-0.344 997 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.344 997 2(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
-0.344 997 2(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:

|-0.344 997 2| = 0.344 997 2


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: 0.344 997 2.

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) 0.344 997 2 × 2 = 0 + 0.689 994 4;
  • 2) 0.689 994 4 × 2 = 1 + 0.379 988 8;
  • 3) 0.379 988 8 × 2 = 0 + 0.759 977 6;
  • 4) 0.759 977 6 × 2 = 1 + 0.519 955 2;
  • 5) 0.519 955 2 × 2 = 1 + 0.039 910 4;
  • 6) 0.039 910 4 × 2 = 0 + 0.079 820 8;
  • 7) 0.079 820 8 × 2 = 0 + 0.159 641 6;
  • 8) 0.159 641 6 × 2 = 0 + 0.319 283 2;
  • 9) 0.319 283 2 × 2 = 0 + 0.638 566 4;
  • 10) 0.638 566 4 × 2 = 1 + 0.277 132 8;
  • 11) 0.277 132 8 × 2 = 0 + 0.554 265 6;
  • 12) 0.554 265 6 × 2 = 1 + 0.108 531 2;
  • 13) 0.108 531 2 × 2 = 0 + 0.217 062 4;
  • 14) 0.217 062 4 × 2 = 0 + 0.434 124 8;
  • 15) 0.434 124 8 × 2 = 0 + 0.868 249 6;
  • 16) 0.868 249 6 × 2 = 1 + 0.736 499 2;
  • 17) 0.736 499 2 × 2 = 1 + 0.472 998 4;
  • 18) 0.472 998 4 × 2 = 0 + 0.945 996 8;
  • 19) 0.945 996 8 × 2 = 1 + 0.891 993 6;
  • 20) 0.891 993 6 × 2 = 1 + 0.783 987 2;
  • 21) 0.783 987 2 × 2 = 1 + 0.567 974 4;
  • 22) 0.567 974 4 × 2 = 1 + 0.135 948 8;
  • 23) 0.135 948 8 × 2 = 0 + 0.271 897 6;
  • 24) 0.271 897 6 × 2 = 0 + 0.543 795 2;
  • 25) 0.543 795 2 × 2 = 1 + 0.087 590 4;
  • 26) 0.087 590 4 × 2 = 0 + 0.175 180 8;
  • 27) 0.175 180 8 × 2 = 0 + 0.350 361 6;
  • 28) 0.350 361 6 × 2 = 0 + 0.700 723 2;
  • 29) 0.700 723 2 × 2 = 1 + 0.401 446 4;
  • 30) 0.401 446 4 × 2 = 0 + 0.802 892 8;
  • 31) 0.802 892 8 × 2 = 1 + 0.605 785 6;
  • 32) 0.605 785 6 × 2 = 1 + 0.211 571 2;
  • 33) 0.211 571 2 × 2 = 0 + 0.423 142 4;
  • 34) 0.423 142 4 × 2 = 0 + 0.846 284 8;
  • 35) 0.846 284 8 × 2 = 1 + 0.692 569 6;
  • 36) 0.692 569 6 × 2 = 1 + 0.385 139 2;
  • 37) 0.385 139 2 × 2 = 0 + 0.770 278 4;
  • 38) 0.770 278 4 × 2 = 1 + 0.540 556 8;
  • 39) 0.540 556 8 × 2 = 1 + 0.081 113 6;
  • 40) 0.081 113 6 × 2 = 0 + 0.162 227 2;
  • 41) 0.162 227 2 × 2 = 0 + 0.324 454 4;
  • 42) 0.324 454 4 × 2 = 0 + 0.648 908 8;
  • 43) 0.648 908 8 × 2 = 1 + 0.297 817 6;
  • 44) 0.297 817 6 × 2 = 0 + 0.595 635 2;
  • 45) 0.595 635 2 × 2 = 1 + 0.191 270 4;
  • 46) 0.191 270 4 × 2 = 0 + 0.382 540 8;
  • 47) 0.382 540 8 × 2 = 0 + 0.765 081 6;
  • 48) 0.765 081 6 × 2 = 1 + 0.530 163 2;
  • 49) 0.530 163 2 × 2 = 1 + 0.060 326 4;
  • 50) 0.060 326 4 × 2 = 0 + 0.120 652 8;
  • 51) 0.120 652 8 × 2 = 0 + 0.241 305 6;
  • 52) 0.241 305 6 × 2 = 0 + 0.482 611 2;
  • 53) 0.482 611 2 × 2 = 0 + 0.965 222 4;
  • 54) 0.965 222 4 × 2 = 1 + 0.930 444 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:


0.344 997 2(10) =

0.0101 1000 0101 0001 1011 1100 1000 1011 0011 0110 0010 1001 1000 01(2)

6. Positive number before normalization:

0.344 997 2(10) =

0.0101 1000 0101 0001 1011 1100 1000 1011 0011 0110 0010 1001 1000 01(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:


0.344 997 2(10) =

0.0101 1000 0101 0001 1011 1100 1000 1011 0011 0110 0010 1001 1000 01(2) =

0.0101 1000 0101 0001 1011 1100 1000 1011 0011 0110 0010 1001 1000 01(2) × 20 =

1.0110 0001 0100 0110 1111 0010 0010 1100 1101 1000 1010 0110 0001(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 0110 1111 0010 0010 1100 1101 1000 1010 0110 0001


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 0110 1111 0010 0010 1100 1101 1000 1010 0110 0001 =

0110 0001 0100 0110 1111 0010 0010 1100 1101 1000 1010 0110 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) =
011 1111 1101

Mantissa (52 bits) =
0110 0001 0100 0110 1111 0010 0010 1100 1101 1000 1010 0110 0001


Decimal number -0.344 997 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1101 - 0110 0001 0100 0110 1111 0010 0010 1100 1101 1000 1010 0110 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