0.100 000 000 000 010 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.100 000 000 000 010 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
0.100 000 000 000 010 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: 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: 0.100 000 000 000 010 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) 0.100 000 000 000 010 4 × 2 = 0 + 0.200 000 000 000 020 8;
  • 2) 0.200 000 000 000 020 8 × 2 = 0 + 0.400 000 000 000 041 6;
  • 3) 0.400 000 000 000 041 6 × 2 = 0 + 0.800 000 000 000 083 2;
  • 4) 0.800 000 000 000 083 2 × 2 = 1 + 0.600 000 000 000 166 4;
  • 5) 0.600 000 000 000 166 4 × 2 = 1 + 0.200 000 000 000 332 8;
  • 6) 0.200 000 000 000 332 8 × 2 = 0 + 0.400 000 000 000 665 6;
  • 7) 0.400 000 000 000 665 6 × 2 = 0 + 0.800 000 000 001 331 2;
  • 8) 0.800 000 000 001 331 2 × 2 = 1 + 0.600 000 000 002 662 4;
  • 9) 0.600 000 000 002 662 4 × 2 = 1 + 0.200 000 000 005 324 8;
  • 10) 0.200 000 000 005 324 8 × 2 = 0 + 0.400 000 000 010 649 6;
  • 11) 0.400 000 000 010 649 6 × 2 = 0 + 0.800 000 000 021 299 2;
  • 12) 0.800 000 000 021 299 2 × 2 = 1 + 0.600 000 000 042 598 4;
  • 13) 0.600 000 000 042 598 4 × 2 = 1 + 0.200 000 000 085 196 8;
  • 14) 0.200 000 000 085 196 8 × 2 = 0 + 0.400 000 000 170 393 6;
  • 15) 0.400 000 000 170 393 6 × 2 = 0 + 0.800 000 000 340 787 2;
  • 16) 0.800 000 000 340 787 2 × 2 = 1 + 0.600 000 000 681 574 4;
  • 17) 0.600 000 000 681 574 4 × 2 = 1 + 0.200 000 001 363 148 8;
  • 18) 0.200 000 001 363 148 8 × 2 = 0 + 0.400 000 002 726 297 6;
  • 19) 0.400 000 002 726 297 6 × 2 = 0 + 0.800 000 005 452 595 2;
  • 20) 0.800 000 005 452 595 2 × 2 = 1 + 0.600 000 010 905 190 4;
  • 21) 0.600 000 010 905 190 4 × 2 = 1 + 0.200 000 021 810 380 8;
  • 22) 0.200 000 021 810 380 8 × 2 = 0 + 0.400 000 043 620 761 6;
  • 23) 0.400 000 043 620 761 6 × 2 = 0 + 0.800 000 087 241 523 2;
  • 24) 0.800 000 087 241 523 2 × 2 = 1 + 0.600 000 174 483 046 4;
  • 25) 0.600 000 174 483 046 4 × 2 = 1 + 0.200 000 348 966 092 8;
  • 26) 0.200 000 348 966 092 8 × 2 = 0 + 0.400 000 697 932 185 6;
  • 27) 0.400 000 697 932 185 6 × 2 = 0 + 0.800 001 395 864 371 2;
  • 28) 0.800 001 395 864 371 2 × 2 = 1 + 0.600 002 791 728 742 4;
  • 29) 0.600 002 791 728 742 4 × 2 = 1 + 0.200 005 583 457 484 8;
  • 30) 0.200 005 583 457 484 8 × 2 = 0 + 0.400 011 166 914 969 6;
  • 31) 0.400 011 166 914 969 6 × 2 = 0 + 0.800 022 333 829 939 2;
  • 32) 0.800 022 333 829 939 2 × 2 = 1 + 0.600 044 667 659 878 4;
  • 33) 0.600 044 667 659 878 4 × 2 = 1 + 0.200 089 335 319 756 8;
  • 34) 0.200 089 335 319 756 8 × 2 = 0 + 0.400 178 670 639 513 6;
  • 35) 0.400 178 670 639 513 6 × 2 = 0 + 0.800 357 341 279 027 2;
  • 36) 0.800 357 341 279 027 2 × 2 = 1 + 0.600 714 682 558 054 4;
  • 37) 0.600 714 682 558 054 4 × 2 = 1 + 0.201 429 365 116 108 8;
  • 38) 0.201 429 365 116 108 8 × 2 = 0 + 0.402 858 730 232 217 6;
  • 39) 0.402 858 730 232 217 6 × 2 = 0 + 0.805 717 460 464 435 2;
  • 40) 0.805 717 460 464 435 2 × 2 = 1 + 0.611 434 920 928 870 4;
  • 41) 0.611 434 920 928 870 4 × 2 = 1 + 0.222 869 841 857 740 8;
  • 42) 0.222 869 841 857 740 8 × 2 = 0 + 0.445 739 683 715 481 6;
  • 43) 0.445 739 683 715 481 6 × 2 = 0 + 0.891 479 367 430 963 2;
  • 44) 0.891 479 367 430 963 2 × 2 = 1 + 0.782 958 734 861 926 4;
  • 45) 0.782 958 734 861 926 4 × 2 = 1 + 0.565 917 469 723 852 8;
  • 46) 0.565 917 469 723 852 8 × 2 = 1 + 0.131 834 939 447 705 6;
  • 47) 0.131 834 939 447 705 6 × 2 = 0 + 0.263 669 878 895 411 2;
  • 48) 0.263 669 878 895 411 2 × 2 = 0 + 0.527 339 757 790 822 4;
  • 49) 0.527 339 757 790 822 4 × 2 = 1 + 0.054 679 515 581 644 8;
  • 50) 0.054 679 515 581 644 8 × 2 = 0 + 0.109 359 031 163 289 6;
  • 51) 0.109 359 031 163 289 6 × 2 = 0 + 0.218 718 062 326 579 2;
  • 52) 0.218 718 062 326 579 2 × 2 = 0 + 0.437 436 124 653 158 4;
  • 53) 0.437 436 124 653 158 4 × 2 = 0 + 0.874 872 249 306 316 8;
  • 54) 0.874 872 249 306 316 8 × 2 = 1 + 0.749 744 498 612 633 6;
  • 55) 0.749 744 498 612 633 6 × 2 = 1 + 0.499 488 997 225 267 2;
  • 56) 0.499 488 997 225 267 2 × 2 = 0 + 0.998 977 994 450 534 4;

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:


0.100 000 000 000 010 4(10) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1100 1000 0110(2)

5. Positive number before normalization:

0.100 000 000 000 010 4(10) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1100 1000 0110(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:


0.100 000 000 000 010 4(10) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1100 1000 0110(2) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1100 1000 0110(2) × 20 =

1.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1100 1000 0110(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 1100 1000 0110


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 1100 1000 0110 =

1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1100 1000 0110


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 1100 1000 0110


Decimal number 0.100 000 000 000 010 4 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 1100 1000 0110

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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