0.519 999 999 999 877 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.519 999 999 999 877(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
0.519 999 999 999 877(10) to 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: 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: 0.519 999 999 999 877.

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) 0.519 999 999 999 877 × 2 = 1 + 0.039 999 999 999 754;
  • 2) 0.039 999 999 999 754 × 2 = 0 + 0.079 999 999 999 508;
  • 3) 0.079 999 999 999 508 × 2 = 0 + 0.159 999 999 999 016;
  • 4) 0.159 999 999 999 016 × 2 = 0 + 0.319 999 999 998 032;
  • 5) 0.319 999 999 998 032 × 2 = 0 + 0.639 999 999 996 064;
  • 6) 0.639 999 999 996 064 × 2 = 1 + 0.279 999 999 992 128;
  • 7) 0.279 999 999 992 128 × 2 = 0 + 0.559 999 999 984 256;
  • 8) 0.559 999 999 984 256 × 2 = 1 + 0.119 999 999 968 512;
  • 9) 0.119 999 999 968 512 × 2 = 0 + 0.239 999 999 937 024;
  • 10) 0.239 999 999 937 024 × 2 = 0 + 0.479 999 999 874 048;
  • 11) 0.479 999 999 874 048 × 2 = 0 + 0.959 999 999 748 096;
  • 12) 0.959 999 999 748 096 × 2 = 1 + 0.919 999 999 496 192;
  • 13) 0.919 999 999 496 192 × 2 = 1 + 0.839 999 998 992 384;
  • 14) 0.839 999 998 992 384 × 2 = 1 + 0.679 999 997 984 768;
  • 15) 0.679 999 997 984 768 × 2 = 1 + 0.359 999 995 969 536;
  • 16) 0.359 999 995 969 536 × 2 = 0 + 0.719 999 991 939 072;
  • 17) 0.719 999 991 939 072 × 2 = 1 + 0.439 999 983 878 144;
  • 18) 0.439 999 983 878 144 × 2 = 0 + 0.879 999 967 756 288;
  • 19) 0.879 999 967 756 288 × 2 = 1 + 0.759 999 935 512 576;
  • 20) 0.759 999 935 512 576 × 2 = 1 + 0.519 999 871 025 152;
  • 21) 0.519 999 871 025 152 × 2 = 1 + 0.039 999 742 050 304;
  • 22) 0.039 999 742 050 304 × 2 = 0 + 0.079 999 484 100 608;
  • 23) 0.079 999 484 100 608 × 2 = 0 + 0.159 998 968 201 216;
  • 24) 0.159 998 968 201 216 × 2 = 0 + 0.319 997 936 402 432;
  • 25) 0.319 997 936 402 432 × 2 = 0 + 0.639 995 872 804 864;
  • 26) 0.639 995 872 804 864 × 2 = 1 + 0.279 991 745 609 728;
  • 27) 0.279 991 745 609 728 × 2 = 0 + 0.559 983 491 219 456;
  • 28) 0.559 983 491 219 456 × 2 = 1 + 0.119 966 982 438 912;
  • 29) 0.119 966 982 438 912 × 2 = 0 + 0.239 933 964 877 824;
  • 30) 0.239 933 964 877 824 × 2 = 0 + 0.479 867 929 755 648;
  • 31) 0.479 867 929 755 648 × 2 = 0 + 0.959 735 859 511 296;
  • 32) 0.959 735 859 511 296 × 2 = 1 + 0.919 471 719 022 592;
  • 33) 0.919 471 719 022 592 × 2 = 1 + 0.838 943 438 045 184;
  • 34) 0.838 943 438 045 184 × 2 = 1 + 0.677 886 876 090 368;
  • 35) 0.677 886 876 090 368 × 2 = 1 + 0.355 773 752 180 736;
  • 36) 0.355 773 752 180 736 × 2 = 0 + 0.711 547 504 361 472;
  • 37) 0.711 547 504 361 472 × 2 = 1 + 0.423 095 008 722 944;
  • 38) 0.423 095 008 722 944 × 2 = 0 + 0.846 190 017 445 888;
  • 39) 0.846 190 017 445 888 × 2 = 1 + 0.692 380 034 891 776;
  • 40) 0.692 380 034 891 776 × 2 = 1 + 0.384 760 069 783 552;
  • 41) 0.384 760 069 783 552 × 2 = 0 + 0.769 520 139 567 104;
  • 42) 0.769 520 139 567 104 × 2 = 1 + 0.539 040 279 134 208;
  • 43) 0.539 040 279 134 208 × 2 = 1 + 0.078 080 558 268 416;
  • 44) 0.078 080 558 268 416 × 2 = 0 + 0.156 161 116 536 832;
  • 45) 0.156 161 116 536 832 × 2 = 0 + 0.312 322 233 073 664;
  • 46) 0.312 322 233 073 664 × 2 = 0 + 0.624 644 466 147 328;
  • 47) 0.624 644 466 147 328 × 2 = 1 + 0.249 288 932 294 656;
  • 48) 0.249 288 932 294 656 × 2 = 0 + 0.498 577 864 589 312;
  • 49) 0.498 577 864 589 312 × 2 = 0 + 0.997 155 729 178 624;
  • 50) 0.997 155 729 178 624 × 2 = 1 + 0.994 311 458 357 248;
  • 51) 0.994 311 458 357 248 × 2 = 1 + 0.988 622 916 714 496;
  • 52) 0.988 622 916 714 496 × 2 = 1 + 0.977 245 833 428 992;
  • 53) 0.977 245 833 428 992 × 2 = 1 + 0.954 491 666 857 984;

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:


0.519 999 999 999 877(10) =

0.1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0110 0010 0111 1(2)

5. Positive number before normalization:

0.519 999 999 999 877(10) =

0.1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0110 0010 0111 1(2)

6. Normalize the binary representation of the number.

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


0.519 999 999 999 877(10) =

0.1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0110 0010 0111 1(2) =

0.1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0110 0010 0111 1(2) × 20 =

1.0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1100 0100 1111(2) × 2-1


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):
1.0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1100 0100 1111


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) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 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) =

1022(10) =

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


Mantissa (normalized) =

1. 0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1100 0100 1111 =

0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1100 0100 1111


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 1110

Mantissa (52 bits) =
0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1100 0100 1111


Decimal number 0.519 999 999 999 877 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1100 0100 1111

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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