64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: -5.447 165 52 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number -5.447 165 52(10) converted and written in 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:

|-5.447 165 52| = 5.447 165 52

2. First, convert to binary (in base 2) the integer part: 5.
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;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.


5(10) =


101(2)


4. Convert to binary (base 2) the fractional part: 0.447 165 52.

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) 0.447 165 52 × 2 = 0 + 0.894 331 04;
  • 2) 0.894 331 04 × 2 = 1 + 0.788 662 08;
  • 3) 0.788 662 08 × 2 = 1 + 0.577 324 16;
  • 4) 0.577 324 16 × 2 = 1 + 0.154 648 32;
  • 5) 0.154 648 32 × 2 = 0 + 0.309 296 64;
  • 6) 0.309 296 64 × 2 = 0 + 0.618 593 28;
  • 7) 0.618 593 28 × 2 = 1 + 0.237 186 56;
  • 8) 0.237 186 56 × 2 = 0 + 0.474 373 12;
  • 9) 0.474 373 12 × 2 = 0 + 0.948 746 24;
  • 10) 0.948 746 24 × 2 = 1 + 0.897 492 48;
  • 11) 0.897 492 48 × 2 = 1 + 0.794 984 96;
  • 12) 0.794 984 96 × 2 = 1 + 0.589 969 92;
  • 13) 0.589 969 92 × 2 = 1 + 0.179 939 84;
  • 14) 0.179 939 84 × 2 = 0 + 0.359 879 68;
  • 15) 0.359 879 68 × 2 = 0 + 0.719 759 36;
  • 16) 0.719 759 36 × 2 = 1 + 0.439 518 72;
  • 17) 0.439 518 72 × 2 = 0 + 0.879 037 44;
  • 18) 0.879 037 44 × 2 = 1 + 0.758 074 88;
  • 19) 0.758 074 88 × 2 = 1 + 0.516 149 76;
  • 20) 0.516 149 76 × 2 = 1 + 0.032 299 52;
  • 21) 0.032 299 52 × 2 = 0 + 0.064 599 04;
  • 22) 0.064 599 04 × 2 = 0 + 0.129 198 08;
  • 23) 0.129 198 08 × 2 = 0 + 0.258 396 16;
  • 24) 0.258 396 16 × 2 = 0 + 0.516 792 32;
  • 25) 0.516 792 32 × 2 = 1 + 0.033 584 64;
  • 26) 0.033 584 64 × 2 = 0 + 0.067 169 28;
  • 27) 0.067 169 28 × 2 = 0 + 0.134 338 56;
  • 28) 0.134 338 56 × 2 = 0 + 0.268 677 12;
  • 29) 0.268 677 12 × 2 = 0 + 0.537 354 24;
  • 30) 0.537 354 24 × 2 = 1 + 0.074 708 48;
  • 31) 0.074 708 48 × 2 = 0 + 0.149 416 96;
  • 32) 0.149 416 96 × 2 = 0 + 0.298 833 92;
  • 33) 0.298 833 92 × 2 = 0 + 0.597 667 84;
  • 34) 0.597 667 84 × 2 = 1 + 0.195 335 68;
  • 35) 0.195 335 68 × 2 = 0 + 0.390 671 36;
  • 36) 0.390 671 36 × 2 = 0 + 0.781 342 72;
  • 37) 0.781 342 72 × 2 = 1 + 0.562 685 44;
  • 38) 0.562 685 44 × 2 = 1 + 0.125 370 88;
  • 39) 0.125 370 88 × 2 = 0 + 0.250 741 76;
  • 40) 0.250 741 76 × 2 = 0 + 0.501 483 52;
  • 41) 0.501 483 52 × 2 = 1 + 0.002 967 04;
  • 42) 0.002 967 04 × 2 = 0 + 0.005 934 08;
  • 43) 0.005 934 08 × 2 = 0 + 0.011 868 16;
  • 44) 0.011 868 16 × 2 = 0 + 0.023 736 32;
  • 45) 0.023 736 32 × 2 = 0 + 0.047 472 64;
  • 46) 0.047 472 64 × 2 = 0 + 0.094 945 28;
  • 47) 0.094 945 28 × 2 = 0 + 0.189 890 56;
  • 48) 0.189 890 56 × 2 = 0 + 0.379 781 12;
  • 49) 0.379 781 12 × 2 = 0 + 0.759 562 24;
  • 50) 0.759 562 24 × 2 = 1 + 0.519 124 48;
  • 51) 0.519 124 48 × 2 = 1 + 0.038 248 96;
  • 52) 0.038 248 96 × 2 = 0 + 0.076 497 92;
  • 53) 0.076 497 92 × 2 = 0 + 0.152 995 84;

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...)


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:


0.447 165 52(10) =


0.0111 0010 0111 1001 0111 0000 1000 0100 0100 1100 1000 0000 0110 0(2)


6. Positive number before normalization:

5.447 165 52(10) =


101.0111 0010 0111 1001 0111 0000 1000 0100 0100 1100 1000 0000 0110 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


5.447 165 52(10) =


101.0111 0010 0111 1001 0111 0000 1000 0100 0100 1100 1000 0000 0110 0(2) =


101.0111 0010 0111 1001 0111 0000 1000 0100 0100 1100 1000 0000 0110 0(2) × 20 =


1.0101 1100 1001 1110 0101 1100 0010 0001 0001 0011 0010 0000 0001 100(2) × 22


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): 2


Mantissa (not normalized):
1.0101 1100 1001 1110 0101 1100 0010 0001 0001 0011 0010 0000 0001 100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


2 + 2(11-1) - 1 =


(2 + 1 023)(10) =


1 025(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;
  • 1 025 ÷ 2 = 512 + 1;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =


1025(10) =


100 0000 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 0101 1100 1001 1110 0101 1100 0010 0001 0001 0011 0010 0000 0001 100 =


0101 1100 1001 1110 0101 1100 0010 0001 0001 0011 0010 0000 0001


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) =
100 0000 0001


Mantissa (52 bits) =
0101 1100 1001 1110 0101 1100 0010 0001 0001 0011 0010 0000 0001


The base ten decimal number -5.447 165 52 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
1 - 100 0000 0001 - 0101 1100 1001 1110 0101 1100 0010 0001 0001 0011 0010 0000 0001

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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