0.100 000 000 000 000 088 817 84 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.100 000 000 000 000 088 817 84(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
0.100 000 000 000 000 088 817 84(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: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.100 000 000 000 000 088 817 84.

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) 0.100 000 000 000 000 088 817 84 × 2 = 0 + 0.200 000 000 000 000 177 635 68;
  • 2) 0.200 000 000 000 000 177 635 68 × 2 = 0 + 0.400 000 000 000 000 355 271 36;
  • 3) 0.400 000 000 000 000 355 271 36 × 2 = 0 + 0.800 000 000 000 000 710 542 72;
  • 4) 0.800 000 000 000 000 710 542 72 × 2 = 1 + 0.600 000 000 000 001 421 085 44;
  • 5) 0.600 000 000 000 001 421 085 44 × 2 = 1 + 0.200 000 000 000 002 842 170 88;
  • 6) 0.200 000 000 000 002 842 170 88 × 2 = 0 + 0.400 000 000 000 005 684 341 76;
  • 7) 0.400 000 000 000 005 684 341 76 × 2 = 0 + 0.800 000 000 000 011 368 683 52;
  • 8) 0.800 000 000 000 011 368 683 52 × 2 = 1 + 0.600 000 000 000 022 737 367 04;
  • 9) 0.600 000 000 000 022 737 367 04 × 2 = 1 + 0.200 000 000 000 045 474 734 08;
  • 10) 0.200 000 000 000 045 474 734 08 × 2 = 0 + 0.400 000 000 000 090 949 468 16;
  • 11) 0.400 000 000 000 090 949 468 16 × 2 = 0 + 0.800 000 000 000 181 898 936 32;
  • 12) 0.800 000 000 000 181 898 936 32 × 2 = 1 + 0.600 000 000 000 363 797 872 64;
  • 13) 0.600 000 000 000 363 797 872 64 × 2 = 1 + 0.200 000 000 000 727 595 745 28;
  • 14) 0.200 000 000 000 727 595 745 28 × 2 = 0 + 0.400 000 000 001 455 191 490 56;
  • 15) 0.400 000 000 001 455 191 490 56 × 2 = 0 + 0.800 000 000 002 910 382 981 12;
  • 16) 0.800 000 000 002 910 382 981 12 × 2 = 1 + 0.600 000 000 005 820 765 962 24;
  • 17) 0.600 000 000 005 820 765 962 24 × 2 = 1 + 0.200 000 000 011 641 531 924 48;
  • 18) 0.200 000 000 011 641 531 924 48 × 2 = 0 + 0.400 000 000 023 283 063 848 96;
  • 19) 0.400 000 000 023 283 063 848 96 × 2 = 0 + 0.800 000 000 046 566 127 697 92;
  • 20) 0.800 000 000 046 566 127 697 92 × 2 = 1 + 0.600 000 000 093 132 255 395 84;
  • 21) 0.600 000 000 093 132 255 395 84 × 2 = 1 + 0.200 000 000 186 264 510 791 68;
  • 22) 0.200 000 000 186 264 510 791 68 × 2 = 0 + 0.400 000 000 372 529 021 583 36;
  • 23) 0.400 000 000 372 529 021 583 36 × 2 = 0 + 0.800 000 000 745 058 043 166 72;
  • 24) 0.800 000 000 745 058 043 166 72 × 2 = 1 + 0.600 000 001 490 116 086 333 44;
  • 25) 0.600 000 001 490 116 086 333 44 × 2 = 1 + 0.200 000 002 980 232 172 666 88;
  • 26) 0.200 000 002 980 232 172 666 88 × 2 = 0 + 0.400 000 005 960 464 345 333 76;
  • 27) 0.400 000 005 960 464 345 333 76 × 2 = 0 + 0.800 000 011 920 928 690 667 52;
  • 28) 0.800 000 011 920 928 690 667 52 × 2 = 1 + 0.600 000 023 841 857 381 335 04;
  • 29) 0.600 000 023 841 857 381 335 04 × 2 = 1 + 0.200 000 047 683 714 762 670 08;
  • 30) 0.200 000 047 683 714 762 670 08 × 2 = 0 + 0.400 000 095 367 429 525 340 16;
  • 31) 0.400 000 095 367 429 525 340 16 × 2 = 0 + 0.800 000 190 734 859 050 680 32;
  • 32) 0.800 000 190 734 859 050 680 32 × 2 = 1 + 0.600 000 381 469 718 101 360 64;
  • 33) 0.600 000 381 469 718 101 360 64 × 2 = 1 + 0.200 000 762 939 436 202 721 28;
  • 34) 0.200 000 762 939 436 202 721 28 × 2 = 0 + 0.400 001 525 878 872 405 442 56;
  • 35) 0.400 001 525 878 872 405 442 56 × 2 = 0 + 0.800 003 051 757 744 810 885 12;
  • 36) 0.800 003 051 757 744 810 885 12 × 2 = 1 + 0.600 006 103 515 489 621 770 24;
  • 37) 0.600 006 103 515 489 621 770 24 × 2 = 1 + 0.200 012 207 030 979 243 540 48;
  • 38) 0.200 012 207 030 979 243 540 48 × 2 = 0 + 0.400 024 414 061 958 487 080 96;
  • 39) 0.400 024 414 061 958 487 080 96 × 2 = 0 + 0.800 048 828 123 916 974 161 92;
  • 40) 0.800 048 828 123 916 974 161 92 × 2 = 1 + 0.600 097 656 247 833 948 323 84;
  • 41) 0.600 097 656 247 833 948 323 84 × 2 = 1 + 0.200 195 312 495 667 896 647 68;
  • 42) 0.200 195 312 495 667 896 647 68 × 2 = 0 + 0.400 390 624 991 335 793 295 36;
  • 43) 0.400 390 624 991 335 793 295 36 × 2 = 0 + 0.800 781 249 982 671 586 590 72;
  • 44) 0.800 781 249 982 671 586 590 72 × 2 = 1 + 0.601 562 499 965 343 173 181 44;
  • 45) 0.601 562 499 965 343 173 181 44 × 2 = 1 + 0.203 124 999 930 686 346 362 88;
  • 46) 0.203 124 999 930 686 346 362 88 × 2 = 0 + 0.406 249 999 861 372 692 725 76;
  • 47) 0.406 249 999 861 372 692 725 76 × 2 = 0 + 0.812 499 999 722 745 385 451 52;
  • 48) 0.812 499 999 722 745 385 451 52 × 2 = 1 + 0.624 999 999 445 490 770 903 04;
  • 49) 0.624 999 999 445 490 770 903 04 × 2 = 1 + 0.249 999 998 890 981 541 806 08;
  • 50) 0.249 999 998 890 981 541 806 08 × 2 = 0 + 0.499 999 997 781 963 083 612 16;
  • 51) 0.499 999 997 781 963 083 612 16 × 2 = 0 + 0.999 999 995 563 926 167 224 32;
  • 52) 0.999 999 995 563 926 167 224 32 × 2 = 1 + 0.999 999 991 127 852 334 448 64;
  • 53) 0.999 999 991 127 852 334 448 64 × 2 = 1 + 0.999 999 982 255 704 668 897 28;
  • 54) 0.999 999 982 255 704 668 897 28 × 2 = 1 + 0.999 999 964 511 409 337 794 56;
  • 55) 0.999 999 964 511 409 337 794 56 × 2 = 1 + 0.999 999 929 022 818 675 589 12;
  • 56) 0.999 999 929 022 818 675 589 12 × 2 = 1 + 0.999 999 858 045 637 351 178 24;

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:


0.100 000 000 000 000 088 817 84(10) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1111(2)

5. Positive number before normalization:

0.100 000 000 000 000 088 817 84(10) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.100 000 000 000 000 088 817 84(10) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1111(2) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1111(2) × 20 =

1.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1111(2) × 2-4


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): -4

Mantissa (not normalized):
1.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1019(10) =

011 1111 1011(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1111 =

1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 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) =
011 1111 1011

Mantissa (52 bits) =
1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1111


Decimal number 0.100 000 000 000 000 088 817 84 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1011 - 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 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