64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 12.499 999 999 999 2 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 12.499 999 999 999 2(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: 12.
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;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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.


12(10) =


1100(2)


3. Convert to binary (base 2) the fractional part: 0.499 999 999 999 2.

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) 0.499 999 999 999 2 × 2 = 0 + 0.999 999 999 998 4;
  • 2) 0.999 999 999 998 4 × 2 = 1 + 0.999 999 999 996 8;
  • 3) 0.999 999 999 996 8 × 2 = 1 + 0.999 999 999 993 6;
  • 4) 0.999 999 999 993 6 × 2 = 1 + 0.999 999 999 987 2;
  • 5) 0.999 999 999 987 2 × 2 = 1 + 0.999 999 999 974 4;
  • 6) 0.999 999 999 974 4 × 2 = 1 + 0.999 999 999 948 8;
  • 7) 0.999 999 999 948 8 × 2 = 1 + 0.999 999 999 897 6;
  • 8) 0.999 999 999 897 6 × 2 = 1 + 0.999 999 999 795 2;
  • 9) 0.999 999 999 795 2 × 2 = 1 + 0.999 999 999 590 4;
  • 10) 0.999 999 999 590 4 × 2 = 1 + 0.999 999 999 180 8;
  • 11) 0.999 999 999 180 8 × 2 = 1 + 0.999 999 998 361 6;
  • 12) 0.999 999 998 361 6 × 2 = 1 + 0.999 999 996 723 2;
  • 13) 0.999 999 996 723 2 × 2 = 1 + 0.999 999 993 446 4;
  • 14) 0.999 999 993 446 4 × 2 = 1 + 0.999 999 986 892 8;
  • 15) 0.999 999 986 892 8 × 2 = 1 + 0.999 999 973 785 6;
  • 16) 0.999 999 973 785 6 × 2 = 1 + 0.999 999 947 571 2;
  • 17) 0.999 999 947 571 2 × 2 = 1 + 0.999 999 895 142 4;
  • 18) 0.999 999 895 142 4 × 2 = 1 + 0.999 999 790 284 8;
  • 19) 0.999 999 790 284 8 × 2 = 1 + 0.999 999 580 569 6;
  • 20) 0.999 999 580 569 6 × 2 = 1 + 0.999 999 161 139 2;
  • 21) 0.999 999 161 139 2 × 2 = 1 + 0.999 998 322 278 4;
  • 22) 0.999 998 322 278 4 × 2 = 1 + 0.999 996 644 556 8;
  • 23) 0.999 996 644 556 8 × 2 = 1 + 0.999 993 289 113 6;
  • 24) 0.999 993 289 113 6 × 2 = 1 + 0.999 986 578 227 2;
  • 25) 0.999 986 578 227 2 × 2 = 1 + 0.999 973 156 454 4;
  • 26) 0.999 973 156 454 4 × 2 = 1 + 0.999 946 312 908 8;
  • 27) 0.999 946 312 908 8 × 2 = 1 + 0.999 892 625 817 6;
  • 28) 0.999 892 625 817 6 × 2 = 1 + 0.999 785 251 635 2;
  • 29) 0.999 785 251 635 2 × 2 = 1 + 0.999 570 503 270 4;
  • 30) 0.999 570 503 270 4 × 2 = 1 + 0.999 141 006 540 8;
  • 31) 0.999 141 006 540 8 × 2 = 1 + 0.998 282 013 081 6;
  • 32) 0.998 282 013 081 6 × 2 = 1 + 0.996 564 026 163 2;
  • 33) 0.996 564 026 163 2 × 2 = 1 + 0.993 128 052 326 4;
  • 34) 0.993 128 052 326 4 × 2 = 1 + 0.986 256 104 652 8;
  • 35) 0.986 256 104 652 8 × 2 = 1 + 0.972 512 209 305 6;
  • 36) 0.972 512 209 305 6 × 2 = 1 + 0.945 024 418 611 2;
  • 37) 0.945 024 418 611 2 × 2 = 1 + 0.890 048 837 222 4;
  • 38) 0.890 048 837 222 4 × 2 = 1 + 0.780 097 674 444 8;
  • 39) 0.780 097 674 444 8 × 2 = 1 + 0.560 195 348 889 6;
  • 40) 0.560 195 348 889 6 × 2 = 1 + 0.120 390 697 779 2;
  • 41) 0.120 390 697 779 2 × 2 = 0 + 0.240 781 395 558 4;
  • 42) 0.240 781 395 558 4 × 2 = 0 + 0.481 562 791 116 8;
  • 43) 0.481 562 791 116 8 × 2 = 0 + 0.963 125 582 233 6;
  • 44) 0.963 125 582 233 6 × 2 = 1 + 0.926 251 164 467 2;
  • 45) 0.926 251 164 467 2 × 2 = 1 + 0.852 502 328 934 4;
  • 46) 0.852 502 328 934 4 × 2 = 1 + 0.705 004 657 868 8;
  • 47) 0.705 004 657 868 8 × 2 = 1 + 0.410 009 315 737 6;
  • 48) 0.410 009 315 737 6 × 2 = 0 + 0.820 018 631 475 2;
  • 49) 0.820 018 631 475 2 × 2 = 1 + 0.640 037 262 950 4;
  • 50) 0.640 037 262 950 4 × 2 = 1 + 0.280 074 525 900 8;
  • 51) 0.280 074 525 900 8 × 2 = 0 + 0.560 149 051 801 6;
  • 52) 0.560 149 051 801 6 × 2 = 1 + 0.120 298 103 603 2;
  • 53) 0.120 298 103 603 2 × 2 = 0 + 0.240 596 207 206 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:


0.499 999 999 999 2(10) =


0.0111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0001 1110 1101 0(2)


5. Positive number before normalization:

12.499 999 999 999 2(10) =


1100.0111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0001 1110 1101 0(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:


12.499 999 999 999 2(10) =


1100.0111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0001 1110 1101 0(2) =


1100.0111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0001 1110 1101 0(2) × 20 =


1.1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 0011 1101 1010(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):
1.1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 0011 1101 1010


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. 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 0011 1101 1010 =


1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 0011 1101


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) =
1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 0011 1101


The base ten decimal number 12.499 999 999 999 2 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 0010 - 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 0011 1101

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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