64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 43 144.598 907 673 7 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 43 144.598 907 673 7(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: 43 144.
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;
  • 43 144 ÷ 2 = 21 572 + 0;
  • 21 572 ÷ 2 = 10 786 + 0;
  • 10 786 ÷ 2 = 5 393 + 0;
  • 5 393 ÷ 2 = 2 696 + 1;
  • 2 696 ÷ 2 = 1 348 + 0;
  • 1 348 ÷ 2 = 674 + 0;
  • 674 ÷ 2 = 337 + 0;
  • 337 ÷ 2 = 168 + 1;
  • 168 ÷ 2 = 84 + 0;
  • 84 ÷ 2 = 42 + 0;
  • 42 ÷ 2 = 21 + 0;
  • 21 ÷ 2 = 10 + 1;
  • 10 ÷ 2 = 5 + 0;
  • 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.


43 144(10) =


1010 1000 1000 1000(2)


3. Convert to binary (base 2) the fractional part: 0.598 907 673 7.

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) 0.598 907 673 7 × 2 = 1 + 0.197 815 347 4;
  • 2) 0.197 815 347 4 × 2 = 0 + 0.395 630 694 8;
  • 3) 0.395 630 694 8 × 2 = 0 + 0.791 261 389 6;
  • 4) 0.791 261 389 6 × 2 = 1 + 0.582 522 779 2;
  • 5) 0.582 522 779 2 × 2 = 1 + 0.165 045 558 4;
  • 6) 0.165 045 558 4 × 2 = 0 + 0.330 091 116 8;
  • 7) 0.330 091 116 8 × 2 = 0 + 0.660 182 233 6;
  • 8) 0.660 182 233 6 × 2 = 1 + 0.320 364 467 2;
  • 9) 0.320 364 467 2 × 2 = 0 + 0.640 728 934 4;
  • 10) 0.640 728 934 4 × 2 = 1 + 0.281 457 868 8;
  • 11) 0.281 457 868 8 × 2 = 0 + 0.562 915 737 6;
  • 12) 0.562 915 737 6 × 2 = 1 + 0.125 831 475 2;
  • 13) 0.125 831 475 2 × 2 = 0 + 0.251 662 950 4;
  • 14) 0.251 662 950 4 × 2 = 0 + 0.503 325 900 8;
  • 15) 0.503 325 900 8 × 2 = 1 + 0.006 651 801 6;
  • 16) 0.006 651 801 6 × 2 = 0 + 0.013 303 603 2;
  • 17) 0.013 303 603 2 × 2 = 0 + 0.026 607 206 4;
  • 18) 0.026 607 206 4 × 2 = 0 + 0.053 214 412 8;
  • 19) 0.053 214 412 8 × 2 = 0 + 0.106 428 825 6;
  • 20) 0.106 428 825 6 × 2 = 0 + 0.212 857 651 2;
  • 21) 0.212 857 651 2 × 2 = 0 + 0.425 715 302 4;
  • 22) 0.425 715 302 4 × 2 = 0 + 0.851 430 604 8;
  • 23) 0.851 430 604 8 × 2 = 1 + 0.702 861 209 6;
  • 24) 0.702 861 209 6 × 2 = 1 + 0.405 722 419 2;
  • 25) 0.405 722 419 2 × 2 = 0 + 0.811 444 838 4;
  • 26) 0.811 444 838 4 × 2 = 1 + 0.622 889 676 8;
  • 27) 0.622 889 676 8 × 2 = 1 + 0.245 779 353 6;
  • 28) 0.245 779 353 6 × 2 = 0 + 0.491 558 707 2;
  • 29) 0.491 558 707 2 × 2 = 0 + 0.983 117 414 4;
  • 30) 0.983 117 414 4 × 2 = 1 + 0.966 234 828 8;
  • 31) 0.966 234 828 8 × 2 = 1 + 0.932 469 657 6;
  • 32) 0.932 469 657 6 × 2 = 1 + 0.864 939 315 2;
  • 33) 0.864 939 315 2 × 2 = 1 + 0.729 878 630 4;
  • 34) 0.729 878 630 4 × 2 = 1 + 0.459 757 260 8;
  • 35) 0.459 757 260 8 × 2 = 0 + 0.919 514 521 6;
  • 36) 0.919 514 521 6 × 2 = 1 + 0.839 029 043 2;
  • 37) 0.839 029 043 2 × 2 = 1 + 0.678 058 086 4;
  • 38) 0.678 058 086 4 × 2 = 1 + 0.356 116 172 8;
  • 39) 0.356 116 172 8 × 2 = 0 + 0.712 232 345 6;
  • 40) 0.712 232 345 6 × 2 = 1 + 0.424 464 691 2;
  • 41) 0.424 464 691 2 × 2 = 0 + 0.848 929 382 4;
  • 42) 0.848 929 382 4 × 2 = 1 + 0.697 858 764 8;
  • 43) 0.697 858 764 8 × 2 = 1 + 0.395 717 529 6;
  • 44) 0.395 717 529 6 × 2 = 0 + 0.791 435 059 2;
  • 45) 0.791 435 059 2 × 2 = 1 + 0.582 870 118 4;
  • 46) 0.582 870 118 4 × 2 = 1 + 0.165 740 236 8;
  • 47) 0.165 740 236 8 × 2 = 0 + 0.331 480 473 6;
  • 48) 0.331 480 473 6 × 2 = 0 + 0.662 960 947 2;
  • 49) 0.662 960 947 2 × 2 = 1 + 0.325 921 894 4;
  • 50) 0.325 921 894 4 × 2 = 0 + 0.651 843 788 8;
  • 51) 0.651 843 788 8 × 2 = 1 + 0.303 687 577 6;
  • 52) 0.303 687 577 6 × 2 = 0 + 0.607 375 155 2;
  • 53) 0.607 375 155 2 × 2 = 1 + 0.214 750 310 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:


0.598 907 673 7(10) =


0.1001 1001 0101 0010 0000 0011 0110 0111 1101 1101 0110 1100 1010 1(2)


5. Positive number before normalization:

43 144.598 907 673 7(10) =


1010 1000 1000 1000.1001 1001 0101 0010 0000 0011 0110 0111 1101 1101 0110 1100 1010 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the left, so that only one non zero digit remains to the left of it:


43 144.598 907 673 7(10) =


1010 1000 1000 1000.1001 1001 0101 0010 0000 0011 0110 0111 1101 1101 0110 1100 1010 1(2) =


1010 1000 1000 1000.1001 1001 0101 0010 0000 0011 0110 0111 1101 1101 0110 1100 1010 1(2) × 20 =


1.0101 0001 0001 0001 0011 0010 1010 0100 0000 0110 1100 1111 1011 1010 1101 1001 0101(2) × 215


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): 15


Mantissa (not normalized):
1.0101 0001 0001 0001 0011 0010 1010 0100 0000 0110 1100 1111 1011 1010 1101 1001 0101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


15 + 2(11-1) - 1 =


(15 + 1 023)(10) =


1 038(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;
  • 1 038 ÷ 2 = 519 + 0;
  • 519 ÷ 2 = 259 + 1;
  • 259 ÷ 2 = 129 + 1;
  • 129 ÷ 2 = 64 + 1;
  • 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) =


1038(10) =


100 0000 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, 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. 0101 0001 0001 0001 0011 0010 1010 0100 0000 0110 1100 1111 1011 1010 1101 1001 0101 =


0101 0001 0001 0001 0011 0010 1010 0100 0000 0110 1100 1111 1011


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 1110


Mantissa (52 bits) =
0101 0001 0001 0001 0011 0010 1010 0100 0000 0110 1100 1111 1011


The base ten decimal number 43 144.598 907 673 7 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 1110 - 0101 0001 0001 0001 0011 0010 1010 0100 0000 0110 1100 1111 1011

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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