-0.100 000 000 004 88 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.100 000 000 004 88(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
-0.100 000 000 004 88(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. Start with the positive version of the number:

|-0.100 000 000 004 88| = 0.100 000 000 004 88


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.100 000 000 004 88.

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) 0.100 000 000 004 88 × 2 = 0 + 0.200 000 000 009 76;
  • 2) 0.200 000 000 009 76 × 2 = 0 + 0.400 000 000 019 52;
  • 3) 0.400 000 000 019 52 × 2 = 0 + 0.800 000 000 039 04;
  • 4) 0.800 000 000 039 04 × 2 = 1 + 0.600 000 000 078 08;
  • 5) 0.600 000 000 078 08 × 2 = 1 + 0.200 000 000 156 16;
  • 6) 0.200 000 000 156 16 × 2 = 0 + 0.400 000 000 312 32;
  • 7) 0.400 000 000 312 32 × 2 = 0 + 0.800 000 000 624 64;
  • 8) 0.800 000 000 624 64 × 2 = 1 + 0.600 000 001 249 28;
  • 9) 0.600 000 001 249 28 × 2 = 1 + 0.200 000 002 498 56;
  • 10) 0.200 000 002 498 56 × 2 = 0 + 0.400 000 004 997 12;
  • 11) 0.400 000 004 997 12 × 2 = 0 + 0.800 000 009 994 24;
  • 12) 0.800 000 009 994 24 × 2 = 1 + 0.600 000 019 988 48;
  • 13) 0.600 000 019 988 48 × 2 = 1 + 0.200 000 039 976 96;
  • 14) 0.200 000 039 976 96 × 2 = 0 + 0.400 000 079 953 92;
  • 15) 0.400 000 079 953 92 × 2 = 0 + 0.800 000 159 907 84;
  • 16) 0.800 000 159 907 84 × 2 = 1 + 0.600 000 319 815 68;
  • 17) 0.600 000 319 815 68 × 2 = 1 + 0.200 000 639 631 36;
  • 18) 0.200 000 639 631 36 × 2 = 0 + 0.400 001 279 262 72;
  • 19) 0.400 001 279 262 72 × 2 = 0 + 0.800 002 558 525 44;
  • 20) 0.800 002 558 525 44 × 2 = 1 + 0.600 005 117 050 88;
  • 21) 0.600 005 117 050 88 × 2 = 1 + 0.200 010 234 101 76;
  • 22) 0.200 010 234 101 76 × 2 = 0 + 0.400 020 468 203 52;
  • 23) 0.400 020 468 203 52 × 2 = 0 + 0.800 040 936 407 04;
  • 24) 0.800 040 936 407 04 × 2 = 1 + 0.600 081 872 814 08;
  • 25) 0.600 081 872 814 08 × 2 = 1 + 0.200 163 745 628 16;
  • 26) 0.200 163 745 628 16 × 2 = 0 + 0.400 327 491 256 32;
  • 27) 0.400 327 491 256 32 × 2 = 0 + 0.800 654 982 512 64;
  • 28) 0.800 654 982 512 64 × 2 = 1 + 0.601 309 965 025 28;
  • 29) 0.601 309 965 025 28 × 2 = 1 + 0.202 619 930 050 56;
  • 30) 0.202 619 930 050 56 × 2 = 0 + 0.405 239 860 101 12;
  • 31) 0.405 239 860 101 12 × 2 = 0 + 0.810 479 720 202 24;
  • 32) 0.810 479 720 202 24 × 2 = 1 + 0.620 959 440 404 48;
  • 33) 0.620 959 440 404 48 × 2 = 1 + 0.241 918 880 808 96;
  • 34) 0.241 918 880 808 96 × 2 = 0 + 0.483 837 761 617 92;
  • 35) 0.483 837 761 617 92 × 2 = 0 + 0.967 675 523 235 84;
  • 36) 0.967 675 523 235 84 × 2 = 1 + 0.935 351 046 471 68;
  • 37) 0.935 351 046 471 68 × 2 = 1 + 0.870 702 092 943 36;
  • 38) 0.870 702 092 943 36 × 2 = 1 + 0.741 404 185 886 72;
  • 39) 0.741 404 185 886 72 × 2 = 1 + 0.482 808 371 773 44;
  • 40) 0.482 808 371 773 44 × 2 = 0 + 0.965 616 743 546 88;
  • 41) 0.965 616 743 546 88 × 2 = 1 + 0.931 233 487 093 76;
  • 42) 0.931 233 487 093 76 × 2 = 1 + 0.862 466 974 187 52;
  • 43) 0.862 466 974 187 52 × 2 = 1 + 0.724 933 948 375 04;
  • 44) 0.724 933 948 375 04 × 2 = 1 + 0.449 867 896 750 08;
  • 45) 0.449 867 896 750 08 × 2 = 0 + 0.899 735 793 500 16;
  • 46) 0.899 735 793 500 16 × 2 = 1 + 0.799 471 587 000 32;
  • 47) 0.799 471 587 000 32 × 2 = 1 + 0.598 943 174 000 64;
  • 48) 0.598 943 174 000 64 × 2 = 1 + 0.197 886 348 001 28;
  • 49) 0.197 886 348 001 28 × 2 = 0 + 0.395 772 696 002 56;
  • 50) 0.395 772 696 002 56 × 2 = 0 + 0.791 545 392 005 12;
  • 51) 0.791 545 392 005 12 × 2 = 1 + 0.583 090 784 010 24;
  • 52) 0.583 090 784 010 24 × 2 = 1 + 0.166 181 568 020 48;
  • 53) 0.166 181 568 020 48 × 2 = 0 + 0.332 363 136 040 96;
  • 54) 0.332 363 136 040 96 × 2 = 0 + 0.664 726 272 081 92;
  • 55) 0.664 726 272 081 92 × 2 = 1 + 0.329 452 544 163 84;
  • 56) 0.329 452 544 163 84 × 2 = 0 + 0.658 905 088 327 68;

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).


5. 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:


0.100 000 000 004 88(10) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1110 1111 0111 0011 0010(2)

6. Positive number before normalization:

0.100 000 000 004 88(10) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1110 1111 0111 0011 0010(2)

7. 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:


0.100 000 000 004 88(10) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1110 1111 0111 0011 0010(2) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1110 1111 0111 0011 0010(2) × 20 =

1.1001 1001 1001 1001 1001 1001 1001 1001 1110 1111 0111 0011 0010(2) × 2-4


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):
1.1001 1001 1001 1001 1001 1001 1001 1001 1110 1111 0111 0011 0010


9. 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)


10. 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;

11. 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)


12. 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 1110 1111 0111 0011 0010 =

1001 1001 1001 1001 1001 1001 1001 1001 1110 1111 0111 0011 0010


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1011

Mantissa (52 bits) =
1001 1001 1001 1001 1001 1001 1001 1001 1110 1111 0111 0011 0010


Decimal number -0.100 000 000 004 88 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 1001 1001 1001 1001 1001 1001 1001 1001 1110 1111 0111 0011 0010

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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