-0.000 038 215 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 038 215(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
-0.000 038 215(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:

|-0.000 038 215| = 0.000 038 215


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: 0.000 038 215.

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) 0.000 038 215 × 2 = 0 + 0.000 076 43;
  • 2) 0.000 076 43 × 2 = 0 + 0.000 152 86;
  • 3) 0.000 152 86 × 2 = 0 + 0.000 305 72;
  • 4) 0.000 305 72 × 2 = 0 + 0.000 611 44;
  • 5) 0.000 611 44 × 2 = 0 + 0.001 222 88;
  • 6) 0.001 222 88 × 2 = 0 + 0.002 445 76;
  • 7) 0.002 445 76 × 2 = 0 + 0.004 891 52;
  • 8) 0.004 891 52 × 2 = 0 + 0.009 783 04;
  • 9) 0.009 783 04 × 2 = 0 + 0.019 566 08;
  • 10) 0.019 566 08 × 2 = 0 + 0.039 132 16;
  • 11) 0.039 132 16 × 2 = 0 + 0.078 264 32;
  • 12) 0.078 264 32 × 2 = 0 + 0.156 528 64;
  • 13) 0.156 528 64 × 2 = 0 + 0.313 057 28;
  • 14) 0.313 057 28 × 2 = 0 + 0.626 114 56;
  • 15) 0.626 114 56 × 2 = 1 + 0.252 229 12;
  • 16) 0.252 229 12 × 2 = 0 + 0.504 458 24;
  • 17) 0.504 458 24 × 2 = 1 + 0.008 916 48;
  • 18) 0.008 916 48 × 2 = 0 + 0.017 832 96;
  • 19) 0.017 832 96 × 2 = 0 + 0.035 665 92;
  • 20) 0.035 665 92 × 2 = 0 + 0.071 331 84;
  • 21) 0.071 331 84 × 2 = 0 + 0.142 663 68;
  • 22) 0.142 663 68 × 2 = 0 + 0.285 327 36;
  • 23) 0.285 327 36 × 2 = 0 + 0.570 654 72;
  • 24) 0.570 654 72 × 2 = 1 + 0.141 309 44;
  • 25) 0.141 309 44 × 2 = 0 + 0.282 618 88;
  • 26) 0.282 618 88 × 2 = 0 + 0.565 237 76;
  • 27) 0.565 237 76 × 2 = 1 + 0.130 475 52;
  • 28) 0.130 475 52 × 2 = 0 + 0.260 951 04;
  • 29) 0.260 951 04 × 2 = 0 + 0.521 902 08;
  • 30) 0.521 902 08 × 2 = 1 + 0.043 804 16;
  • 31) 0.043 804 16 × 2 = 0 + 0.087 608 32;
  • 32) 0.087 608 32 × 2 = 0 + 0.175 216 64;
  • 33) 0.175 216 64 × 2 = 0 + 0.350 433 28;
  • 34) 0.350 433 28 × 2 = 0 + 0.700 866 56;
  • 35) 0.700 866 56 × 2 = 1 + 0.401 733 12;
  • 36) 0.401 733 12 × 2 = 0 + 0.803 466 24;
  • 37) 0.803 466 24 × 2 = 1 + 0.606 932 48;
  • 38) 0.606 932 48 × 2 = 1 + 0.213 864 96;
  • 39) 0.213 864 96 × 2 = 0 + 0.427 729 92;
  • 40) 0.427 729 92 × 2 = 0 + 0.855 459 84;
  • 41) 0.855 459 84 × 2 = 1 + 0.710 919 68;
  • 42) 0.710 919 68 × 2 = 1 + 0.421 839 36;
  • 43) 0.421 839 36 × 2 = 0 + 0.843 678 72;
  • 44) 0.843 678 72 × 2 = 1 + 0.687 357 44;
  • 45) 0.687 357 44 × 2 = 1 + 0.374 714 88;
  • 46) 0.374 714 88 × 2 = 0 + 0.749 429 76;
  • 47) 0.749 429 76 × 2 = 1 + 0.498 859 52;
  • 48) 0.498 859 52 × 2 = 0 + 0.997 719 04;
  • 49) 0.997 719 04 × 2 = 1 + 0.995 438 08;
  • 50) 0.995 438 08 × 2 = 1 + 0.990 876 16;
  • 51) 0.990 876 16 × 2 = 1 + 0.981 752 32;
  • 52) 0.981 752 32 × 2 = 1 + 0.963 504 64;
  • 53) 0.963 504 64 × 2 = 1 + 0.927 009 28;
  • 54) 0.927 009 28 × 2 = 1 + 0.854 018 56;
  • 55) 0.854 018 56 × 2 = 1 + 0.708 037 12;
  • 56) 0.708 037 12 × 2 = 1 + 0.416 074 24;
  • 57) 0.416 074 24 × 2 = 0 + 0.832 148 48;
  • 58) 0.832 148 48 × 2 = 1 + 0.664 296 96;
  • 59) 0.664 296 96 × 2 = 1 + 0.328 593 92;
  • 60) 0.328 593 92 × 2 = 0 + 0.657 187 84;
  • 61) 0.657 187 84 × 2 = 1 + 0.314 375 68;
  • 62) 0.314 375 68 × 2 = 0 + 0.628 751 36;
  • 63) 0.628 751 36 × 2 = 1 + 0.257 502 72;
  • 64) 0.257 502 72 × 2 = 0 + 0.515 005 44;
  • 65) 0.515 005 44 × 2 = 1 + 0.030 010 88;
  • 66) 0.030 010 88 × 2 = 0 + 0.060 021 76;
  • 67) 0.060 021 76 × 2 = 0 + 0.120 043 52;

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:


0.000 038 215(10) =

0.0000 0000 0000 0010 1000 0001 0010 0100 0010 1100 1101 1010 1111 1111 0110 1010 100(2)

6. Positive number before normalization:

0.000 038 215(10) =

0.0000 0000 0000 0010 1000 0001 0010 0100 0010 1100 1101 1010 1111 1111 0110 1010 100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 038 215(10) =

0.0000 0000 0000 0010 1000 0001 0010 0100 0010 1100 1101 1010 1111 1111 0110 1010 100(2) =

0.0000 0000 0000 0010 1000 0001 0010 0100 0010 1100 1101 1010 1111 1111 0110 1010 100(2) × 20 =

1.0100 0000 1001 0010 0001 0110 0110 1101 0111 1111 1011 0101 0100(2) × 2-15


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): -15

Mantissa (not normalized):
1.0100 0000 1001 0010 0001 0110 0110 1101 0111 1111 1011 0101 0100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-15 + 2(11-1) - 1 =

(-15 + 1 023)(10) =

1 008(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;
  • 1 008 ÷ 2 = 504 + 0;
  • 504 ÷ 2 = 252 + 0;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1008(10) =

011 1111 0000(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. 0100 0000 1001 0010 0001 0110 0110 1101 0111 1111 1011 0101 0100 =

0100 0000 1001 0010 0001 0110 0110 1101 0111 1111 1011 0101 0100


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 0000

Mantissa (52 bits) =
0100 0000 1001 0010 0001 0110 0110 1101 0111 1111 1011 0101 0100


Decimal number -0.000 038 215 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0100 0000 1001 0010 0001 0110 0110 1101 0111 1111 1011 0101 0100

» Convert decimal -0.000 038 192 to 64 bit double precision IEEE 754 binary floating point representation standard

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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