135.555 555 555 555 555 555 555 791 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 135.555 555 555 555 555 555 555 791(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
135.555 555 555 555 555 555 555 791(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: 135.
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;
  • 135 ÷ 2 = 67 + 1;
  • 67 ÷ 2 = 33 + 1;
  • 33 ÷ 2 = 16 + 1;
  • 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.

135(10) =

1000 0111(2)


3. Convert to binary (base 2) the fractional part: 0.555 555 555 555 555 555 555 791.

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) 0.555 555 555 555 555 555 555 791 × 2 = 1 + 0.111 111 111 111 111 111 111 582;
  • 2) 0.111 111 111 111 111 111 111 582 × 2 = 0 + 0.222 222 222 222 222 222 223 164;
  • 3) 0.222 222 222 222 222 222 223 164 × 2 = 0 + 0.444 444 444 444 444 444 446 328;
  • 4) 0.444 444 444 444 444 444 446 328 × 2 = 0 + 0.888 888 888 888 888 888 892 656;
  • 5) 0.888 888 888 888 888 888 892 656 × 2 = 1 + 0.777 777 777 777 777 777 785 312;
  • 6) 0.777 777 777 777 777 777 785 312 × 2 = 1 + 0.555 555 555 555 555 555 570 624;
  • 7) 0.555 555 555 555 555 555 570 624 × 2 = 1 + 0.111 111 111 111 111 111 141 248;
  • 8) 0.111 111 111 111 111 111 141 248 × 2 = 0 + 0.222 222 222 222 222 222 282 496;
  • 9) 0.222 222 222 222 222 222 282 496 × 2 = 0 + 0.444 444 444 444 444 444 564 992;
  • 10) 0.444 444 444 444 444 444 564 992 × 2 = 0 + 0.888 888 888 888 888 889 129 984;
  • 11) 0.888 888 888 888 888 889 129 984 × 2 = 1 + 0.777 777 777 777 777 778 259 968;
  • 12) 0.777 777 777 777 777 778 259 968 × 2 = 1 + 0.555 555 555 555 555 556 519 936;
  • 13) 0.555 555 555 555 555 556 519 936 × 2 = 1 + 0.111 111 111 111 111 113 039 872;
  • 14) 0.111 111 111 111 111 113 039 872 × 2 = 0 + 0.222 222 222 222 222 226 079 744;
  • 15) 0.222 222 222 222 222 226 079 744 × 2 = 0 + 0.444 444 444 444 444 452 159 488;
  • 16) 0.444 444 444 444 444 452 159 488 × 2 = 0 + 0.888 888 888 888 888 904 318 976;
  • 17) 0.888 888 888 888 888 904 318 976 × 2 = 1 + 0.777 777 777 777 777 808 637 952;
  • 18) 0.777 777 777 777 777 808 637 952 × 2 = 1 + 0.555 555 555 555 555 617 275 904;
  • 19) 0.555 555 555 555 555 617 275 904 × 2 = 1 + 0.111 111 111 111 111 234 551 808;
  • 20) 0.111 111 111 111 111 234 551 808 × 2 = 0 + 0.222 222 222 222 222 469 103 616;
  • 21) 0.222 222 222 222 222 469 103 616 × 2 = 0 + 0.444 444 444 444 444 938 207 232;
  • 22) 0.444 444 444 444 444 938 207 232 × 2 = 0 + 0.888 888 888 888 889 876 414 464;
  • 23) 0.888 888 888 888 889 876 414 464 × 2 = 1 + 0.777 777 777 777 779 752 828 928;
  • 24) 0.777 777 777 777 779 752 828 928 × 2 = 1 + 0.555 555 555 555 559 505 657 856;
  • 25) 0.555 555 555 555 559 505 657 856 × 2 = 1 + 0.111 111 111 111 119 011 315 712;
  • 26) 0.111 111 111 111 119 011 315 712 × 2 = 0 + 0.222 222 222 222 238 022 631 424;
  • 27) 0.222 222 222 222 238 022 631 424 × 2 = 0 + 0.444 444 444 444 476 045 262 848;
  • 28) 0.444 444 444 444 476 045 262 848 × 2 = 0 + 0.888 888 888 888 952 090 525 696;
  • 29) 0.888 888 888 888 952 090 525 696 × 2 = 1 + 0.777 777 777 777 904 181 051 392;
  • 30) 0.777 777 777 777 904 181 051 392 × 2 = 1 + 0.555 555 555 555 808 362 102 784;
  • 31) 0.555 555 555 555 808 362 102 784 × 2 = 1 + 0.111 111 111 111 616 724 205 568;
  • 32) 0.111 111 111 111 616 724 205 568 × 2 = 0 + 0.222 222 222 223 233 448 411 136;
  • 33) 0.222 222 222 223 233 448 411 136 × 2 = 0 + 0.444 444 444 446 466 896 822 272;
  • 34) 0.444 444 444 446 466 896 822 272 × 2 = 0 + 0.888 888 888 892 933 793 644 544;
  • 35) 0.888 888 888 892 933 793 644 544 × 2 = 1 + 0.777 777 777 785 867 587 289 088;
  • 36) 0.777 777 777 785 867 587 289 088 × 2 = 1 + 0.555 555 555 571 735 174 578 176;
  • 37) 0.555 555 555 571 735 174 578 176 × 2 = 1 + 0.111 111 111 143 470 349 156 352;
  • 38) 0.111 111 111 143 470 349 156 352 × 2 = 0 + 0.222 222 222 286 940 698 312 704;
  • 39) 0.222 222 222 286 940 698 312 704 × 2 = 0 + 0.444 444 444 573 881 396 625 408;
  • 40) 0.444 444 444 573 881 396 625 408 × 2 = 0 + 0.888 888 889 147 762 793 250 816;
  • 41) 0.888 888 889 147 762 793 250 816 × 2 = 1 + 0.777 777 778 295 525 586 501 632;
  • 42) 0.777 777 778 295 525 586 501 632 × 2 = 1 + 0.555 555 556 591 051 173 003 264;
  • 43) 0.555 555 556 591 051 173 003 264 × 2 = 1 + 0.111 111 113 182 102 346 006 528;
  • 44) 0.111 111 113 182 102 346 006 528 × 2 = 0 + 0.222 222 226 364 204 692 013 056;
  • 45) 0.222 222 226 364 204 692 013 056 × 2 = 0 + 0.444 444 452 728 409 384 026 112;
  • 46) 0.444 444 452 728 409 384 026 112 × 2 = 0 + 0.888 888 905 456 818 768 052 224;
  • 47) 0.888 888 905 456 818 768 052 224 × 2 = 1 + 0.777 777 810 913 637 536 104 448;
  • 48) 0.777 777 810 913 637 536 104 448 × 2 = 1 + 0.555 555 621 827 275 072 208 896;
  • 49) 0.555 555 621 827 275 072 208 896 × 2 = 1 + 0.111 111 243 654 550 144 417 792;
  • 50) 0.111 111 243 654 550 144 417 792 × 2 = 0 + 0.222 222 487 309 100 288 835 584;
  • 51) 0.222 222 487 309 100 288 835 584 × 2 = 0 + 0.444 444 974 618 200 577 671 168;
  • 52) 0.444 444 974 618 200 577 671 168 × 2 = 0 + 0.888 889 949 236 401 155 342 336;
  • 53) 0.888 889 949 236 401 155 342 336 × 2 = 1 + 0.777 779 898 472 802 310 684 672;

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:


0.555 555 555 555 555 555 555 791(10) =

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

5. Positive number before normalization:

135.555 555 555 555 555 555 555 791(10) =

1000 0111.1000 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 7 positions to the left, so that only one non zero digit remains to the left of it:


135.555 555 555 555 555 555 555 791(10) =

1000 0111.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1(2) =

1000 0111.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1(2) × 20 =

1.0000 1111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001(2) × 27


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): 7

Mantissa (not normalized):
1.0000 1111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

7 + 2(11-1) - 1 =

(7 + 1 023)(10) =

1 030(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;
  • 1 030 ÷ 2 = 515 + 0;
  • 515 ÷ 2 = 257 + 1;
  • 257 ÷ 2 = 128 + 1;
  • 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) =

1030(10) =

100 0000 0110(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 1111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 =

0000 1111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100


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 0110

Mantissa (52 bits) =
0000 1111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100


Decimal number 135.555 555 555 555 555 555 555 791 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0110 - 0000 1111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100

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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