-0.235 000 000 000 13 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.235 000 000 000 13(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
-0.235 000 000 000 13(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:

|-0.235 000 000 000 13| = 0.235 000 000 000 13


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: 0.235 000 000 000 13.

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) 0.235 000 000 000 13 × 2 = 0 + 0.470 000 000 000 26;
  • 2) 0.470 000 000 000 26 × 2 = 0 + 0.940 000 000 000 52;
  • 3) 0.940 000 000 000 52 × 2 = 1 + 0.880 000 000 001 04;
  • 4) 0.880 000 000 001 04 × 2 = 1 + 0.760 000 000 002 08;
  • 5) 0.760 000 000 002 08 × 2 = 1 + 0.520 000 000 004 16;
  • 6) 0.520 000 000 004 16 × 2 = 1 + 0.040 000 000 008 32;
  • 7) 0.040 000 000 008 32 × 2 = 0 + 0.080 000 000 016 64;
  • 8) 0.080 000 000 016 64 × 2 = 0 + 0.160 000 000 033 28;
  • 9) 0.160 000 000 033 28 × 2 = 0 + 0.320 000 000 066 56;
  • 10) 0.320 000 000 066 56 × 2 = 0 + 0.640 000 000 133 12;
  • 11) 0.640 000 000 133 12 × 2 = 1 + 0.280 000 000 266 24;
  • 12) 0.280 000 000 266 24 × 2 = 0 + 0.560 000 000 532 48;
  • 13) 0.560 000 000 532 48 × 2 = 1 + 0.120 000 001 064 96;
  • 14) 0.120 000 001 064 96 × 2 = 0 + 0.240 000 002 129 92;
  • 15) 0.240 000 002 129 92 × 2 = 0 + 0.480 000 004 259 84;
  • 16) 0.480 000 004 259 84 × 2 = 0 + 0.960 000 008 519 68;
  • 17) 0.960 000 008 519 68 × 2 = 1 + 0.920 000 017 039 36;
  • 18) 0.920 000 017 039 36 × 2 = 1 + 0.840 000 034 078 72;
  • 19) 0.840 000 034 078 72 × 2 = 1 + 0.680 000 068 157 44;
  • 20) 0.680 000 068 157 44 × 2 = 1 + 0.360 000 136 314 88;
  • 21) 0.360 000 136 314 88 × 2 = 0 + 0.720 000 272 629 76;
  • 22) 0.720 000 272 629 76 × 2 = 1 + 0.440 000 545 259 52;
  • 23) 0.440 000 545 259 52 × 2 = 0 + 0.880 001 090 519 04;
  • 24) 0.880 001 090 519 04 × 2 = 1 + 0.760 002 181 038 08;
  • 25) 0.760 002 181 038 08 × 2 = 1 + 0.520 004 362 076 16;
  • 26) 0.520 004 362 076 16 × 2 = 1 + 0.040 008 724 152 32;
  • 27) 0.040 008 724 152 32 × 2 = 0 + 0.080 017 448 304 64;
  • 28) 0.080 017 448 304 64 × 2 = 0 + 0.160 034 896 609 28;
  • 29) 0.160 034 896 609 28 × 2 = 0 + 0.320 069 793 218 56;
  • 30) 0.320 069 793 218 56 × 2 = 0 + 0.640 139 586 437 12;
  • 31) 0.640 139 586 437 12 × 2 = 1 + 0.280 279 172 874 24;
  • 32) 0.280 279 172 874 24 × 2 = 0 + 0.560 558 345 748 48;
  • 33) 0.560 558 345 748 48 × 2 = 1 + 0.121 116 691 496 96;
  • 34) 0.121 116 691 496 96 × 2 = 0 + 0.242 233 382 993 92;
  • 35) 0.242 233 382 993 92 × 2 = 0 + 0.484 466 765 987 84;
  • 36) 0.484 466 765 987 84 × 2 = 0 + 0.968 933 531 975 68;
  • 37) 0.968 933 531 975 68 × 2 = 1 + 0.937 867 063 951 36;
  • 38) 0.937 867 063 951 36 × 2 = 1 + 0.875 734 127 902 72;
  • 39) 0.875 734 127 902 72 × 2 = 1 + 0.751 468 255 805 44;
  • 40) 0.751 468 255 805 44 × 2 = 1 + 0.502 936 511 610 88;
  • 41) 0.502 936 511 610 88 × 2 = 1 + 0.005 873 023 221 76;
  • 42) 0.005 873 023 221 76 × 2 = 0 + 0.011 746 046 443 52;
  • 43) 0.011 746 046 443 52 × 2 = 0 + 0.023 492 092 887 04;
  • 44) 0.023 492 092 887 04 × 2 = 0 + 0.046 984 185 774 08;
  • 45) 0.046 984 185 774 08 × 2 = 0 + 0.093 968 371 548 16;
  • 46) 0.093 968 371 548 16 × 2 = 0 + 0.187 936 743 096 32;
  • 47) 0.187 936 743 096 32 × 2 = 0 + 0.375 873 486 192 64;
  • 48) 0.375 873 486 192 64 × 2 = 0 + 0.751 746 972 385 28;
  • 49) 0.751 746 972 385 28 × 2 = 1 + 0.503 493 944 770 56;
  • 50) 0.503 493 944 770 56 × 2 = 1 + 0.006 987 889 541 12;
  • 51) 0.006 987 889 541 12 × 2 = 0 + 0.013 975 779 082 24;
  • 52) 0.013 975 779 082 24 × 2 = 0 + 0.027 951 558 164 48;
  • 53) 0.027 951 558 164 48 × 2 = 0 + 0.055 903 116 328 96;
  • 54) 0.055 903 116 328 96 × 2 = 0 + 0.111 806 232 657 92;
  • 55) 0.111 806 232 657 92 × 2 = 0 + 0.223 612 465 315 84;

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:


0.235 000 000 000 13(10) =

0.0011 1100 0010 1000 1111 0101 1100 0010 1000 1111 1000 0000 1100 000(2)

6. Positive number before normalization:

0.235 000 000 000 13(10) =

0.0011 1100 0010 1000 1111 0101 1100 0010 1000 1111 1000 0000 1100 000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.235 000 000 000 13(10) =

0.0011 1100 0010 1000 1111 0101 1100 0010 1000 1111 1000 0000 1100 000(2) =

0.0011 1100 0010 1000 1111 0101 1100 0010 1000 1111 1000 0000 1100 000(2) × 20 =

1.1110 0001 0100 0111 1010 1110 0001 0100 0111 1100 0000 0110 0000(2) × 2-3


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): -3

Mantissa (not normalized):
1.1110 0001 0100 0111 1010 1110 0001 0100 0111 1100 0000 0110 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-3 + 2(11-1) - 1 =

(-3 + 1 023)(10) =

1 020(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;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 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) =

1020(10) =

011 1111 1100(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. 1110 0001 0100 0111 1010 1110 0001 0100 0111 1100 0000 0110 0000 =

1110 0001 0100 0111 1010 1110 0001 0100 0111 1100 0000 0110 0000


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 1100

Mantissa (52 bits) =
1110 0001 0100 0111 1010 1110 0001 0100 0111 1100 0000 0110 0000


Decimal number -0.235 000 000 000 13 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 1110 0001 0100 0111 1010 1110 0001 0100 0111 1100 0000 0110 0000

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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