-0.057 025 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.057 025 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
-0.057 025 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:

|-0.057 025 8| = 0.057 025 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: 0.057 025 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) 0.057 025 8 × 2 = 0 + 0.114 051 6;
  • 2) 0.114 051 6 × 2 = 0 + 0.228 103 2;
  • 3) 0.228 103 2 × 2 = 0 + 0.456 206 4;
  • 4) 0.456 206 4 × 2 = 0 + 0.912 412 8;
  • 5) 0.912 412 8 × 2 = 1 + 0.824 825 6;
  • 6) 0.824 825 6 × 2 = 1 + 0.649 651 2;
  • 7) 0.649 651 2 × 2 = 1 + 0.299 302 4;
  • 8) 0.299 302 4 × 2 = 0 + 0.598 604 8;
  • 9) 0.598 604 8 × 2 = 1 + 0.197 209 6;
  • 10) 0.197 209 6 × 2 = 0 + 0.394 419 2;
  • 11) 0.394 419 2 × 2 = 0 + 0.788 838 4;
  • 12) 0.788 838 4 × 2 = 1 + 0.577 676 8;
  • 13) 0.577 676 8 × 2 = 1 + 0.155 353 6;
  • 14) 0.155 353 6 × 2 = 0 + 0.310 707 2;
  • 15) 0.310 707 2 × 2 = 0 + 0.621 414 4;
  • 16) 0.621 414 4 × 2 = 1 + 0.242 828 8;
  • 17) 0.242 828 8 × 2 = 0 + 0.485 657 6;
  • 18) 0.485 657 6 × 2 = 0 + 0.971 315 2;
  • 19) 0.971 315 2 × 2 = 1 + 0.942 630 4;
  • 20) 0.942 630 4 × 2 = 1 + 0.885 260 8;
  • 21) 0.885 260 8 × 2 = 1 + 0.770 521 6;
  • 22) 0.770 521 6 × 2 = 1 + 0.541 043 2;
  • 23) 0.541 043 2 × 2 = 1 + 0.082 086 4;
  • 24) 0.082 086 4 × 2 = 0 + 0.164 172 8;
  • 25) 0.164 172 8 × 2 = 0 + 0.328 345 6;
  • 26) 0.328 345 6 × 2 = 0 + 0.656 691 2;
  • 27) 0.656 691 2 × 2 = 1 + 0.313 382 4;
  • 28) 0.313 382 4 × 2 = 0 + 0.626 764 8;
  • 29) 0.626 764 8 × 2 = 1 + 0.253 529 6;
  • 30) 0.253 529 6 × 2 = 0 + 0.507 059 2;
  • 31) 0.507 059 2 × 2 = 1 + 0.014 118 4;
  • 32) 0.014 118 4 × 2 = 0 + 0.028 236 8;
  • 33) 0.028 236 8 × 2 = 0 + 0.056 473 6;
  • 34) 0.056 473 6 × 2 = 0 + 0.112 947 2;
  • 35) 0.112 947 2 × 2 = 0 + 0.225 894 4;
  • 36) 0.225 894 4 × 2 = 0 + 0.451 788 8;
  • 37) 0.451 788 8 × 2 = 0 + 0.903 577 6;
  • 38) 0.903 577 6 × 2 = 1 + 0.807 155 2;
  • 39) 0.807 155 2 × 2 = 1 + 0.614 310 4;
  • 40) 0.614 310 4 × 2 = 1 + 0.228 620 8;
  • 41) 0.228 620 8 × 2 = 0 + 0.457 241 6;
  • 42) 0.457 241 6 × 2 = 0 + 0.914 483 2;
  • 43) 0.914 483 2 × 2 = 1 + 0.828 966 4;
  • 44) 0.828 966 4 × 2 = 1 + 0.657 932 8;
  • 45) 0.657 932 8 × 2 = 1 + 0.315 865 6;
  • 46) 0.315 865 6 × 2 = 0 + 0.631 731 2;
  • 47) 0.631 731 2 × 2 = 1 + 0.263 462 4;
  • 48) 0.263 462 4 × 2 = 0 + 0.526 924 8;
  • 49) 0.526 924 8 × 2 = 1 + 0.053 849 6;
  • 50) 0.053 849 6 × 2 = 0 + 0.107 699 2;
  • 51) 0.107 699 2 × 2 = 0 + 0.215 398 4;
  • 52) 0.215 398 4 × 2 = 0 + 0.430 796 8;
  • 53) 0.430 796 8 × 2 = 0 + 0.861 593 6;
  • 54) 0.861 593 6 × 2 = 1 + 0.723 187 2;
  • 55) 0.723 187 2 × 2 = 1 + 0.446 374 4;
  • 56) 0.446 374 4 × 2 = 0 + 0.892 748 8;
  • 57) 0.892 748 8 × 2 = 1 + 0.785 497 6;

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:


0.057 025 8(10) =

0.0000 1110 1001 1001 0011 1110 0010 1010 0000 0111 0011 1010 1000 0110 1(2)

6. Positive number before normalization:

0.057 025 8(10) =

0.0000 1110 1001 1001 0011 1110 0010 1010 0000 0111 0011 1010 1000 0110 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the right, so that only one non zero digit remains to the left of it:


0.057 025 8(10) =

0.0000 1110 1001 1001 0011 1110 0010 1010 0000 0111 0011 1010 1000 0110 1(2) =

0.0000 1110 1001 1001 0011 1110 0010 1010 0000 0111 0011 1010 1000 0110 1(2) × 20 =

1.1101 0011 0010 0111 1100 0101 0100 0000 1110 0111 0101 0000 1101(2) × 2-5


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): -5

Mantissa (not normalized):
1.1101 0011 0010 0111 1100 0101 0100 0000 1110 0111 0101 0000 1101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-5 + 2(11-1) - 1 =

(-5 + 1 023)(10) =

1 018(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;
  • 1 018 ÷ 2 = 509 + 0;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1018(10) =

011 1111 1010(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. 1101 0011 0010 0111 1100 0101 0100 0000 1110 0111 0101 0000 1101 =

1101 0011 0010 0111 1100 0101 0100 0000 1110 0111 0101 0000 1101


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 1010

Mantissa (52 bits) =
1101 0011 0010 0111 1100 0101 0100 0000 1110 0111 0101 0000 1101


Decimal number -0.057 025 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 1101 0011 0010 0111 1100 0101 0100 0000 1110 0111 0101 0000 1101

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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