-0.000 282 009 16 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 009 16(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
-0.000 282 009 16(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:

|-0.000 282 009 16| = 0.000 282 009 16


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: 0.000 282 009 16.

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) 0.000 282 009 16 × 2 = 0 + 0.000 564 018 32;
  • 2) 0.000 564 018 32 × 2 = 0 + 0.001 128 036 64;
  • 3) 0.001 128 036 64 × 2 = 0 + 0.002 256 073 28;
  • 4) 0.002 256 073 28 × 2 = 0 + 0.004 512 146 56;
  • 5) 0.004 512 146 56 × 2 = 0 + 0.009 024 293 12;
  • 6) 0.009 024 293 12 × 2 = 0 + 0.018 048 586 24;
  • 7) 0.018 048 586 24 × 2 = 0 + 0.036 097 172 48;
  • 8) 0.036 097 172 48 × 2 = 0 + 0.072 194 344 96;
  • 9) 0.072 194 344 96 × 2 = 0 + 0.144 388 689 92;
  • 10) 0.144 388 689 92 × 2 = 0 + 0.288 777 379 84;
  • 11) 0.288 777 379 84 × 2 = 0 + 0.577 554 759 68;
  • 12) 0.577 554 759 68 × 2 = 1 + 0.155 109 519 36;
  • 13) 0.155 109 519 36 × 2 = 0 + 0.310 219 038 72;
  • 14) 0.310 219 038 72 × 2 = 0 + 0.620 438 077 44;
  • 15) 0.620 438 077 44 × 2 = 1 + 0.240 876 154 88;
  • 16) 0.240 876 154 88 × 2 = 0 + 0.481 752 309 76;
  • 17) 0.481 752 309 76 × 2 = 0 + 0.963 504 619 52;
  • 18) 0.963 504 619 52 × 2 = 1 + 0.927 009 239 04;
  • 19) 0.927 009 239 04 × 2 = 1 + 0.854 018 478 08;
  • 20) 0.854 018 478 08 × 2 = 1 + 0.708 036 956 16;
  • 21) 0.708 036 956 16 × 2 = 1 + 0.416 073 912 32;
  • 22) 0.416 073 912 32 × 2 = 0 + 0.832 147 824 64;
  • 23) 0.832 147 824 64 × 2 = 1 + 0.664 295 649 28;
  • 24) 0.664 295 649 28 × 2 = 1 + 0.328 591 298 56;
  • 25) 0.328 591 298 56 × 2 = 0 + 0.657 182 597 12;
  • 26) 0.657 182 597 12 × 2 = 1 + 0.314 365 194 24;
  • 27) 0.314 365 194 24 × 2 = 0 + 0.628 730 388 48;
  • 28) 0.628 730 388 48 × 2 = 1 + 0.257 460 776 96;
  • 29) 0.257 460 776 96 × 2 = 0 + 0.514 921 553 92;
  • 30) 0.514 921 553 92 × 2 = 1 + 0.029 843 107 84;
  • 31) 0.029 843 107 84 × 2 = 0 + 0.059 686 215 68;
  • 32) 0.059 686 215 68 × 2 = 0 + 0.119 372 431 36;
  • 33) 0.119 372 431 36 × 2 = 0 + 0.238 744 862 72;
  • 34) 0.238 744 862 72 × 2 = 0 + 0.477 489 725 44;
  • 35) 0.477 489 725 44 × 2 = 0 + 0.954 979 450 88;
  • 36) 0.954 979 450 88 × 2 = 1 + 0.909 958 901 76;
  • 37) 0.909 958 901 76 × 2 = 1 + 0.819 917 803 52;
  • 38) 0.819 917 803 52 × 2 = 1 + 0.639 835 607 04;
  • 39) 0.639 835 607 04 × 2 = 1 + 0.279 671 214 08;
  • 40) 0.279 671 214 08 × 2 = 0 + 0.559 342 428 16;
  • 41) 0.559 342 428 16 × 2 = 1 + 0.118 684 856 32;
  • 42) 0.118 684 856 32 × 2 = 0 + 0.237 369 712 64;
  • 43) 0.237 369 712 64 × 2 = 0 + 0.474 739 425 28;
  • 44) 0.474 739 425 28 × 2 = 0 + 0.949 478 850 56;
  • 45) 0.949 478 850 56 × 2 = 1 + 0.898 957 701 12;
  • 46) 0.898 957 701 12 × 2 = 1 + 0.797 915 402 24;
  • 47) 0.797 915 402 24 × 2 = 1 + 0.595 830 804 48;
  • 48) 0.595 830 804 48 × 2 = 1 + 0.191 661 608 96;
  • 49) 0.191 661 608 96 × 2 = 0 + 0.383 323 217 92;
  • 50) 0.383 323 217 92 × 2 = 0 + 0.766 646 435 84;
  • 51) 0.766 646 435 84 × 2 = 1 + 0.533 292 871 68;
  • 52) 0.533 292 871 68 × 2 = 1 + 0.066 585 743 36;
  • 53) 0.066 585 743 36 × 2 = 0 + 0.133 171 486 72;
  • 54) 0.133 171 486 72 × 2 = 0 + 0.266 342 973 44;
  • 55) 0.266 342 973 44 × 2 = 0 + 0.532 685 946 88;
  • 56) 0.532 685 946 88 × 2 = 1 + 0.065 371 893 76;
  • 57) 0.065 371 893 76 × 2 = 0 + 0.130 743 787 52;
  • 58) 0.130 743 787 52 × 2 = 0 + 0.261 487 575 04;
  • 59) 0.261 487 575 04 × 2 = 0 + 0.522 975 150 08;
  • 60) 0.522 975 150 08 × 2 = 1 + 0.045 950 300 16;
  • 61) 0.045 950 300 16 × 2 = 0 + 0.091 900 600 32;
  • 62) 0.091 900 600 32 × 2 = 0 + 0.183 801 200 64;
  • 63) 0.183 801 200 64 × 2 = 0 + 0.367 602 401 28;
  • 64) 0.367 602 401 28 × 2 = 0 + 0.735 204 802 56;

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:


0.000 282 009 16(10) =

0.0000 0000 0001 0010 0111 1011 0101 0100 0001 1110 1000 1111 0011 0001 0001 0000(2)

6. Positive number before normalization:

0.000 282 009 16(10) =

0.0000 0000 0001 0010 0111 1011 0101 0100 0001 1110 1000 1111 0011 0001 0001 0000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 009 16(10) =

0.0000 0000 0001 0010 0111 1011 0101 0100 0001 1110 1000 1111 0011 0001 0001 0000(2) =

0.0000 0000 0001 0010 0111 1011 0101 0100 0001 1110 1000 1111 0011 0001 0001 0000(2) × 20 =

1.0010 0111 1011 0101 0100 0001 1110 1000 1111 0011 0001 0001 0000(2) × 2-12


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): -12

Mantissa (not normalized):
1.0010 0111 1011 0101 0100 0001 1110 1000 1111 0011 0001 0001 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 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) =

1011(10) =

011 1111 0011(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. 0010 0111 1011 0101 0100 0001 1110 1000 1111 0011 0001 0001 0000 =

0010 0111 1011 0101 0100 0001 1110 1000 1111 0011 0001 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 0011

Mantissa (52 bits) =
0010 0111 1011 0101 0100 0001 1110 1000 1111 0011 0001 0001 0000


Decimal number -0.000 282 009 16 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 0100 0001 1110 1000 1111 0011 0001 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