2.222 222 222 222 361 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.222 222 222 222 361(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
2.222 222 222 222 361(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: 0.222 222 222 222 361.

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) 0.222 222 222 222 361 × 2 = 0 + 0.444 444 444 444 722;
  • 2) 0.444 444 444 444 722 × 2 = 0 + 0.888 888 888 889 444;
  • 3) 0.888 888 888 889 444 × 2 = 1 + 0.777 777 777 778 888;
  • 4) 0.777 777 777 778 888 × 2 = 1 + 0.555 555 555 557 776;
  • 5) 0.555 555 555 557 776 × 2 = 1 + 0.111 111 111 115 552;
  • 6) 0.111 111 111 115 552 × 2 = 0 + 0.222 222 222 231 104;
  • 7) 0.222 222 222 231 104 × 2 = 0 + 0.444 444 444 462 208;
  • 8) 0.444 444 444 462 208 × 2 = 0 + 0.888 888 888 924 416;
  • 9) 0.888 888 888 924 416 × 2 = 1 + 0.777 777 777 848 832;
  • 10) 0.777 777 777 848 832 × 2 = 1 + 0.555 555 555 697 664;
  • 11) 0.555 555 555 697 664 × 2 = 1 + 0.111 111 111 395 328;
  • 12) 0.111 111 111 395 328 × 2 = 0 + 0.222 222 222 790 656;
  • 13) 0.222 222 222 790 656 × 2 = 0 + 0.444 444 445 581 312;
  • 14) 0.444 444 445 581 312 × 2 = 0 + 0.888 888 891 162 624;
  • 15) 0.888 888 891 162 624 × 2 = 1 + 0.777 777 782 325 248;
  • 16) 0.777 777 782 325 248 × 2 = 1 + 0.555 555 564 650 496;
  • 17) 0.555 555 564 650 496 × 2 = 1 + 0.111 111 129 300 992;
  • 18) 0.111 111 129 300 992 × 2 = 0 + 0.222 222 258 601 984;
  • 19) 0.222 222 258 601 984 × 2 = 0 + 0.444 444 517 203 968;
  • 20) 0.444 444 517 203 968 × 2 = 0 + 0.888 889 034 407 936;
  • 21) 0.888 889 034 407 936 × 2 = 1 + 0.777 778 068 815 872;
  • 22) 0.777 778 068 815 872 × 2 = 1 + 0.555 556 137 631 744;
  • 23) 0.555 556 137 631 744 × 2 = 1 + 0.111 112 275 263 488;
  • 24) 0.111 112 275 263 488 × 2 = 0 + 0.222 224 550 526 976;
  • 25) 0.222 224 550 526 976 × 2 = 0 + 0.444 449 101 053 952;
  • 26) 0.444 449 101 053 952 × 2 = 0 + 0.888 898 202 107 904;
  • 27) 0.888 898 202 107 904 × 2 = 1 + 0.777 796 404 215 808;
  • 28) 0.777 796 404 215 808 × 2 = 1 + 0.555 592 808 431 616;
  • 29) 0.555 592 808 431 616 × 2 = 1 + 0.111 185 616 863 232;
  • 30) 0.111 185 616 863 232 × 2 = 0 + 0.222 371 233 726 464;
  • 31) 0.222 371 233 726 464 × 2 = 0 + 0.444 742 467 452 928;
  • 32) 0.444 742 467 452 928 × 2 = 0 + 0.889 484 934 905 856;
  • 33) 0.889 484 934 905 856 × 2 = 1 + 0.778 969 869 811 712;
  • 34) 0.778 969 869 811 712 × 2 = 1 + 0.557 939 739 623 424;
  • 35) 0.557 939 739 623 424 × 2 = 1 + 0.115 879 479 246 848;
  • 36) 0.115 879 479 246 848 × 2 = 0 + 0.231 758 958 493 696;
  • 37) 0.231 758 958 493 696 × 2 = 0 + 0.463 517 916 987 392;
  • 38) 0.463 517 916 987 392 × 2 = 0 + 0.927 035 833 974 784;
  • 39) 0.927 035 833 974 784 × 2 = 1 + 0.854 071 667 949 568;
  • 40) 0.854 071 667 949 568 × 2 = 1 + 0.708 143 335 899 136;
  • 41) 0.708 143 335 899 136 × 2 = 1 + 0.416 286 671 798 272;
  • 42) 0.416 286 671 798 272 × 2 = 0 + 0.832 573 343 596 544;
  • 43) 0.832 573 343 596 544 × 2 = 1 + 0.665 146 687 193 088;
  • 44) 0.665 146 687 193 088 × 2 = 1 + 0.330 293 374 386 176;
  • 45) 0.330 293 374 386 176 × 2 = 0 + 0.660 586 748 772 352;
  • 46) 0.660 586 748 772 352 × 2 = 1 + 0.321 173 497 544 704;
  • 47) 0.321 173 497 544 704 × 2 = 0 + 0.642 346 995 089 408;
  • 48) 0.642 346 995 089 408 × 2 = 1 + 0.284 693 990 178 816;
  • 49) 0.284 693 990 178 816 × 2 = 0 + 0.569 387 980 357 632;
  • 50) 0.569 387 980 357 632 × 2 = 1 + 0.138 775 960 715 264;
  • 51) 0.138 775 960 715 264 × 2 = 0 + 0.277 551 921 430 528;
  • 52) 0.277 551 921 430 528 × 2 = 0 + 0.555 103 842 861 056;
  • 53) 0.555 103 842 861 056 × 2 = 1 + 0.110 207 685 722 112;

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:


0.222 222 222 222 361(10) =

0.0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1011 0101 0100 1(2)

5. Positive number before normalization:

2.222 222 222 222 361(10) =

10.0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1011 0101 0100 1(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:


2.222 222 222 222 361(10) =

10.0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1011 0101 0100 1(2) =

10.0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1011 0101 0100 1(2) × 20 =

1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1101 1010 1010 01(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):
1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1101 1010 1010 01


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. 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1101 1010 1010 01 =

0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1101 1010 1010


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) =
0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1101 1010 1010


Decimal number 2.222 222 222 222 361 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1101 1010 1010

Share

» Convert decimal 2.222 222 222 222 421 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 08, 2026 [August]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: August - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. Gemini Notebook 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