55.111 111 111 112 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 55.111 111 111 112 4(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
55.111 111 111 112 4(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: 55.
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;
  • 55 ÷ 2 = 27 + 1;
  • 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.

55(10) =

11 0111(2)


3. Convert to binary (base 2) the fractional part: 0.111 111 111 112 4.

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) 0.111 111 111 112 4 × 2 = 0 + 0.222 222 222 224 8;
  • 2) 0.222 222 222 224 8 × 2 = 0 + 0.444 444 444 449 6;
  • 3) 0.444 444 444 449 6 × 2 = 0 + 0.888 888 888 899 2;
  • 4) 0.888 888 888 899 2 × 2 = 1 + 0.777 777 777 798 4;
  • 5) 0.777 777 777 798 4 × 2 = 1 + 0.555 555 555 596 8;
  • 6) 0.555 555 555 596 8 × 2 = 1 + 0.111 111 111 193 6;
  • 7) 0.111 111 111 193 6 × 2 = 0 + 0.222 222 222 387 2;
  • 8) 0.222 222 222 387 2 × 2 = 0 + 0.444 444 444 774 4;
  • 9) 0.444 444 444 774 4 × 2 = 0 + 0.888 888 889 548 8;
  • 10) 0.888 888 889 548 8 × 2 = 1 + 0.777 777 779 097 6;
  • 11) 0.777 777 779 097 6 × 2 = 1 + 0.555 555 558 195 2;
  • 12) 0.555 555 558 195 2 × 2 = 1 + 0.111 111 116 390 4;
  • 13) 0.111 111 116 390 4 × 2 = 0 + 0.222 222 232 780 8;
  • 14) 0.222 222 232 780 8 × 2 = 0 + 0.444 444 465 561 6;
  • 15) 0.444 444 465 561 6 × 2 = 0 + 0.888 888 931 123 2;
  • 16) 0.888 888 931 123 2 × 2 = 1 + 0.777 777 862 246 4;
  • 17) 0.777 777 862 246 4 × 2 = 1 + 0.555 555 724 492 8;
  • 18) 0.555 555 724 492 8 × 2 = 1 + 0.111 111 448 985 6;
  • 19) 0.111 111 448 985 6 × 2 = 0 + 0.222 222 897 971 2;
  • 20) 0.222 222 897 971 2 × 2 = 0 + 0.444 445 795 942 4;
  • 21) 0.444 445 795 942 4 × 2 = 0 + 0.888 891 591 884 8;
  • 22) 0.888 891 591 884 8 × 2 = 1 + 0.777 783 183 769 6;
  • 23) 0.777 783 183 769 6 × 2 = 1 + 0.555 566 367 539 2;
  • 24) 0.555 566 367 539 2 × 2 = 1 + 0.111 132 735 078 4;
  • 25) 0.111 132 735 078 4 × 2 = 0 + 0.222 265 470 156 8;
  • 26) 0.222 265 470 156 8 × 2 = 0 + 0.444 530 940 313 6;
  • 27) 0.444 530 940 313 6 × 2 = 0 + 0.889 061 880 627 2;
  • 28) 0.889 061 880 627 2 × 2 = 1 + 0.778 123 761 254 4;
  • 29) 0.778 123 761 254 4 × 2 = 1 + 0.556 247 522 508 8;
  • 30) 0.556 247 522 508 8 × 2 = 1 + 0.112 495 045 017 6;
  • 31) 0.112 495 045 017 6 × 2 = 0 + 0.224 990 090 035 2;
  • 32) 0.224 990 090 035 2 × 2 = 0 + 0.449 980 180 070 4;
  • 33) 0.449 980 180 070 4 × 2 = 0 + 0.899 960 360 140 8;
  • 34) 0.899 960 360 140 8 × 2 = 1 + 0.799 920 720 281 6;
  • 35) 0.799 920 720 281 6 × 2 = 1 + 0.599 841 440 563 2;
  • 36) 0.599 841 440 563 2 × 2 = 1 + 0.199 682 881 126 4;
  • 37) 0.199 682 881 126 4 × 2 = 0 + 0.399 365 762 252 8;
  • 38) 0.399 365 762 252 8 × 2 = 0 + 0.798 731 524 505 6;
  • 39) 0.798 731 524 505 6 × 2 = 1 + 0.597 463 049 011 2;
  • 40) 0.597 463 049 011 2 × 2 = 1 + 0.194 926 098 022 4;
  • 41) 0.194 926 098 022 4 × 2 = 0 + 0.389 852 196 044 8;
  • 42) 0.389 852 196 044 8 × 2 = 0 + 0.779 704 392 089 6;
  • 43) 0.779 704 392 089 6 × 2 = 1 + 0.559 408 784 179 2;
  • 44) 0.559 408 784 179 2 × 2 = 1 + 0.118 817 568 358 4;
  • 45) 0.118 817 568 358 4 × 2 = 0 + 0.237 635 136 716 8;
  • 46) 0.237 635 136 716 8 × 2 = 0 + 0.475 270 273 433 6;
  • 47) 0.475 270 273 433 6 × 2 = 0 + 0.950 540 546 867 2;
  • 48) 0.950 540 546 867 2 × 2 = 1 + 0.901 081 093 734 4;
  • 49) 0.901 081 093 734 4 × 2 = 1 + 0.802 162 187 468 8;
  • 50) 0.802 162 187 468 8 × 2 = 1 + 0.604 324 374 937 6;
  • 51) 0.604 324 374 937 6 × 2 = 1 + 0.208 648 749 875 2;
  • 52) 0.208 648 749 875 2 × 2 = 0 + 0.417 297 499 750 4;
  • 53) 0.417 297 499 750 4 × 2 = 0 + 0.834 594 999 500 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:


0.111 111 111 112 4(10) =

0.0001 1100 0111 0001 1100 0111 0001 1100 0111 0011 0011 0001 1110 0(2)

5. Positive number before normalization:

55.111 111 111 112 4(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0111 0011 0011 0001 1110 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the left, so that only one non zero digit remains to the left of it:


55.111 111 111 112 4(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0111 0011 0011 0001 1110 0(2) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0111 0011 0011 0001 1110 0(2) × 20 =

1.1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1001 1001 1000 1111 00(2) × 25


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): 5

Mantissa (not normalized):
1.1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1001 1001 1000 1111 00


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

5 + 2(11-1) - 1 =

(5 + 1 023)(10) =

1 028(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;
  • 1 028 ÷ 2 = 514 + 0;
  • 514 ÷ 2 = 257 + 0;
  • 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) =

1028(10) =

100 0000 0100(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 1001 1001 1000 11 1100 =

1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1001 1001 1000


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 0100

Mantissa (52 bits) =
1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1001 1001 1000


Decimal number 55.111 111 111 112 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0100 - 1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1001 1001 1000

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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