-0.003 223 62 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.003 223 62(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
-0.003 223 62(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:

|-0.003 223 62| = 0.003 223 62


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: 0.003 223 62.

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) 0.003 223 62 × 2 = 0 + 0.006 447 24;
  • 2) 0.006 447 24 × 2 = 0 + 0.012 894 48;
  • 3) 0.012 894 48 × 2 = 0 + 0.025 788 96;
  • 4) 0.025 788 96 × 2 = 0 + 0.051 577 92;
  • 5) 0.051 577 92 × 2 = 0 + 0.103 155 84;
  • 6) 0.103 155 84 × 2 = 0 + 0.206 311 68;
  • 7) 0.206 311 68 × 2 = 0 + 0.412 623 36;
  • 8) 0.412 623 36 × 2 = 0 + 0.825 246 72;
  • 9) 0.825 246 72 × 2 = 1 + 0.650 493 44;
  • 10) 0.650 493 44 × 2 = 1 + 0.300 986 88;
  • 11) 0.300 986 88 × 2 = 0 + 0.601 973 76;
  • 12) 0.601 973 76 × 2 = 1 + 0.203 947 52;
  • 13) 0.203 947 52 × 2 = 0 + 0.407 895 04;
  • 14) 0.407 895 04 × 2 = 0 + 0.815 790 08;
  • 15) 0.815 790 08 × 2 = 1 + 0.631 580 16;
  • 16) 0.631 580 16 × 2 = 1 + 0.263 160 32;
  • 17) 0.263 160 32 × 2 = 0 + 0.526 320 64;
  • 18) 0.526 320 64 × 2 = 1 + 0.052 641 28;
  • 19) 0.052 641 28 × 2 = 0 + 0.105 282 56;
  • 20) 0.105 282 56 × 2 = 0 + 0.210 565 12;
  • 21) 0.210 565 12 × 2 = 0 + 0.421 130 24;
  • 22) 0.421 130 24 × 2 = 0 + 0.842 260 48;
  • 23) 0.842 260 48 × 2 = 1 + 0.684 520 96;
  • 24) 0.684 520 96 × 2 = 1 + 0.369 041 92;
  • 25) 0.369 041 92 × 2 = 0 + 0.738 083 84;
  • 26) 0.738 083 84 × 2 = 1 + 0.476 167 68;
  • 27) 0.476 167 68 × 2 = 0 + 0.952 335 36;
  • 28) 0.952 335 36 × 2 = 1 + 0.904 670 72;
  • 29) 0.904 670 72 × 2 = 1 + 0.809 341 44;
  • 30) 0.809 341 44 × 2 = 1 + 0.618 682 88;
  • 31) 0.618 682 88 × 2 = 1 + 0.237 365 76;
  • 32) 0.237 365 76 × 2 = 0 + 0.474 731 52;
  • 33) 0.474 731 52 × 2 = 0 + 0.949 463 04;
  • 34) 0.949 463 04 × 2 = 1 + 0.898 926 08;
  • 35) 0.898 926 08 × 2 = 1 + 0.797 852 16;
  • 36) 0.797 852 16 × 2 = 1 + 0.595 704 32;
  • 37) 0.595 704 32 × 2 = 1 + 0.191 408 64;
  • 38) 0.191 408 64 × 2 = 0 + 0.382 817 28;
  • 39) 0.382 817 28 × 2 = 0 + 0.765 634 56;
  • 40) 0.765 634 56 × 2 = 1 + 0.531 269 12;
  • 41) 0.531 269 12 × 2 = 1 + 0.062 538 24;
  • 42) 0.062 538 24 × 2 = 0 + 0.125 076 48;
  • 43) 0.125 076 48 × 2 = 0 + 0.250 152 96;
  • 44) 0.250 152 96 × 2 = 0 + 0.500 305 92;
  • 45) 0.500 305 92 × 2 = 1 + 0.000 611 84;
  • 46) 0.000 611 84 × 2 = 0 + 0.001 223 68;
  • 47) 0.001 223 68 × 2 = 0 + 0.002 447 36;
  • 48) 0.002 447 36 × 2 = 0 + 0.004 894 72;
  • 49) 0.004 894 72 × 2 = 0 + 0.009 789 44;
  • 50) 0.009 789 44 × 2 = 0 + 0.019 578 88;
  • 51) 0.019 578 88 × 2 = 0 + 0.039 157 76;
  • 52) 0.039 157 76 × 2 = 0 + 0.078 315 52;
  • 53) 0.078 315 52 × 2 = 0 + 0.156 631 04;
  • 54) 0.156 631 04 × 2 = 0 + 0.313 262 08;
  • 55) 0.313 262 08 × 2 = 0 + 0.626 524 16;
  • 56) 0.626 524 16 × 2 = 1 + 0.253 048 32;
  • 57) 0.253 048 32 × 2 = 0 + 0.506 096 64;
  • 58) 0.506 096 64 × 2 = 1 + 0.012 193 28;
  • 59) 0.012 193 28 × 2 = 0 + 0.024 386 56;
  • 60) 0.024 386 56 × 2 = 0 + 0.048 773 12;
  • 61) 0.048 773 12 × 2 = 0 + 0.097 546 24;

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:


0.003 223 62(10) =

0.0000 0000 1101 0011 0100 0011 0101 1110 0111 1001 1000 1000 0000 0001 0100 0(2)

6. Positive number before normalization:

0.003 223 62(10) =

0.0000 0000 1101 0011 0100 0011 0101 1110 0111 1001 1000 1000 0000 0001 0100 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 9 positions to the right, so that only one non zero digit remains to the left of it:


0.003 223 62(10) =

0.0000 0000 1101 0011 0100 0011 0101 1110 0111 1001 1000 1000 0000 0001 0100 0(2) =

0.0000 0000 1101 0011 0100 0011 0101 1110 0111 1001 1000 1000 0000 0001 0100 0(2) × 20 =

1.1010 0110 1000 0110 1011 1100 1111 0011 0001 0000 0000 0010 1000(2) × 2-9


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): -9

Mantissa (not normalized):
1.1010 0110 1000 0110 1011 1100 1111 0011 0001 0000 0000 0010 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-9 + 2(11-1) - 1 =

(-9 + 1 023)(10) =

1 014(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;
  • 1 014 ÷ 2 = 507 + 0;
  • 507 ÷ 2 = 253 + 1;
  • 253 ÷ 2 = 126 + 1;
  • 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) =

1014(10) =

011 1111 0110(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. 1010 0110 1000 0110 1011 1100 1111 0011 0001 0000 0000 0010 1000 =

1010 0110 1000 0110 1011 1100 1111 0011 0001 0000 0000 0010 1000


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 0110

Mantissa (52 bits) =
1010 0110 1000 0110 1011 1100 1111 0011 0001 0000 0000 0010 1000


Decimal number -0.003 223 62 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0110 - 1010 0110 1000 0110 1011 1100 1111 0011 0001 0000 0000 0010 1000

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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