-0.344 999 998 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.344 999 998 7(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
-0.344 999 998 7(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:

|-0.344 999 998 7| = 0.344 999 998 7


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: 0.344 999 998 7.

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) 0.344 999 998 7 × 2 = 0 + 0.689 999 997 4;
  • 2) 0.689 999 997 4 × 2 = 1 + 0.379 999 994 8;
  • 3) 0.379 999 994 8 × 2 = 0 + 0.759 999 989 6;
  • 4) 0.759 999 989 6 × 2 = 1 + 0.519 999 979 2;
  • 5) 0.519 999 979 2 × 2 = 1 + 0.039 999 958 4;
  • 6) 0.039 999 958 4 × 2 = 0 + 0.079 999 916 8;
  • 7) 0.079 999 916 8 × 2 = 0 + 0.159 999 833 6;
  • 8) 0.159 999 833 6 × 2 = 0 + 0.319 999 667 2;
  • 9) 0.319 999 667 2 × 2 = 0 + 0.639 999 334 4;
  • 10) 0.639 999 334 4 × 2 = 1 + 0.279 998 668 8;
  • 11) 0.279 998 668 8 × 2 = 0 + 0.559 997 337 6;
  • 12) 0.559 997 337 6 × 2 = 1 + 0.119 994 675 2;
  • 13) 0.119 994 675 2 × 2 = 0 + 0.239 989 350 4;
  • 14) 0.239 989 350 4 × 2 = 0 + 0.479 978 700 8;
  • 15) 0.479 978 700 8 × 2 = 0 + 0.959 957 401 6;
  • 16) 0.959 957 401 6 × 2 = 1 + 0.919 914 803 2;
  • 17) 0.919 914 803 2 × 2 = 1 + 0.839 829 606 4;
  • 18) 0.839 829 606 4 × 2 = 1 + 0.679 659 212 8;
  • 19) 0.679 659 212 8 × 2 = 1 + 0.359 318 425 6;
  • 20) 0.359 318 425 6 × 2 = 0 + 0.718 636 851 2;
  • 21) 0.718 636 851 2 × 2 = 1 + 0.437 273 702 4;
  • 22) 0.437 273 702 4 × 2 = 0 + 0.874 547 404 8;
  • 23) 0.874 547 404 8 × 2 = 1 + 0.749 094 809 6;
  • 24) 0.749 094 809 6 × 2 = 1 + 0.498 189 619 2;
  • 25) 0.498 189 619 2 × 2 = 0 + 0.996 379 238 4;
  • 26) 0.996 379 238 4 × 2 = 1 + 0.992 758 476 8;
  • 27) 0.992 758 476 8 × 2 = 1 + 0.985 516 953 6;
  • 28) 0.985 516 953 6 × 2 = 1 + 0.971 033 907 2;
  • 29) 0.971 033 907 2 × 2 = 1 + 0.942 067 814 4;
  • 30) 0.942 067 814 4 × 2 = 1 + 0.884 135 628 8;
  • 31) 0.884 135 628 8 × 2 = 1 + 0.768 271 257 6;
  • 32) 0.768 271 257 6 × 2 = 1 + 0.536 542 515 2;
  • 33) 0.536 542 515 2 × 2 = 1 + 0.073 085 030 4;
  • 34) 0.073 085 030 4 × 2 = 0 + 0.146 170 060 8;
  • 35) 0.146 170 060 8 × 2 = 0 + 0.292 340 121 6;
  • 36) 0.292 340 121 6 × 2 = 0 + 0.584 680 243 2;
  • 37) 0.584 680 243 2 × 2 = 1 + 0.169 360 486 4;
  • 38) 0.169 360 486 4 × 2 = 0 + 0.338 720 972 8;
  • 39) 0.338 720 972 8 × 2 = 0 + 0.677 441 945 6;
  • 40) 0.677 441 945 6 × 2 = 1 + 0.354 883 891 2;
  • 41) 0.354 883 891 2 × 2 = 0 + 0.709 767 782 4;
  • 42) 0.709 767 782 4 × 2 = 1 + 0.419 535 564 8;
  • 43) 0.419 535 564 8 × 2 = 0 + 0.839 071 129 6;
  • 44) 0.839 071 129 6 × 2 = 1 + 0.678 142 259 2;
  • 45) 0.678 142 259 2 × 2 = 1 + 0.356 284 518 4;
  • 46) 0.356 284 518 4 × 2 = 0 + 0.712 569 036 8;
  • 47) 0.712 569 036 8 × 2 = 1 + 0.425 138 073 6;
  • 48) 0.425 138 073 6 × 2 = 0 + 0.850 276 147 2;
  • 49) 0.850 276 147 2 × 2 = 1 + 0.700 552 294 4;
  • 50) 0.700 552 294 4 × 2 = 1 + 0.401 104 588 8;
  • 51) 0.401 104 588 8 × 2 = 0 + 0.802 209 177 6;
  • 52) 0.802 209 177 6 × 2 = 1 + 0.604 418 355 2;
  • 53) 0.604 418 355 2 × 2 = 1 + 0.208 836 710 4;
  • 54) 0.208 836 710 4 × 2 = 0 + 0.417 673 420 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:


0.344 999 998 7(10) =

0.0101 1000 0101 0001 1110 1011 0111 1111 1000 1001 0101 1010 1101 10(2)

6. Positive number before normalization:

0.344 999 998 7(10) =

0.0101 1000 0101 0001 1110 1011 0111 1111 1000 1001 0101 1010 1101 10(2)

7. Normalize the binary representation of the number.

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


0.344 999 998 7(10) =

0.0101 1000 0101 0001 1110 1011 0111 1111 1000 1001 0101 1010 1101 10(2) =

0.0101 1000 0101 0001 1110 1011 0111 1111 1000 1001 0101 1010 1101 10(2) × 20 =

1.0110 0001 0100 0111 1010 1101 1111 1110 0010 0101 0110 1011 0110(2) × 2-2


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):
1.0110 0001 0100 0111 1010 1101 1111 1110 0010 0101 0110 1011 0110


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) =

1 021(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;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 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) =

1021(10) =

011 1111 1101(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. 0110 0001 0100 0111 1010 1101 1111 1110 0010 0101 0110 1011 0110 =

0110 0001 0100 0111 1010 1101 1111 1110 0010 0101 0110 1011 0110


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 1101

Mantissa (52 bits) =
0110 0001 0100 0111 1010 1101 1111 1110 0010 0101 0110 1011 0110


Decimal number -0.344 999 998 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1101 - 0110 0001 0100 0111 1010 1101 1111 1110 0010 0101 0110 1011 0110

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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