-0.001 196 76 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.001 196 76(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
-0.001 196 76(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:

|-0.001 196 76| = 0.001 196 76


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: 0.001 196 76.

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) 0.001 196 76 × 2 = 0 + 0.002 393 52;
  • 2) 0.002 393 52 × 2 = 0 + 0.004 787 04;
  • 3) 0.004 787 04 × 2 = 0 + 0.009 574 08;
  • 4) 0.009 574 08 × 2 = 0 + 0.019 148 16;
  • 5) 0.019 148 16 × 2 = 0 + 0.038 296 32;
  • 6) 0.038 296 32 × 2 = 0 + 0.076 592 64;
  • 7) 0.076 592 64 × 2 = 0 + 0.153 185 28;
  • 8) 0.153 185 28 × 2 = 0 + 0.306 370 56;
  • 9) 0.306 370 56 × 2 = 0 + 0.612 741 12;
  • 10) 0.612 741 12 × 2 = 1 + 0.225 482 24;
  • 11) 0.225 482 24 × 2 = 0 + 0.450 964 48;
  • 12) 0.450 964 48 × 2 = 0 + 0.901 928 96;
  • 13) 0.901 928 96 × 2 = 1 + 0.803 857 92;
  • 14) 0.803 857 92 × 2 = 1 + 0.607 715 84;
  • 15) 0.607 715 84 × 2 = 1 + 0.215 431 68;
  • 16) 0.215 431 68 × 2 = 0 + 0.430 863 36;
  • 17) 0.430 863 36 × 2 = 0 + 0.861 726 72;
  • 18) 0.861 726 72 × 2 = 1 + 0.723 453 44;
  • 19) 0.723 453 44 × 2 = 1 + 0.446 906 88;
  • 20) 0.446 906 88 × 2 = 0 + 0.893 813 76;
  • 21) 0.893 813 76 × 2 = 1 + 0.787 627 52;
  • 22) 0.787 627 52 × 2 = 1 + 0.575 255 04;
  • 23) 0.575 255 04 × 2 = 1 + 0.150 510 08;
  • 24) 0.150 510 08 × 2 = 0 + 0.301 020 16;
  • 25) 0.301 020 16 × 2 = 0 + 0.602 040 32;
  • 26) 0.602 040 32 × 2 = 1 + 0.204 080 64;
  • 27) 0.204 080 64 × 2 = 0 + 0.408 161 28;
  • 28) 0.408 161 28 × 2 = 0 + 0.816 322 56;
  • 29) 0.816 322 56 × 2 = 1 + 0.632 645 12;
  • 30) 0.632 645 12 × 2 = 1 + 0.265 290 24;
  • 31) 0.265 290 24 × 2 = 0 + 0.530 580 48;
  • 32) 0.530 580 48 × 2 = 1 + 0.061 160 96;
  • 33) 0.061 160 96 × 2 = 0 + 0.122 321 92;
  • 34) 0.122 321 92 × 2 = 0 + 0.244 643 84;
  • 35) 0.244 643 84 × 2 = 0 + 0.489 287 68;
  • 36) 0.489 287 68 × 2 = 0 + 0.978 575 36;
  • 37) 0.978 575 36 × 2 = 1 + 0.957 150 72;
  • 38) 0.957 150 72 × 2 = 1 + 0.914 301 44;
  • 39) 0.914 301 44 × 2 = 1 + 0.828 602 88;
  • 40) 0.828 602 88 × 2 = 1 + 0.657 205 76;
  • 41) 0.657 205 76 × 2 = 1 + 0.314 411 52;
  • 42) 0.314 411 52 × 2 = 0 + 0.628 823 04;
  • 43) 0.628 823 04 × 2 = 1 + 0.257 646 08;
  • 44) 0.257 646 08 × 2 = 0 + 0.515 292 16;
  • 45) 0.515 292 16 × 2 = 1 + 0.030 584 32;
  • 46) 0.030 584 32 × 2 = 0 + 0.061 168 64;
  • 47) 0.061 168 64 × 2 = 0 + 0.122 337 28;
  • 48) 0.122 337 28 × 2 = 0 + 0.244 674 56;
  • 49) 0.244 674 56 × 2 = 0 + 0.489 349 12;
  • 50) 0.489 349 12 × 2 = 0 + 0.978 698 24;
  • 51) 0.978 698 24 × 2 = 1 + 0.957 396 48;
  • 52) 0.957 396 48 × 2 = 1 + 0.914 792 96;
  • 53) 0.914 792 96 × 2 = 1 + 0.829 585 92;
  • 54) 0.829 585 92 × 2 = 1 + 0.659 171 84;
  • 55) 0.659 171 84 × 2 = 1 + 0.318 343 68;
  • 56) 0.318 343 68 × 2 = 0 + 0.636 687 36;
  • 57) 0.636 687 36 × 2 = 1 + 0.273 374 72;
  • 58) 0.273 374 72 × 2 = 0 + 0.546 749 44;
  • 59) 0.546 749 44 × 2 = 1 + 0.093 498 88;
  • 60) 0.093 498 88 × 2 = 0 + 0.186 997 76;
  • 61) 0.186 997 76 × 2 = 0 + 0.373 995 52;
  • 62) 0.373 995 52 × 2 = 0 + 0.747 991 04;

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:


0.001 196 76(10) =

0.0000 0000 0100 1110 0110 1110 0100 1101 0000 1111 1010 1000 0011 1110 1010 00(2)

6. Positive number before normalization:

0.001 196 76(10) =

0.0000 0000 0100 1110 0110 1110 0100 1101 0000 1111 1010 1000 0011 1110 1010 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:


0.001 196 76(10) =

0.0000 0000 0100 1110 0110 1110 0100 1101 0000 1111 1010 1000 0011 1110 1010 00(2) =

0.0000 0000 0100 1110 0110 1110 0100 1101 0000 1111 1010 1000 0011 1110 1010 00(2) × 20 =

1.0011 1001 1011 1001 0011 0100 0011 1110 1010 0000 1111 1010 1000(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 1011 1001 0011 0100 0011 1110 1010 0000 1111 1010 1000


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 1011 1001 0011 0100 0011 1110 1010 0000 1111 1010 1000 =

0011 1001 1011 1001 0011 0100 0011 1110 1010 0000 1111 1010 1000


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 1011 1001 0011 0100 0011 1110 1010 0000 1111 1010 1000


Decimal number -0.001 196 76 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0101 - 0011 1001 1011 1001 0011 0100 0011 1110 1010 0000 1111 1010 1000

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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