-0.001 197 33 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.001 197 33(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
-0.001 197 33(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:

|-0.001 197 33| = 0.001 197 33


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: 0.001 197 33.

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) 0.001 197 33 × 2 = 0 + 0.002 394 66;
  • 2) 0.002 394 66 × 2 = 0 + 0.004 789 32;
  • 3) 0.004 789 32 × 2 = 0 + 0.009 578 64;
  • 4) 0.009 578 64 × 2 = 0 + 0.019 157 28;
  • 5) 0.019 157 28 × 2 = 0 + 0.038 314 56;
  • 6) 0.038 314 56 × 2 = 0 + 0.076 629 12;
  • 7) 0.076 629 12 × 2 = 0 + 0.153 258 24;
  • 8) 0.153 258 24 × 2 = 0 + 0.306 516 48;
  • 9) 0.306 516 48 × 2 = 0 + 0.613 032 96;
  • 10) 0.613 032 96 × 2 = 1 + 0.226 065 92;
  • 11) 0.226 065 92 × 2 = 0 + 0.452 131 84;
  • 12) 0.452 131 84 × 2 = 0 + 0.904 263 68;
  • 13) 0.904 263 68 × 2 = 1 + 0.808 527 36;
  • 14) 0.808 527 36 × 2 = 1 + 0.617 054 72;
  • 15) 0.617 054 72 × 2 = 1 + 0.234 109 44;
  • 16) 0.234 109 44 × 2 = 0 + 0.468 218 88;
  • 17) 0.468 218 88 × 2 = 0 + 0.936 437 76;
  • 18) 0.936 437 76 × 2 = 1 + 0.872 875 52;
  • 19) 0.872 875 52 × 2 = 1 + 0.745 751 04;
  • 20) 0.745 751 04 × 2 = 1 + 0.491 502 08;
  • 21) 0.491 502 08 × 2 = 0 + 0.983 004 16;
  • 22) 0.983 004 16 × 2 = 1 + 0.966 008 32;
  • 23) 0.966 008 32 × 2 = 1 + 0.932 016 64;
  • 24) 0.932 016 64 × 2 = 1 + 0.864 033 28;
  • 25) 0.864 033 28 × 2 = 1 + 0.728 066 56;
  • 26) 0.728 066 56 × 2 = 1 + 0.456 133 12;
  • 27) 0.456 133 12 × 2 = 0 + 0.912 266 24;
  • 28) 0.912 266 24 × 2 = 1 + 0.824 532 48;
  • 29) 0.824 532 48 × 2 = 1 + 0.649 064 96;
  • 30) 0.649 064 96 × 2 = 1 + 0.298 129 92;
  • 31) 0.298 129 92 × 2 = 0 + 0.596 259 84;
  • 32) 0.596 259 84 × 2 = 1 + 0.192 519 68;
  • 33) 0.192 519 68 × 2 = 0 + 0.385 039 36;
  • 34) 0.385 039 36 × 2 = 0 + 0.770 078 72;
  • 35) 0.770 078 72 × 2 = 1 + 0.540 157 44;
  • 36) 0.540 157 44 × 2 = 1 + 0.080 314 88;
  • 37) 0.080 314 88 × 2 = 0 + 0.160 629 76;
  • 38) 0.160 629 76 × 2 = 0 + 0.321 259 52;
  • 39) 0.321 259 52 × 2 = 0 + 0.642 519 04;
  • 40) 0.642 519 04 × 2 = 1 + 0.285 038 08;
  • 41) 0.285 038 08 × 2 = 0 + 0.570 076 16;
  • 42) 0.570 076 16 × 2 = 1 + 0.140 152 32;
  • 43) 0.140 152 32 × 2 = 0 + 0.280 304 64;
  • 44) 0.280 304 64 × 2 = 0 + 0.560 609 28;
  • 45) 0.560 609 28 × 2 = 1 + 0.121 218 56;
  • 46) 0.121 218 56 × 2 = 0 + 0.242 437 12;
  • 47) 0.242 437 12 × 2 = 0 + 0.484 874 24;
  • 48) 0.484 874 24 × 2 = 0 + 0.969 748 48;
  • 49) 0.969 748 48 × 2 = 1 + 0.939 496 96;
  • 50) 0.939 496 96 × 2 = 1 + 0.878 993 92;
  • 51) 0.878 993 92 × 2 = 1 + 0.757 987 84;
  • 52) 0.757 987 84 × 2 = 1 + 0.515 975 68;
  • 53) 0.515 975 68 × 2 = 1 + 0.031 951 36;
  • 54) 0.031 951 36 × 2 = 0 + 0.063 902 72;
  • 55) 0.063 902 72 × 2 = 0 + 0.127 805 44;
  • 56) 0.127 805 44 × 2 = 0 + 0.255 610 88;
  • 57) 0.255 610 88 × 2 = 0 + 0.511 221 76;
  • 58) 0.511 221 76 × 2 = 1 + 0.022 443 52;
  • 59) 0.022 443 52 × 2 = 0 + 0.044 887 04;
  • 60) 0.044 887 04 × 2 = 0 + 0.089 774 08;
  • 61) 0.089 774 08 × 2 = 0 + 0.179 548 16;
  • 62) 0.179 548 16 × 2 = 0 + 0.359 096 32;

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:


0.001 197 33(10) =

0.0000 0000 0100 1110 0111 0111 1101 1101 0011 0001 0100 1000 1111 1000 0100 00(2)

6. Positive number before normalization:

0.001 197 33(10) =

0.0000 0000 0100 1110 0111 0111 1101 1101 0011 0001 0100 1000 1111 1000 0100 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 10 positions to the right, so that only one non zero digit remains to the left of it:


0.001 197 33(10) =

0.0000 0000 0100 1110 0111 0111 1101 1101 0011 0001 0100 1000 1111 1000 0100 00(2) =

0.0000 0000 0100 1110 0111 0111 1101 1101 0011 0001 0100 1000 1111 1000 0100 00(2) × 20 =

1.0011 1001 1101 1111 0111 0100 1100 0101 0010 0011 1110 0001 0000(2) × 2-10


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): -10

Mantissa (not normalized):
1.0011 1001 1101 1111 0111 0100 1100 0101 0010 0011 1110 0001 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-10 + 2(11-1) - 1 =

(-10 + 1 023)(10) =

1 013(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;
  • 1 013 ÷ 2 = 506 + 1;
  • 506 ÷ 2 = 253 + 0;
  • 253 ÷ 2 = 126 + 1;
  • 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) =

1013(10) =

011 1111 0101(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. 0011 1001 1101 1111 0111 0100 1100 0101 0010 0011 1110 0001 0000 =

0011 1001 1101 1111 0111 0100 1100 0101 0010 0011 1110 0001 0000


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 0101

Mantissa (52 bits) =
0011 1001 1101 1111 0111 0100 1100 0101 0010 0011 1110 0001 0000


Decimal number -0.001 197 33 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0101 - 0011 1001 1101 1111 0111 0100 1100 0101 0010 0011 1110 0001 0000

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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