55.111 111 111 105 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 55.111 111 111 105 3(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
55.111 111 111 105 3(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: 0.111 111 111 105 3.

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) 0.111 111 111 105 3 × 2 = 0 + 0.222 222 222 210 6;
  • 2) 0.222 222 222 210 6 × 2 = 0 + 0.444 444 444 421 2;
  • 3) 0.444 444 444 421 2 × 2 = 0 + 0.888 888 888 842 4;
  • 4) 0.888 888 888 842 4 × 2 = 1 + 0.777 777 777 684 8;
  • 5) 0.777 777 777 684 8 × 2 = 1 + 0.555 555 555 369 6;
  • 6) 0.555 555 555 369 6 × 2 = 1 + 0.111 111 110 739 2;
  • 7) 0.111 111 110 739 2 × 2 = 0 + 0.222 222 221 478 4;
  • 8) 0.222 222 221 478 4 × 2 = 0 + 0.444 444 442 956 8;
  • 9) 0.444 444 442 956 8 × 2 = 0 + 0.888 888 885 913 6;
  • 10) 0.888 888 885 913 6 × 2 = 1 + 0.777 777 771 827 2;
  • 11) 0.777 777 771 827 2 × 2 = 1 + 0.555 555 543 654 4;
  • 12) 0.555 555 543 654 4 × 2 = 1 + 0.111 111 087 308 8;
  • 13) 0.111 111 087 308 8 × 2 = 0 + 0.222 222 174 617 6;
  • 14) 0.222 222 174 617 6 × 2 = 0 + 0.444 444 349 235 2;
  • 15) 0.444 444 349 235 2 × 2 = 0 + 0.888 888 698 470 4;
  • 16) 0.888 888 698 470 4 × 2 = 1 + 0.777 777 396 940 8;
  • 17) 0.777 777 396 940 8 × 2 = 1 + 0.555 554 793 881 6;
  • 18) 0.555 554 793 881 6 × 2 = 1 + 0.111 109 587 763 2;
  • 19) 0.111 109 587 763 2 × 2 = 0 + 0.222 219 175 526 4;
  • 20) 0.222 219 175 526 4 × 2 = 0 + 0.444 438 351 052 8;
  • 21) 0.444 438 351 052 8 × 2 = 0 + 0.888 876 702 105 6;
  • 22) 0.888 876 702 105 6 × 2 = 1 + 0.777 753 404 211 2;
  • 23) 0.777 753 404 211 2 × 2 = 1 + 0.555 506 808 422 4;
  • 24) 0.555 506 808 422 4 × 2 = 1 + 0.111 013 616 844 8;
  • 25) 0.111 013 616 844 8 × 2 = 0 + 0.222 027 233 689 6;
  • 26) 0.222 027 233 689 6 × 2 = 0 + 0.444 054 467 379 2;
  • 27) 0.444 054 467 379 2 × 2 = 0 + 0.888 108 934 758 4;
  • 28) 0.888 108 934 758 4 × 2 = 1 + 0.776 217 869 516 8;
  • 29) 0.776 217 869 516 8 × 2 = 1 + 0.552 435 739 033 6;
  • 30) 0.552 435 739 033 6 × 2 = 1 + 0.104 871 478 067 2;
  • 31) 0.104 871 478 067 2 × 2 = 0 + 0.209 742 956 134 4;
  • 32) 0.209 742 956 134 4 × 2 = 0 + 0.419 485 912 268 8;
  • 33) 0.419 485 912 268 8 × 2 = 0 + 0.838 971 824 537 6;
  • 34) 0.838 971 824 537 6 × 2 = 1 + 0.677 943 649 075 2;
  • 35) 0.677 943 649 075 2 × 2 = 1 + 0.355 887 298 150 4;
  • 36) 0.355 887 298 150 4 × 2 = 0 + 0.711 774 596 300 8;
  • 37) 0.711 774 596 300 8 × 2 = 1 + 0.423 549 192 601 6;
  • 38) 0.423 549 192 601 6 × 2 = 0 + 0.847 098 385 203 2;
  • 39) 0.847 098 385 203 2 × 2 = 1 + 0.694 196 770 406 4;
  • 40) 0.694 196 770 406 4 × 2 = 1 + 0.388 393 540 812 8;
  • 41) 0.388 393 540 812 8 × 2 = 0 + 0.776 787 081 625 6;
  • 42) 0.776 787 081 625 6 × 2 = 1 + 0.553 574 163 251 2;
  • 43) 0.553 574 163 251 2 × 2 = 1 + 0.107 148 326 502 4;
  • 44) 0.107 148 326 502 4 × 2 = 0 + 0.214 296 653 004 8;
  • 45) 0.214 296 653 004 8 × 2 = 0 + 0.428 593 306 009 6;
  • 46) 0.428 593 306 009 6 × 2 = 0 + 0.857 186 612 019 2;
  • 47) 0.857 186 612 019 2 × 2 = 1 + 0.714 373 224 038 4;
  • 48) 0.714 373 224 038 4 × 2 = 1 + 0.428 746 448 076 8;
  • 49) 0.428 746 448 076 8 × 2 = 0 + 0.857 492 896 153 6;
  • 50) 0.857 492 896 153 6 × 2 = 1 + 0.714 985 792 307 2;
  • 51) 0.714 985 792 307 2 × 2 = 1 + 0.429 971 584 614 4;
  • 52) 0.429 971 584 614 4 × 2 = 0 + 0.859 943 169 228 8;
  • 53) 0.859 943 169 228 8 × 2 = 1 + 0.719 886 338 457 6;

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:


0.111 111 111 105 3(10) =

0.0001 1100 0111 0001 1100 0111 0001 1100 0110 1011 0110 0011 0110 1(2)

5. Positive number before normalization:

55.111 111 111 105 3(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0110 1011 0110 0011 0110 1(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:


55.111 111 111 105 3(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0110 1011 0110 0011 0110 1(2) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0110 1011 0110 0011 0110 1(2) × 20 =

1.1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 0101 1011 0001 1011 01(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 0101 1011 0001 1011 01


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 0101 1011 0001 10 1101 =

1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 0101 1011 0001


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 0101 1011 0001


Decimal number 55.111 111 111 105 3 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 0101 1011 0001

Share

» Convert decimal 55.111 111 111 096 4 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 07, 2026 [July]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: July - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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