-0.000 253 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 253 8(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
-0.000 253 8(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:

|-0.000 253 8| = 0.000 253 8


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: 0.000 253 8.

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) 0.000 253 8 × 2 = 0 + 0.000 507 6;
  • 2) 0.000 507 6 × 2 = 0 + 0.001 015 2;
  • 3) 0.001 015 2 × 2 = 0 + 0.002 030 4;
  • 4) 0.002 030 4 × 2 = 0 + 0.004 060 8;
  • 5) 0.004 060 8 × 2 = 0 + 0.008 121 6;
  • 6) 0.008 121 6 × 2 = 0 + 0.016 243 2;
  • 7) 0.016 243 2 × 2 = 0 + 0.032 486 4;
  • 8) 0.032 486 4 × 2 = 0 + 0.064 972 8;
  • 9) 0.064 972 8 × 2 = 0 + 0.129 945 6;
  • 10) 0.129 945 6 × 2 = 0 + 0.259 891 2;
  • 11) 0.259 891 2 × 2 = 0 + 0.519 782 4;
  • 12) 0.519 782 4 × 2 = 1 + 0.039 564 8;
  • 13) 0.039 564 8 × 2 = 0 + 0.079 129 6;
  • 14) 0.079 129 6 × 2 = 0 + 0.158 259 2;
  • 15) 0.158 259 2 × 2 = 0 + 0.316 518 4;
  • 16) 0.316 518 4 × 2 = 0 + 0.633 036 8;
  • 17) 0.633 036 8 × 2 = 1 + 0.266 073 6;
  • 18) 0.266 073 6 × 2 = 0 + 0.532 147 2;
  • 19) 0.532 147 2 × 2 = 1 + 0.064 294 4;
  • 20) 0.064 294 4 × 2 = 0 + 0.128 588 8;
  • 21) 0.128 588 8 × 2 = 0 + 0.257 177 6;
  • 22) 0.257 177 6 × 2 = 0 + 0.514 355 2;
  • 23) 0.514 355 2 × 2 = 1 + 0.028 710 4;
  • 24) 0.028 710 4 × 2 = 0 + 0.057 420 8;
  • 25) 0.057 420 8 × 2 = 0 + 0.114 841 6;
  • 26) 0.114 841 6 × 2 = 0 + 0.229 683 2;
  • 27) 0.229 683 2 × 2 = 0 + 0.459 366 4;
  • 28) 0.459 366 4 × 2 = 0 + 0.918 732 8;
  • 29) 0.918 732 8 × 2 = 1 + 0.837 465 6;
  • 30) 0.837 465 6 × 2 = 1 + 0.674 931 2;
  • 31) 0.674 931 2 × 2 = 1 + 0.349 862 4;
  • 32) 0.349 862 4 × 2 = 0 + 0.699 724 8;
  • 33) 0.699 724 8 × 2 = 1 + 0.399 449 6;
  • 34) 0.399 449 6 × 2 = 0 + 0.798 899 2;
  • 35) 0.798 899 2 × 2 = 1 + 0.597 798 4;
  • 36) 0.597 798 4 × 2 = 1 + 0.195 596 8;
  • 37) 0.195 596 8 × 2 = 0 + 0.391 193 6;
  • 38) 0.391 193 6 × 2 = 0 + 0.782 387 2;
  • 39) 0.782 387 2 × 2 = 1 + 0.564 774 4;
  • 40) 0.564 774 4 × 2 = 1 + 0.129 548 8;
  • 41) 0.129 548 8 × 2 = 0 + 0.259 097 6;
  • 42) 0.259 097 6 × 2 = 0 + 0.518 195 2;
  • 43) 0.518 195 2 × 2 = 1 + 0.036 390 4;
  • 44) 0.036 390 4 × 2 = 0 + 0.072 780 8;
  • 45) 0.072 780 8 × 2 = 0 + 0.145 561 6;
  • 46) 0.145 561 6 × 2 = 0 + 0.291 123 2;
  • 47) 0.291 123 2 × 2 = 0 + 0.582 246 4;
  • 48) 0.582 246 4 × 2 = 1 + 0.164 492 8;
  • 49) 0.164 492 8 × 2 = 0 + 0.328 985 6;
  • 50) 0.328 985 6 × 2 = 0 + 0.657 971 2;
  • 51) 0.657 971 2 × 2 = 1 + 0.315 942 4;
  • 52) 0.315 942 4 × 2 = 0 + 0.631 884 8;
  • 53) 0.631 884 8 × 2 = 1 + 0.263 769 6;
  • 54) 0.263 769 6 × 2 = 0 + 0.527 539 2;
  • 55) 0.527 539 2 × 2 = 1 + 0.055 078 4;
  • 56) 0.055 078 4 × 2 = 0 + 0.110 156 8;
  • 57) 0.110 156 8 × 2 = 0 + 0.220 313 6;
  • 58) 0.220 313 6 × 2 = 0 + 0.440 627 2;
  • 59) 0.440 627 2 × 2 = 0 + 0.881 254 4;
  • 60) 0.881 254 4 × 2 = 1 + 0.762 508 8;
  • 61) 0.762 508 8 × 2 = 1 + 0.525 017 6;
  • 62) 0.525 017 6 × 2 = 1 + 0.050 035 2;
  • 63) 0.050 035 2 × 2 = 0 + 0.100 070 4;
  • 64) 0.100 070 4 × 2 = 0 + 0.200 140 8;

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:


0.000 253 8(10) =

0.0000 0000 0001 0000 1010 0010 0000 1110 1011 0011 0010 0001 0010 1010 0001 1100(2)

6. Positive number before normalization:

0.000 253 8(10) =

0.0000 0000 0001 0000 1010 0010 0000 1110 1011 0011 0010 0001 0010 1010 0001 1100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 253 8(10) =

0.0000 0000 0001 0000 1010 0010 0000 1110 1011 0011 0010 0001 0010 1010 0001 1100(2) =

0.0000 0000 0001 0000 1010 0010 0000 1110 1011 0011 0010 0001 0010 1010 0001 1100(2) × 20 =

1.0000 1010 0010 0000 1110 1011 0011 0010 0001 0010 1010 0001 1100(2) × 2-12


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): -12

Mantissa (not normalized):
1.0000 1010 0010 0000 1110 1011 0011 0010 0001 0010 1010 0001 1100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 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) =

1011(10) =

011 1111 0011(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. 0000 1010 0010 0000 1110 1011 0011 0010 0001 0010 1010 0001 1100 =

0000 1010 0010 0000 1110 1011 0011 0010 0001 0010 1010 0001 1100


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 0011

Mantissa (52 bits) =
0000 1010 0010 0000 1110 1011 0011 0010 0001 0010 1010 0001 1100


Decimal number -0.000 253 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0000 1010 0010 0000 1110 1011 0011 0010 0001 0010 1010 0001 1100

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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