-0.000 072 97 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 072 97(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
-0.000 072 97(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:

|-0.000 072 97| = 0.000 072 97


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: 0.000 072 97.

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) 0.000 072 97 × 2 = 0 + 0.000 145 94;
  • 2) 0.000 145 94 × 2 = 0 + 0.000 291 88;
  • 3) 0.000 291 88 × 2 = 0 + 0.000 583 76;
  • 4) 0.000 583 76 × 2 = 0 + 0.001 167 52;
  • 5) 0.001 167 52 × 2 = 0 + 0.002 335 04;
  • 6) 0.002 335 04 × 2 = 0 + 0.004 670 08;
  • 7) 0.004 670 08 × 2 = 0 + 0.009 340 16;
  • 8) 0.009 340 16 × 2 = 0 + 0.018 680 32;
  • 9) 0.018 680 32 × 2 = 0 + 0.037 360 64;
  • 10) 0.037 360 64 × 2 = 0 + 0.074 721 28;
  • 11) 0.074 721 28 × 2 = 0 + 0.149 442 56;
  • 12) 0.149 442 56 × 2 = 0 + 0.298 885 12;
  • 13) 0.298 885 12 × 2 = 0 + 0.597 770 24;
  • 14) 0.597 770 24 × 2 = 1 + 0.195 540 48;
  • 15) 0.195 540 48 × 2 = 0 + 0.391 080 96;
  • 16) 0.391 080 96 × 2 = 0 + 0.782 161 92;
  • 17) 0.782 161 92 × 2 = 1 + 0.564 323 84;
  • 18) 0.564 323 84 × 2 = 1 + 0.128 647 68;
  • 19) 0.128 647 68 × 2 = 0 + 0.257 295 36;
  • 20) 0.257 295 36 × 2 = 0 + 0.514 590 72;
  • 21) 0.514 590 72 × 2 = 1 + 0.029 181 44;
  • 22) 0.029 181 44 × 2 = 0 + 0.058 362 88;
  • 23) 0.058 362 88 × 2 = 0 + 0.116 725 76;
  • 24) 0.116 725 76 × 2 = 0 + 0.233 451 52;
  • 25) 0.233 451 52 × 2 = 0 + 0.466 903 04;
  • 26) 0.466 903 04 × 2 = 0 + 0.933 806 08;
  • 27) 0.933 806 08 × 2 = 1 + 0.867 612 16;
  • 28) 0.867 612 16 × 2 = 1 + 0.735 224 32;
  • 29) 0.735 224 32 × 2 = 1 + 0.470 448 64;
  • 30) 0.470 448 64 × 2 = 0 + 0.940 897 28;
  • 31) 0.940 897 28 × 2 = 1 + 0.881 794 56;
  • 32) 0.881 794 56 × 2 = 1 + 0.763 589 12;
  • 33) 0.763 589 12 × 2 = 1 + 0.527 178 24;
  • 34) 0.527 178 24 × 2 = 1 + 0.054 356 48;
  • 35) 0.054 356 48 × 2 = 0 + 0.108 712 96;
  • 36) 0.108 712 96 × 2 = 0 + 0.217 425 92;
  • 37) 0.217 425 92 × 2 = 0 + 0.434 851 84;
  • 38) 0.434 851 84 × 2 = 0 + 0.869 703 68;
  • 39) 0.869 703 68 × 2 = 1 + 0.739 407 36;
  • 40) 0.739 407 36 × 2 = 1 + 0.478 814 72;
  • 41) 0.478 814 72 × 2 = 0 + 0.957 629 44;
  • 42) 0.957 629 44 × 2 = 1 + 0.915 258 88;
  • 43) 0.915 258 88 × 2 = 1 + 0.830 517 76;
  • 44) 0.830 517 76 × 2 = 1 + 0.661 035 52;
  • 45) 0.661 035 52 × 2 = 1 + 0.322 071 04;
  • 46) 0.322 071 04 × 2 = 0 + 0.644 142 08;
  • 47) 0.644 142 08 × 2 = 1 + 0.288 284 16;
  • 48) 0.288 284 16 × 2 = 0 + 0.576 568 32;
  • 49) 0.576 568 32 × 2 = 1 + 0.153 136 64;
  • 50) 0.153 136 64 × 2 = 0 + 0.306 273 28;
  • 51) 0.306 273 28 × 2 = 0 + 0.612 546 56;
  • 52) 0.612 546 56 × 2 = 1 + 0.225 093 12;
  • 53) 0.225 093 12 × 2 = 0 + 0.450 186 24;
  • 54) 0.450 186 24 × 2 = 0 + 0.900 372 48;
  • 55) 0.900 372 48 × 2 = 1 + 0.800 744 96;
  • 56) 0.800 744 96 × 2 = 1 + 0.601 489 92;
  • 57) 0.601 489 92 × 2 = 1 + 0.202 979 84;
  • 58) 0.202 979 84 × 2 = 0 + 0.405 959 68;
  • 59) 0.405 959 68 × 2 = 0 + 0.811 919 36;
  • 60) 0.811 919 36 × 2 = 1 + 0.623 838 72;
  • 61) 0.623 838 72 × 2 = 1 + 0.247 677 44;
  • 62) 0.247 677 44 × 2 = 0 + 0.495 354 88;
  • 63) 0.495 354 88 × 2 = 0 + 0.990 709 76;
  • 64) 0.990 709 76 × 2 = 1 + 0.981 419 52;
  • 65) 0.981 419 52 × 2 = 1 + 0.962 839 04;
  • 66) 0.962 839 04 × 2 = 1 + 0.925 678 08;

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:


0.000 072 97(10) =

0.0000 0000 0000 0100 1100 1000 0011 1011 1100 0011 0111 1010 1001 0011 1001 1001 11(2)

6. Positive number before normalization:

0.000 072 97(10) =

0.0000 0000 0000 0100 1100 1000 0011 1011 1100 0011 0111 1010 1001 0011 1001 1001 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 14 positions to the right, so that only one non zero digit remains to the left of it:


0.000 072 97(10) =

0.0000 0000 0000 0100 1100 1000 0011 1011 1100 0011 0111 1010 1001 0011 1001 1001 11(2) =

0.0000 0000 0000 0100 1100 1000 0011 1011 1100 0011 0111 1010 1001 0011 1001 1001 11(2) × 20 =

1.0011 0010 0000 1110 1111 0000 1101 1110 1010 0100 1110 0110 0111(2) × 2-14


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): -14

Mantissa (not normalized):
1.0011 0010 0000 1110 1111 0000 1101 1110 1010 0100 1110 0110 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-14 + 2(11-1) - 1 =

(-14 + 1 023)(10) =

1 009(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;
  • 1 009 ÷ 2 = 504 + 1;
  • 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) =

1009(10) =

011 1111 0001(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. 0011 0010 0000 1110 1111 0000 1101 1110 1010 0100 1110 0110 0111 =

0011 0010 0000 1110 1111 0000 1101 1110 1010 0100 1110 0110 0111


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 0001

Mantissa (52 bits) =
0011 0010 0000 1110 1111 0000 1101 1110 1010 0100 1110 0110 0111


Decimal number -0.000 072 97 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0001 - 0011 0010 0000 1110 1111 0000 1101 1110 1010 0100 1110 0110 0111

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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