 0.859 096 883 2;
  • 52) 0.859 096 883 2 × 2 = 1 + 0.718 193 766 4;
  • 53) 0.718 193 766 4 × 2 = 1 + 0.436 387 532 8;

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:


0.560 879 608 4(10) =


0.1000 1111 1001 0101 1100 1110 0101 0111 0001 0010 0011 1111 1101 1(2)

5. Positive number before normalization:

2.560 879 608 4(10) =


10.1000 1111 1001 0101 1100 1110 0101 0111 0001 0010 0011 1111 1101 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


2.560 879 608 4(10) =


10.1000 1111 1001 0101 1100 1110 0101 0111 0001 0010 0011 1111 1101 1(2) =


10.1000 1111 1001 0101 1100 1110 0101 0111 0001 0010 0011 1111 1101 1(2) × 20 =


1.0100 0111 1100 1010 1110 0111 0010 1011 1000 1001 0001 1111 1110 11(2) × 21


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):
1.0100 0111 1100 1010 1110 0111 0010 1011 1000 1001 0001 1111 1110 11


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) =


1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 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) =


1024(10) =


100 0000 0000(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. 0100 0111 1100 1010 1110 0111 0010 1011 1000 1001 0001 1111 1110 11 =


0100 0111 1100 1010 1110 0111 0010 1011 1000 1001 0001 1111 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) =
100 0000 0000


Mantissa (52 bits) =
0100 0111 1100 1010 1110 0111 0010 1011 1000 1001 0001 1111 1110


The base ten decimal number 2.560 879 608 4 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0100 0111 1100 1010 1110 0111 0010 1011 1000 1001 0001 1111 1110

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