27.555 555 555 555 555 555 555 555 555 555 555 555 551 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 27.555 555 555 555 555 555 555 555 555 555 555 555 551 2(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
27.555 555 555 555 555 555 555 555 555 555 555 555 551 2(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: 27.
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;
  • 27 ÷ 2 = 13 + 1;
  • 13 ÷ 2 = 6 + 1;
  • 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.

27(10) =

1 1011(2)


3. Convert to binary (base 2) the fractional part: 0.555 555 555 555 555 555 555 555 555 555 555 555 551 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) 0.555 555 555 555 555 555 555 555 555 555 555 555 551 2 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 102 4;
  • 2) 0.111 111 111 111 111 111 111 111 111 111 111 111 102 4 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 204 8;
  • 3) 0.222 222 222 222 222 222 222 222 222 222 222 222 204 8 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 409 6;
  • 4) 0.444 444 444 444 444 444 444 444 444 444 444 444 409 6 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 819 2;
  • 5) 0.888 888 888 888 888 888 888 888 888 888 888 888 819 2 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 638 4;
  • 6) 0.777 777 777 777 777 777 777 777 777 777 777 777 638 4 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 276 8;
  • 7) 0.555 555 555 555 555 555 555 555 555 555 555 555 276 8 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 110 553 6;
  • 8) 0.111 111 111 111 111 111 111 111 111 111 111 110 553 6 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 221 107 2;
  • 9) 0.222 222 222 222 222 222 222 222 222 222 222 221 107 2 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 442 214 4;
  • 10) 0.444 444 444 444 444 444 444 444 444 444 444 442 214 4 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 884 428 8;
  • 11) 0.888 888 888 888 888 888 888 888 888 888 888 884 428 8 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 768 857 6;
  • 12) 0.777 777 777 777 777 777 777 777 777 777 777 768 857 6 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 537 715 2;
  • 13) 0.555 555 555 555 555 555 555 555 555 555 555 537 715 2 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 075 430 4;
  • 14) 0.111 111 111 111 111 111 111 111 111 111 111 075 430 4 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 150 860 8;
  • 15) 0.222 222 222 222 222 222 222 222 222 222 222 150 860 8 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 301 721 6;
  • 16) 0.444 444 444 444 444 444 444 444 444 444 444 301 721 6 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 603 443 2;
  • 17) 0.888 888 888 888 888 888 888 888 888 888 888 603 443 2 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 206 886 4;
  • 18) 0.777 777 777 777 777 777 777 777 777 777 777 206 886 4 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 554 413 772 8;
  • 19) 0.555 555 555 555 555 555 555 555 555 555 554 413 772 8 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 108 827 545 6;
  • 20) 0.111 111 111 111 111 111 111 111 111 111 108 827 545 6 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 217 655 091 2;
  • 21) 0.222 222 222 222 222 222 222 222 222 222 217 655 091 2 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 435 310 182 4;
  • 22) 0.444 444 444 444 444 444 444 444 444 444 435 310 182 4 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 870 620 364 8;
  • 23) 0.888 888 888 888 888 888 888 888 888 888 870 620 364 8 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 741 240 729 6;
  • 24) 0.777 777 777 777 777 777 777 777 777 777 741 240 729 6 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 482 481 459 2;
  • 25) 0.555 555 555 555 555 555 555 555 555 555 482 481 459 2 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 110 964 962 918 4;
  • 26) 0.111 111 111 111 111 111 111 111 111 110 964 962 918 4 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 221 929 925 836 8;
  • 27) 0.222 222 222 222 222 222 222 222 222 221 929 925 836 8 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 443 859 851 673 6;
  • 28) 0.444 444 444 444 444 444 444 444 444 443 859 851 673 6 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 887 719 703 347 2;
  • 29) 0.888 888 888 888 888 888 888 888 888 887 719 703 347 2 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 775 439 406 694 4;
  • 30) 0.777 777 777 777 777 777 777 777 777 775 439 406 694 4 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 550 878 813 388 8;
  • 31) 0.555 555 555 555 555 555 555 555 555 550 878 813 388 8 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 101 757 626 777 6;
  • 32) 0.111 111 111 111 111 111 111 111 111 101 757 626 777 6 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 203 515 253 555 2;
  • 33) 0.222 222 222 222 222 222 222 222 222 203 515 253 555 2 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 407 030 507 110 4;
  • 34) 0.444 444 444 444 444 444 444 444 444 407 030 507 110 4 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 814 061 014 220 8;
  • 35) 0.888 888 888 888 888 888 888 888 888 814 061 014 220 8 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 628 122 028 441 6;
  • 36) 0.777 777 777 777 777 777 777 777 777 628 122 028 441 6 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 256 244 056 883 2;
  • 37) 0.555 555 555 555 555 555 555 555 555 256 244 056 883 2 × 2 = 1 + 0.111 111 111 111 111 111 111 111 110 512 488 113 766 4;
  • 38) 0.111 111 111 111 111 111 111 111 110 512 488 113 766 4 × 2 = 0 + 0.222 222 222 222 222 222 222 222 221 024 976 227 532 8;
  • 39) 0.222 222 222 222 222 222 222 222 221 024 976 227 532 8 × 2 = 0 + 0.444 444 444 444 444 444 444 444 442 049 952 455 065 6;
  • 40) 0.444 444 444 444 444 444 444 444 442 049 952 455 065 6 × 2 = 0 + 0.888 888 888 888 888 888 888 888 884 099 904 910 131 2;
  • 41) 0.888 888 888 888 888 888 888 888 884 099 904 910 131 2 × 2 = 1 + 0.777 777 777 777 777 777 777 777 768 199 809 820 262 4;
  • 42) 0.777 777 777 777 777 777 777 777 768 199 809 820 262 4 × 2 = 1 + 0.555 555 555 555 555 555 555 555 536 399 619 640 524 8;
  • 43) 0.555 555 555 555 555 555 555 555 536 399 619 640 524 8 × 2 = 1 + 0.111 111 111 111 111 111 111 111 072 799 239 281 049 6;
  • 44) 0.111 111 111 111 111 111 111 111 072 799 239 281 049 6 × 2 = 0 + 0.222 222 222 222 222 222 222 222 145 598 478 562 099 2;
  • 45) 0.222 222 222 222 222 222 222 222 145 598 478 562 099 2 × 2 = 0 + 0.444 444 444 444 444 444 444 444 291 196 957 124 198 4;
  • 46) 0.444 444 444 444 444 444 444 444 291 196 957 124 198 4 × 2 = 0 + 0.888 888 888 888 888 888 888 888 582 393 914 248 396 8;
  • 47) 0.888 888 888 888 888 888 888 888 582 393 914 248 396 8 × 2 = 1 + 0.777 777 777 777 777 777 777 777 164 787 828 496 793 6;
  • 48) 0.777 777 777 777 777 777 777 777 164 787 828 496 793 6 × 2 = 1 + 0.555 555 555 555 555 555 555 554 329 575 656 993 587 2;
  • 49) 0.555 555 555 555 555 555 555 554 329 575 656 993 587 2 × 2 = 1 + 0.111 111 111 111 111 111 111 108 659 151 313 987 174 4;
  • 50) 0.111 111 111 111 111 111 111 108 659 151 313 987 174 4 × 2 = 0 + 0.222 222 222 222 222 222 222 217 318 302 627 974 348 8;
  • 51) 0.222 222 222 222 222 222 222 217 318 302 627 974 348 8 × 2 = 0 + 0.444 444 444 444 444 444 444 434 636 605 255 948 697 6;
  • 52) 0.444 444 444 444 444 444 444 434 636 605 255 948 697 6 × 2 = 0 + 0.888 888 888 888 888 888 888 869 273 210 511 897 395 2;
  • 53) 0.888 888 888 888 888 888 888 869 273 210 511 897 395 2 × 2 = 1 + 0.777 777 777 777 777 777 777 738 546 421 023 794 790 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 - 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:


0.555 555 555 555 555 555 555 555 555 555 555 555 551 2(10) =

0.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1(2)

5. Positive number before normalization:

27.555 555 555 555 555 555 555 555 555 555 555 555 551 2(10) =

1 1011.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1(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:


27.555 555 555 555 555 555 555 555 555 555 555 555 551 2(10) =

1 1011.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1(2) =

1 1011.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1(2) × 20 =

1.1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1(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):
1.1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1


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. 1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1 0001 =

1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011


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) =
1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011


Decimal number 27.555 555 555 555 555 555 555 555 555 555 555 555 551 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0011 - 1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011

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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