2.555 555 567 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.555 555 567 9(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
2.555 555 567 9(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: 2.
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;
  • 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.

2(10) =

10(2)


3. Convert to binary (base 2) the fractional part: 0.555 555 567 9.

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) 0.555 555 567 9 × 2 = 1 + 0.111 111 135 8;
  • 2) 0.111 111 135 8 × 2 = 0 + 0.222 222 271 6;
  • 3) 0.222 222 271 6 × 2 = 0 + 0.444 444 543 2;
  • 4) 0.444 444 543 2 × 2 = 0 + 0.888 889 086 4;
  • 5) 0.888 889 086 4 × 2 = 1 + 0.777 778 172 8;
  • 6) 0.777 778 172 8 × 2 = 1 + 0.555 556 345 6;
  • 7) 0.555 556 345 6 × 2 = 1 + 0.111 112 691 2;
  • 8) 0.111 112 691 2 × 2 = 0 + 0.222 225 382 4;
  • 9) 0.222 225 382 4 × 2 = 0 + 0.444 450 764 8;
  • 10) 0.444 450 764 8 × 2 = 0 + 0.888 901 529 6;
  • 11) 0.888 901 529 6 × 2 = 1 + 0.777 803 059 2;
  • 12) 0.777 803 059 2 × 2 = 1 + 0.555 606 118 4;
  • 13) 0.555 606 118 4 × 2 = 1 + 0.111 212 236 8;
  • 14) 0.111 212 236 8 × 2 = 0 + 0.222 424 473 6;
  • 15) 0.222 424 473 6 × 2 = 0 + 0.444 848 947 2;
  • 16) 0.444 848 947 2 × 2 = 0 + 0.889 697 894 4;
  • 17) 0.889 697 894 4 × 2 = 1 + 0.779 395 788 8;
  • 18) 0.779 395 788 8 × 2 = 1 + 0.558 791 577 6;
  • 19) 0.558 791 577 6 × 2 = 1 + 0.117 583 155 2;
  • 20) 0.117 583 155 2 × 2 = 0 + 0.235 166 310 4;
  • 21) 0.235 166 310 4 × 2 = 0 + 0.470 332 620 8;
  • 22) 0.470 332 620 8 × 2 = 0 + 0.940 665 241 6;
  • 23) 0.940 665 241 6 × 2 = 1 + 0.881 330 483 2;
  • 24) 0.881 330 483 2 × 2 = 1 + 0.762 660 966 4;
  • 25) 0.762 660 966 4 × 2 = 1 + 0.525 321 932 8;
  • 26) 0.525 321 932 8 × 2 = 1 + 0.050 643 865 6;
  • 27) 0.050 643 865 6 × 2 = 0 + 0.101 287 731 2;
  • 28) 0.101 287 731 2 × 2 = 0 + 0.202 575 462 4;
  • 29) 0.202 575 462 4 × 2 = 0 + 0.405 150 924 8;
  • 30) 0.405 150 924 8 × 2 = 0 + 0.810 301 849 6;
  • 31) 0.810 301 849 6 × 2 = 1 + 0.620 603 699 2;
  • 32) 0.620 603 699 2 × 2 = 1 + 0.241 207 398 4;
  • 33) 0.241 207 398 4 × 2 = 0 + 0.482 414 796 8;
  • 34) 0.482 414 796 8 × 2 = 0 + 0.964 829 593 6;
  • 35) 0.964 829 593 6 × 2 = 1 + 0.929 659 187 2;
  • 36) 0.929 659 187 2 × 2 = 1 + 0.859 318 374 4;
  • 37) 0.859 318 374 4 × 2 = 1 + 0.718 636 748 8;
  • 38) 0.718 636 748 8 × 2 = 1 + 0.437 273 497 6;
  • 39) 0.437 273 497 6 × 2 = 0 + 0.874 546 995 2;
  • 40) 0.874 546 995 2 × 2 = 1 + 0.749 093 990 4;
  • 41) 0.749 093 990 4 × 2 = 1 + 0.498 187 980 8;
  • 42) 0.498 187 980 8 × 2 = 0 + 0.996 375 961 6;
  • 43) 0.996 375 961 6 × 2 = 1 + 0.992 751 923 2;
  • 44) 0.992 751 923 2 × 2 = 1 + 0.985 503 846 4;
  • 45) 0.985 503 846 4 × 2 = 1 + 0.971 007 692 8;
  • 46) 0.971 007 692 8 × 2 = 1 + 0.942 015 385 6;
  • 47) 0.942 015 385 6 × 2 = 1 + 0.884 030 771 2;
  • 48) 0.884 030 771 2 × 2 = 1 + 0.768 061 542 4;
  • 49) 0.768 061 542 4 × 2 = 1 + 0.536 123 084 8;
  • 50) 0.536 123 084 8 × 2 = 1 + 0.072 246 169 6;
  • 51) 0.072 246 169 6 × 2 = 0 + 0.144 492 339 2;
  • 52) 0.144 492 339 2 × 2 = 0 + 0.288 984 678 4;
  • 53) 0.288 984 678 4 × 2 = 0 + 0.577 969 356 8;

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:


0.555 555 567 9(10) =

0.1000 1110 0011 1000 1110 0011 1100 0011 0011 1101 1011 1111 1100 0(2)

5. Positive number before normalization:

2.555 555 567 9(10) =

10.1000 1110 0011 1000 1110 0011 1100 0011 0011 1101 1011 1111 1100 0(2)

6. Normalize the binary representation of the number.

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


2.555 555 567 9(10) =

10.1000 1110 0011 1000 1110 0011 1100 0011 0011 1101 1011 1111 1100 0(2) =

10.1000 1110 0011 1000 1110 0011 1100 0011 0011 1101 1011 1111 1100 0(2) × 20 =

1.0100 0111 0001 1100 0111 0001 1110 0001 1001 1110 1101 1111 1110 00(2) × 21


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):
1.0100 0111 0001 1100 0111 0001 1110 0001 1001 1110 1101 1111 1110 00


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) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 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) =

1024(10) =

100 0000 0000(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. 0100 0111 0001 1100 0111 0001 1110 0001 1001 1110 1101 1111 1110 00 =

0100 0111 0001 1100 0111 0001 1110 0001 1001 1110 1101 1111 1110


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 0000

Mantissa (52 bits) =
0100 0111 0001 1100 0111 0001 1110 0001 1001 1110 1101 1111 1110


Decimal number 2.555 555 567 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0100 0111 0001 1100 0111 0001 1110 0001 1001 1110 1101 1111 1110

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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