-0.000 073 37 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 073 37(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
-0.000 073 37(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:

|-0.000 073 37| = 0.000 073 37


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: 0.000 073 37.

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) 0.000 073 37 × 2 = 0 + 0.000 146 74;
  • 2) 0.000 146 74 × 2 = 0 + 0.000 293 48;
  • 3) 0.000 293 48 × 2 = 0 + 0.000 586 96;
  • 4) 0.000 586 96 × 2 = 0 + 0.001 173 92;
  • 5) 0.001 173 92 × 2 = 0 + 0.002 347 84;
  • 6) 0.002 347 84 × 2 = 0 + 0.004 695 68;
  • 7) 0.004 695 68 × 2 = 0 + 0.009 391 36;
  • 8) 0.009 391 36 × 2 = 0 + 0.018 782 72;
  • 9) 0.018 782 72 × 2 = 0 + 0.037 565 44;
  • 10) 0.037 565 44 × 2 = 0 + 0.075 130 88;
  • 11) 0.075 130 88 × 2 = 0 + 0.150 261 76;
  • 12) 0.150 261 76 × 2 = 0 + 0.300 523 52;
  • 13) 0.300 523 52 × 2 = 0 + 0.601 047 04;
  • 14) 0.601 047 04 × 2 = 1 + 0.202 094 08;
  • 15) 0.202 094 08 × 2 = 0 + 0.404 188 16;
  • 16) 0.404 188 16 × 2 = 0 + 0.808 376 32;
  • 17) 0.808 376 32 × 2 = 1 + 0.616 752 64;
  • 18) 0.616 752 64 × 2 = 1 + 0.233 505 28;
  • 19) 0.233 505 28 × 2 = 0 + 0.467 010 56;
  • 20) 0.467 010 56 × 2 = 0 + 0.934 021 12;
  • 21) 0.934 021 12 × 2 = 1 + 0.868 042 24;
  • 22) 0.868 042 24 × 2 = 1 + 0.736 084 48;
  • 23) 0.736 084 48 × 2 = 1 + 0.472 168 96;
  • 24) 0.472 168 96 × 2 = 0 + 0.944 337 92;
  • 25) 0.944 337 92 × 2 = 1 + 0.888 675 84;
  • 26) 0.888 675 84 × 2 = 1 + 0.777 351 68;
  • 27) 0.777 351 68 × 2 = 1 + 0.554 703 36;
  • 28) 0.554 703 36 × 2 = 1 + 0.109 406 72;
  • 29) 0.109 406 72 × 2 = 0 + 0.218 813 44;
  • 30) 0.218 813 44 × 2 = 0 + 0.437 626 88;
  • 31) 0.437 626 88 × 2 = 0 + 0.875 253 76;
  • 32) 0.875 253 76 × 2 = 1 + 0.750 507 52;
  • 33) 0.750 507 52 × 2 = 1 + 0.501 015 04;
  • 34) 0.501 015 04 × 2 = 1 + 0.002 030 08;
  • 35) 0.002 030 08 × 2 = 0 + 0.004 060 16;
  • 36) 0.004 060 16 × 2 = 0 + 0.008 120 32;
  • 37) 0.008 120 32 × 2 = 0 + 0.016 240 64;
  • 38) 0.016 240 64 × 2 = 0 + 0.032 481 28;
  • 39) 0.032 481 28 × 2 = 0 + 0.064 962 56;
  • 40) 0.064 962 56 × 2 = 0 + 0.129 925 12;
  • 41) 0.129 925 12 × 2 = 0 + 0.259 850 24;
  • 42) 0.259 850 24 × 2 = 0 + 0.519 700 48;
  • 43) 0.519 700 48 × 2 = 1 + 0.039 400 96;
  • 44) 0.039 400 96 × 2 = 0 + 0.078 801 92;
  • 45) 0.078 801 92 × 2 = 0 + 0.157 603 84;
  • 46) 0.157 603 84 × 2 = 0 + 0.315 207 68;
  • 47) 0.315 207 68 × 2 = 0 + 0.630 415 36;
  • 48) 0.630 415 36 × 2 = 1 + 0.260 830 72;
  • 49) 0.260 830 72 × 2 = 0 + 0.521 661 44;
  • 50) 0.521 661 44 × 2 = 1 + 0.043 322 88;
  • 51) 0.043 322 88 × 2 = 0 + 0.086 645 76;
  • 52) 0.086 645 76 × 2 = 0 + 0.173 291 52;
  • 53) 0.173 291 52 × 2 = 0 + 0.346 583 04;
  • 54) 0.346 583 04 × 2 = 0 + 0.693 166 08;
  • 55) 0.693 166 08 × 2 = 1 + 0.386 332 16;
  • 56) 0.386 332 16 × 2 = 0 + 0.772 664 32;
  • 57) 0.772 664 32 × 2 = 1 + 0.545 328 64;
  • 58) 0.545 328 64 × 2 = 1 + 0.090 657 28;
  • 59) 0.090 657 28 × 2 = 0 + 0.181 314 56;
  • 60) 0.181 314 56 × 2 = 0 + 0.362 629 12;
  • 61) 0.362 629 12 × 2 = 0 + 0.725 258 24;
  • 62) 0.725 258 24 × 2 = 1 + 0.450 516 48;
  • 63) 0.450 516 48 × 2 = 0 + 0.901 032 96;
  • 64) 0.901 032 96 × 2 = 1 + 0.802 065 92;
  • 65) 0.802 065 92 × 2 = 1 + 0.604 131 84;
  • 66) 0.604 131 84 × 2 = 1 + 0.208 263 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:


0.000 073 37(10) =

0.0000 0000 0000 0100 1100 1110 1111 0001 1100 0000 0010 0001 0100 0010 1100 0101 11(2)

6. Positive number before normalization:

0.000 073 37(10) =

0.0000 0000 0000 0100 1100 1110 1111 0001 1100 0000 0010 0001 0100 0010 1100 0101 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:


0.000 073 37(10) =

0.0000 0000 0000 0100 1100 1110 1111 0001 1100 0000 0010 0001 0100 0010 1100 0101 11(2) =

0.0000 0000 0000 0100 1100 1110 1111 0001 1100 0000 0010 0001 0100 0010 1100 0101 11(2) × 20 =

1.0011 0011 1011 1100 0111 0000 0000 1000 0101 0000 1011 0001 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 0011 1011 1100 0111 0000 0000 1000 0101 0000 1011 0001 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 0011 1011 1100 0111 0000 0000 1000 0101 0000 1011 0001 0111 =

0011 0011 1011 1100 0111 0000 0000 1000 0101 0000 1011 0001 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 0011 1011 1100 0111 0000 0000 1000 0101 0000 1011 0001 0111


Decimal number -0.000 073 37 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0001 - 0011 0011 1011 1100 0111 0000 0000 1000 0101 0000 1011 0001 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