2.225 073 858 507 201 381 64 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.225 073 858 507 201 381 64(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
2.225 073 858 507 201 381 64(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: 0.225 073 858 507 201 381 64.

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) 0.225 073 858 507 201 381 64 × 2 = 0 + 0.450 147 717 014 402 763 28;
  • 2) 0.450 147 717 014 402 763 28 × 2 = 0 + 0.900 295 434 028 805 526 56;
  • 3) 0.900 295 434 028 805 526 56 × 2 = 1 + 0.800 590 868 057 611 053 12;
  • 4) 0.800 590 868 057 611 053 12 × 2 = 1 + 0.601 181 736 115 222 106 24;
  • 5) 0.601 181 736 115 222 106 24 × 2 = 1 + 0.202 363 472 230 444 212 48;
  • 6) 0.202 363 472 230 444 212 48 × 2 = 0 + 0.404 726 944 460 888 424 96;
  • 7) 0.404 726 944 460 888 424 96 × 2 = 0 + 0.809 453 888 921 776 849 92;
  • 8) 0.809 453 888 921 776 849 92 × 2 = 1 + 0.618 907 777 843 553 699 84;
  • 9) 0.618 907 777 843 553 699 84 × 2 = 1 + 0.237 815 555 687 107 399 68;
  • 10) 0.237 815 555 687 107 399 68 × 2 = 0 + 0.475 631 111 374 214 799 36;
  • 11) 0.475 631 111 374 214 799 36 × 2 = 0 + 0.951 262 222 748 429 598 72;
  • 12) 0.951 262 222 748 429 598 72 × 2 = 1 + 0.902 524 445 496 859 197 44;
  • 13) 0.902 524 445 496 859 197 44 × 2 = 1 + 0.805 048 890 993 718 394 88;
  • 14) 0.805 048 890 993 718 394 88 × 2 = 1 + 0.610 097 781 987 436 789 76;
  • 15) 0.610 097 781 987 436 789 76 × 2 = 1 + 0.220 195 563 974 873 579 52;
  • 16) 0.220 195 563 974 873 579 52 × 2 = 0 + 0.440 391 127 949 747 159 04;
  • 17) 0.440 391 127 949 747 159 04 × 2 = 0 + 0.880 782 255 899 494 318 08;
  • 18) 0.880 782 255 899 494 318 08 × 2 = 1 + 0.761 564 511 798 988 636 16;
  • 19) 0.761 564 511 798 988 636 16 × 2 = 1 + 0.523 129 023 597 977 272 32;
  • 20) 0.523 129 023 597 977 272 32 × 2 = 1 + 0.046 258 047 195 954 544 64;
  • 21) 0.046 258 047 195 954 544 64 × 2 = 0 + 0.092 516 094 391 909 089 28;
  • 22) 0.092 516 094 391 909 089 28 × 2 = 0 + 0.185 032 188 783 818 178 56;
  • 23) 0.185 032 188 783 818 178 56 × 2 = 0 + 0.370 064 377 567 636 357 12;
  • 24) 0.370 064 377 567 636 357 12 × 2 = 0 + 0.740 128 755 135 272 714 24;
  • 25) 0.740 128 755 135 272 714 24 × 2 = 1 + 0.480 257 510 270 545 428 48;
  • 26) 0.480 257 510 270 545 428 48 × 2 = 0 + 0.960 515 020 541 090 856 96;
  • 27) 0.960 515 020 541 090 856 96 × 2 = 1 + 0.921 030 041 082 181 713 92;
  • 28) 0.921 030 041 082 181 713 92 × 2 = 1 + 0.842 060 082 164 363 427 84;
  • 29) 0.842 060 082 164 363 427 84 × 2 = 1 + 0.684 120 164 328 726 855 68;
  • 30) 0.684 120 164 328 726 855 68 × 2 = 1 + 0.368 240 328 657 453 711 36;
  • 31) 0.368 240 328 657 453 711 36 × 2 = 0 + 0.736 480 657 314 907 422 72;
  • 32) 0.736 480 657 314 907 422 72 × 2 = 1 + 0.472 961 314 629 814 845 44;
  • 33) 0.472 961 314 629 814 845 44 × 2 = 0 + 0.945 922 629 259 629 690 88;
  • 34) 0.945 922 629 259 629 690 88 × 2 = 1 + 0.891 845 258 519 259 381 76;
  • 35) 0.891 845 258 519 259 381 76 × 2 = 1 + 0.783 690 517 038 518 763 52;
  • 36) 0.783 690 517 038 518 763 52 × 2 = 1 + 0.567 381 034 077 037 527 04;
  • 37) 0.567 381 034 077 037 527 04 × 2 = 1 + 0.134 762 068 154 075 054 08;
  • 38) 0.134 762 068 154 075 054 08 × 2 = 0 + 0.269 524 136 308 150 108 16;
  • 39) 0.269 524 136 308 150 108 16 × 2 = 0 + 0.539 048 272 616 300 216 32;
  • 40) 0.539 048 272 616 300 216 32 × 2 = 1 + 0.078 096 545 232 600 432 64;
  • 41) 0.078 096 545 232 600 432 64 × 2 = 0 + 0.156 193 090 465 200 865 28;
  • 42) 0.156 193 090 465 200 865 28 × 2 = 0 + 0.312 386 180 930 401 730 56;
  • 43) 0.312 386 180 930 401 730 56 × 2 = 0 + 0.624 772 361 860 803 461 12;
  • 44) 0.624 772 361 860 803 461 12 × 2 = 1 + 0.249 544 723 721 606 922 24;
  • 45) 0.249 544 723 721 606 922 24 × 2 = 0 + 0.499 089 447 443 213 844 48;
  • 46) 0.499 089 447 443 213 844 48 × 2 = 0 + 0.998 178 894 886 427 688 96;
  • 47) 0.998 178 894 886 427 688 96 × 2 = 1 + 0.996 357 789 772 855 377 92;
  • 48) 0.996 357 789 772 855 377 92 × 2 = 1 + 0.992 715 579 545 710 755 84;
  • 49) 0.992 715 579 545 710 755 84 × 2 = 1 + 0.985 431 159 091 421 511 68;
  • 50) 0.985 431 159 091 421 511 68 × 2 = 1 + 0.970 862 318 182 843 023 36;
  • 51) 0.970 862 318 182 843 023 36 × 2 = 1 + 0.941 724 636 365 686 046 72;
  • 52) 0.941 724 636 365 686 046 72 × 2 = 1 + 0.883 449 272 731 372 093 44;
  • 53) 0.883 449 272 731 372 093 44 × 2 = 1 + 0.766 898 545 462 744 186 88;

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:


0.225 073 858 507 201 381 64(10) =

0.0011 1001 1001 1110 0111 0000 1011 1101 0111 1001 0001 0011 1111 1(2)

5. Positive number before normalization:

2.225 073 858 507 201 381 64(10) =

10.0011 1001 1001 1110 0111 0000 1011 1101 0111 1001 0001 0011 1111 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:


2.225 073 858 507 201 381 64(10) =

10.0011 1001 1001 1110 0111 0000 1011 1101 0111 1001 0001 0011 1111 1(2) =

10.0011 1001 1001 1110 0111 0000 1011 1101 0111 1001 0001 0011 1111 1(2) × 20 =

1.0001 1100 1100 1111 0011 1000 0101 1110 1011 1100 1000 1001 1111 11(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 1100 1111 0011 1000 0101 1110 1011 1100 1000 1001 1111 11


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 1100 1111 0011 1000 0101 1110 1011 1100 1000 1001 1111 11 =

0001 1100 1100 1111 0011 1000 0101 1110 1011 1100 1000 1001 1111


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 1100 1111 0011 1000 0101 1110 1011 1100 1000 1001 1111


Decimal number 2.225 073 858 507 201 381 64 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0001 1100 1100 1111 0011 1000 0101 1110 1011 1100 1000 1001 1111

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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