Decimal to 64 Bit IEEE 754 Binary: Convert Number 0.019 999 999 999 999 93 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 0.019 999 999 999 999 93(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: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.019 999 999 999 999 93.

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) 0.019 999 999 999 999 93 × 2 = 0 + 0.039 999 999 999 999 86;
  • 2) 0.039 999 999 999 999 86 × 2 = 0 + 0.079 999 999 999 999 72;
  • 3) 0.079 999 999 999 999 72 × 2 = 0 + 0.159 999 999 999 999 44;
  • 4) 0.159 999 999 999 999 44 × 2 = 0 + 0.319 999 999 999 998 88;
  • 5) 0.319 999 999 999 998 88 × 2 = 0 + 0.639 999 999 999 997 76;
  • 6) 0.639 999 999 999 997 76 × 2 = 1 + 0.279 999 999 999 995 52;
  • 7) 0.279 999 999 999 995 52 × 2 = 0 + 0.559 999 999 999 991 04;
  • 8) 0.559 999 999 999 991 04 × 2 = 1 + 0.119 999 999 999 982 08;
  • 9) 0.119 999 999 999 982 08 × 2 = 0 + 0.239 999 999 999 964 16;
  • 10) 0.239 999 999 999 964 16 × 2 = 0 + 0.479 999 999 999 928 32;
  • 11) 0.479 999 999 999 928 32 × 2 = 0 + 0.959 999 999 999 856 64;
  • 12) 0.959 999 999 999 856 64 × 2 = 1 + 0.919 999 999 999 713 28;
  • 13) 0.919 999 999 999 713 28 × 2 = 1 + 0.839 999 999 999 426 56;
  • 14) 0.839 999 999 999 426 56 × 2 = 1 + 0.679 999 999 998 853 12;
  • 15) 0.679 999 999 998 853 12 × 2 = 1 + 0.359 999 999 997 706 24;
  • 16) 0.359 999 999 997 706 24 × 2 = 0 + 0.719 999 999 995 412 48;
  • 17) 0.719 999 999 995 412 48 × 2 = 1 + 0.439 999 999 990 824 96;
  • 18) 0.439 999 999 990 824 96 × 2 = 0 + 0.879 999 999 981 649 92;
  • 19) 0.879 999 999 981 649 92 × 2 = 1 + 0.759 999 999 963 299 84;
  • 20) 0.759 999 999 963 299 84 × 2 = 1 + 0.519 999 999 926 599 68;
  • 21) 0.519 999 999 926 599 68 × 2 = 1 + 0.039 999 999 853 199 36;
  • 22) 0.039 999 999 853 199 36 × 2 = 0 + 0.079 999 999 706 398 72;
  • 23) 0.079 999 999 706 398 72 × 2 = 0 + 0.159 999 999 412 797 44;
  • 24) 0.159 999 999 412 797 44 × 2 = 0 + 0.319 999 998 825 594 88;
  • 25) 0.319 999 998 825 594 88 × 2 = 0 + 0.639 999 997 651 189 76;
  • 26) 0.639 999 997 651 189 76 × 2 = 1 + 0.279 999 995 302 379 52;
  • 27) 0.279 999 995 302 379 52 × 2 = 0 + 0.559 999 990 604 759 04;
  • 28) 0.559 999 990 604 759 04 × 2 = 1 + 0.119 999 981 209 518 08;
  • 29) 0.119 999 981 209 518 08 × 2 = 0 + 0.239 999 962 419 036 16;
  • 30) 0.239 999 962 419 036 16 × 2 = 0 + 0.479 999 924 838 072 32;
  • 31) 0.479 999 924 838 072 32 × 2 = 0 + 0.959 999 849 676 144 64;
  • 32) 0.959 999 849 676 144 64 × 2 = 1 + 0.919 999 699 352 289 28;
  • 33) 0.919 999 699 352 289 28 × 2 = 1 + 0.839 999 398 704 578 56;
  • 34) 0.839 999 398 704 578 56 × 2 = 1 + 0.679 998 797 409 157 12;
  • 35) 0.679 998 797 409 157 12 × 2 = 1 + 0.359 997 594 818 314 24;
  • 36) 0.359 997 594 818 314 24 × 2 = 0 + 0.719 995 189 636 628 48;
  • 37) 0.719 995 189 636 628 48 × 2 = 1 + 0.439 990 379 273 256 96;
  • 38) 0.439 990 379 273 256 96 × 2 = 0 + 0.879 980 758 546 513 92;
  • 39) 0.879 980 758 546 513 92 × 2 = 1 + 0.759 961 517 093 027 84;
  • 40) 0.759 961 517 093 027 84 × 2 = 1 + 0.519 923 034 186 055 68;
  • 41) 0.519 923 034 186 055 68 × 2 = 1 + 0.039 846 068 372 111 36;
  • 42) 0.039 846 068 372 111 36 × 2 = 0 + 0.079 692 136 744 222 72;
  • 43) 0.079 692 136 744 222 72 × 2 = 0 + 0.159 384 273 488 445 44;
  • 44) 0.159 384 273 488 445 44 × 2 = 0 + 0.318 768 546 976 890 88;
  • 45) 0.318 768 546 976 890 88 × 2 = 0 + 0.637 537 093 953 781 76;
  • 46) 0.637 537 093 953 781 76 × 2 = 1 + 0.275 074 187 907 563 52;
  • 47) 0.275 074 187 907 563 52 × 2 = 0 + 0.550 148 375 815 127 04;
  • 48) 0.550 148 375 815 127 04 × 2 = 1 + 0.100 296 751 630 254 08;
  • 49) 0.100 296 751 630 254 08 × 2 = 0 + 0.200 593 503 260 508 16;
  • 50) 0.200 593 503 260 508 16 × 2 = 0 + 0.401 187 006 521 016 32;
  • 51) 0.401 187 006 521 016 32 × 2 = 0 + 0.802 374 013 042 032 64;
  • 52) 0.802 374 013 042 032 64 × 2 = 1 + 0.604 748 026 084 065 28;
  • 53) 0.604 748 026 084 065 28 × 2 = 1 + 0.209 496 052 168 130 56;
  • 54) 0.209 496 052 168 130 56 × 2 = 0 + 0.418 992 104 336 261 12;
  • 55) 0.418 992 104 336 261 12 × 2 = 0 + 0.837 984 208 672 522 24;
  • 56) 0.837 984 208 672 522 24 × 2 = 1 + 0.675 968 417 345 044 48;
  • 57) 0.675 968 417 345 044 48 × 2 = 1 + 0.351 936 834 690 088 96;
  • 58) 0.351 936 834 690 088 96 × 2 = 0 + 0.703 873 669 380 177 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:


0.019 999 999 999 999 93(10) =

0.0000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1001 10(2)

5. Positive number before normalization:

0.019 999 999 999 999 93(10) =

0.0000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1001 10(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.019 999 999 999 999 93(10) =

0.0000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1001 10(2) =

0.0000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1001 10(2) × 20 =

1.0100 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0110 0110(2) × 2-6


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): -6

Mantissa (not normalized):
1.0100 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0110 0110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1017(10) =

011 1111 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0100 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0110 0110 =

0100 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0110 0110


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) =
011 1111 1001

Mantissa (52 bits) =
0100 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0110 0110


The base ten decimal number 0.019 999 999 999 999 93 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1001 - 0100 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0110 0110

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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