Decimal to 64 Bit IEEE 754 Binary: Convert Number 16.929 817 199 76 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 16.929 817 199 76(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: 16.
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;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 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.

16(10) =


1 0000(2)


3. Convert to binary (base 2) the fractional part: 0.929 817 199 76.

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) 0.929 817 199 76 × 2 = 1 + 0.859 634 399 52;
  • 2) 0.859 634 399 52 × 2 = 1 + 0.719 268 799 04;
  • 3) 0.719 268 799 04 × 2 = 1 + 0.438 537 598 08;
  • 4) 0.438 537 598 08 × 2 = 0 + 0.877 075 196 16;
  • 5) 0.877 075 196 16 × 2 = 1 + 0.754 150 392 32;
  • 6) 0.754 150 392 32 × 2 = 1 + 0.508 300 784 64;
  • 7) 0.508 300 784 64 × 2 = 1 + 0.016 601 569 28;
  • 8) 0.016 601 569 28 × 2 = 0 + 0.033 203 138 56;
  • 9) 0.033 203 138 56 × 2 = 0 + 0.066 406 277 12;
  • 10) 0.066 406 277 12 × 2 = 0 + 0.132 812 554 24;
  • 11) 0.132 812 554 24 × 2 = 0 + 0.265 625 108 48;
  • 12) 0.265 625 108 48 × 2 = 0 + 0.531 250 216 96;
  • 13) 0.531 250 216 96 × 2 = 1 + 0.062 500 433 92;
  • 14) 0.062 500 433 92 × 2 = 0 + 0.125 000 867 84;
  • 15) 0.125 000 867 84 × 2 = 0 + 0.250 001 735 68;
  • 16) 0.250 001 735 68 × 2 = 0 + 0.500 003 471 36;
  • 17) 0.500 003 471 36 × 2 = 1 + 0.000 006 942 72;
  • 18) 0.000 006 942 72 × 2 = 0 + 0.000 013 885 44;
  • 19) 0.000 013 885 44 × 2 = 0 + 0.000 027 770 88;
  • 20) 0.000 027 770 88 × 2 = 0 + 0.000 055 541 76;
  • 21) 0.000 055 541 76 × 2 = 0 + 0.000 111 083 52;
  • 22) 0.000 111 083 52 × 2 = 0 + 0.000 222 167 04;
  • 23) 0.000 222 167 04 × 2 = 0 + 0.000 444 334 08;
  • 24) 0.000 444 334 08 × 2 = 0 + 0.000 888 668 16;
  • 25) 0.000 888 668 16 × 2 = 0 + 0.001 777 336 32;
  • 26) 0.001 777 336 32 × 2 = 0 + 0.003 554 672 64;
  • 27) 0.003 554 672 64 × 2 = 0 + 0.007 109 345 28;
  • 28) 0.007 109 345 28 × 2 = 0 + 0.014 218 690 56;
  • 29) 0.014 218 690 56 × 2 = 0 + 0.028 437 381 12;
  • 30) 0.028 437 381 12 × 2 = 0 + 0.056 874 762 24;
  • 31) 0.056 874 762 24 × 2 = 0 + 0.113 749 524 48;
  • 32) 0.113 749 524 48 × 2 = 0 + 0.227 499 048 96;
  • 33) 0.227 499 048 96 × 2 = 0 + 0.454 998 097 92;
  • 34) 0.454 998 097 92 × 2 = 0 + 0.909 996 195 84;
  • 35) 0.909 996 195 84 × 2 = 1 + 0.819 992 391 68;
  • 36) 0.819 992 391 68 × 2 = 1 + 0.639 984 783 36;
  • 37) 0.639 984 783 36 × 2 = 1 + 0.279 969 566 72;
  • 38) 0.279 969 566 72 × 2 = 0 + 0.559 939 133 44;
  • 39) 0.559 939 133 44 × 2 = 1 + 0.119 878 266 88;
  • 40) 0.119 878 266 88 × 2 = 0 + 0.239 756 533 76;
  • 41) 0.239 756 533 76 × 2 = 0 + 0.479 513 067 52;
  • 42) 0.479 513 067 52 × 2 = 0 + 0.959 026 135 04;
  • 43) 0.959 026 135 04 × 2 = 1 + 0.918 052 270 08;
  • 44) 0.918 052 270 08 × 2 = 1 + 0.836 104 540 16;
  • 45) 0.836 104 540 16 × 2 = 1 + 0.672 209 080 32;
  • 46) 0.672 209 080 32 × 2 = 1 + 0.344 418 160 64;
  • 47) 0.344 418 160 64 × 2 = 0 + 0.688 836 321 28;
  • 48) 0.688 836 321 28 × 2 = 1 + 0.377 672 642 56;
  • 49) 0.377 672 642 56 × 2 = 0 + 0.755 345 285 12;
  • 50) 0.755 345 285 12 × 2 = 1 + 0.510 690 570 24;
  • 51) 0.510 690 570 24 × 2 = 1 + 0.021 381 140 48;
  • 52) 0.021 381 140 48 × 2 = 0 + 0.042 762 280 96;
  • 53) 0.042 762 280 96 × 2 = 0 + 0.085 524 561 92;

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:


0.929 817 199 76(10) =


0.1110 1110 0000 1000 1000 0000 0000 0000 0011 1010 0011 1101 0110 0(2)

5. Positive number before normalization:

16.929 817 199 76(10) =


1 0000.1110 1110 0000 1000 1000 0000 0000 0000 0011 1010 0011 1101 0110 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the left, so that only one non zero digit remains to the left of it:


16.929 817 199 76(10) =


1 0000.1110 1110 0000 1000 1000 0000 0000 0000 0011 1010 0011 1101 0110 0(2) =


1 0000.1110 1110 0000 1000 1000 0000 0000 0000 0011 1010 0011 1101 0110 0(2) × 20 =


1.0000 1110 1110 0000 1000 1000 0000 0000 0000 0011 1010 0011 1101 0110 0(2) × 24


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): 4


Mantissa (not normalized):
1.0000 1110 1110 0000 1000 1000 0000 0000 0000 0011 1010 0011 1101 0110 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


4 + 2(11-1) - 1 =


(4 + 1 023)(10) =


1 027(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;
  • 1 027 ÷ 2 = 513 + 1;
  • 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) =


1027(10) =


100 0000 0011(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. 0000 1110 1110 0000 1000 1000 0000 0000 0000 0011 1010 0011 1101 0 1100 =


0000 1110 1110 0000 1000 1000 0000 0000 0000 0011 1010 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 0011


Mantissa (52 bits) =
0000 1110 1110 0000 1000 1000 0000 0000 0000 0011 1010 0011 1101


The base ten decimal number 16.929 817 199 76 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0011 - 0000 1110 1110 0000 1000 1000 0000 0000 0000 0011 1010 0011 1101

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