-0.000 282 008 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 008 3(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
-0.000 282 008 3(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:

|-0.000 282 008 3| = 0.000 282 008 3


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: 0.000 282 008 3.

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) 0.000 282 008 3 × 2 = 0 + 0.000 564 016 6;
  • 2) 0.000 564 016 6 × 2 = 0 + 0.001 128 033 2;
  • 3) 0.001 128 033 2 × 2 = 0 + 0.002 256 066 4;
  • 4) 0.002 256 066 4 × 2 = 0 + 0.004 512 132 8;
  • 5) 0.004 512 132 8 × 2 = 0 + 0.009 024 265 6;
  • 6) 0.009 024 265 6 × 2 = 0 + 0.018 048 531 2;
  • 7) 0.018 048 531 2 × 2 = 0 + 0.036 097 062 4;
  • 8) 0.036 097 062 4 × 2 = 0 + 0.072 194 124 8;
  • 9) 0.072 194 124 8 × 2 = 0 + 0.144 388 249 6;
  • 10) 0.144 388 249 6 × 2 = 0 + 0.288 776 499 2;
  • 11) 0.288 776 499 2 × 2 = 0 + 0.577 552 998 4;
  • 12) 0.577 552 998 4 × 2 = 1 + 0.155 105 996 8;
  • 13) 0.155 105 996 8 × 2 = 0 + 0.310 211 993 6;
  • 14) 0.310 211 993 6 × 2 = 0 + 0.620 423 987 2;
  • 15) 0.620 423 987 2 × 2 = 1 + 0.240 847 974 4;
  • 16) 0.240 847 974 4 × 2 = 0 + 0.481 695 948 8;
  • 17) 0.481 695 948 8 × 2 = 0 + 0.963 391 897 6;
  • 18) 0.963 391 897 6 × 2 = 1 + 0.926 783 795 2;
  • 19) 0.926 783 795 2 × 2 = 1 + 0.853 567 590 4;
  • 20) 0.853 567 590 4 × 2 = 1 + 0.707 135 180 8;
  • 21) 0.707 135 180 8 × 2 = 1 + 0.414 270 361 6;
  • 22) 0.414 270 361 6 × 2 = 0 + 0.828 540 723 2;
  • 23) 0.828 540 723 2 × 2 = 1 + 0.657 081 446 4;
  • 24) 0.657 081 446 4 × 2 = 1 + 0.314 162 892 8;
  • 25) 0.314 162 892 8 × 2 = 0 + 0.628 325 785 6;
  • 26) 0.628 325 785 6 × 2 = 1 + 0.256 651 571 2;
  • 27) 0.256 651 571 2 × 2 = 0 + 0.513 303 142 4;
  • 28) 0.513 303 142 4 × 2 = 1 + 0.026 606 284 8;
  • 29) 0.026 606 284 8 × 2 = 0 + 0.053 212 569 6;
  • 30) 0.053 212 569 6 × 2 = 0 + 0.106 425 139 2;
  • 31) 0.106 425 139 2 × 2 = 0 + 0.212 850 278 4;
  • 32) 0.212 850 278 4 × 2 = 0 + 0.425 700 556 8;
  • 33) 0.425 700 556 8 × 2 = 0 + 0.851 401 113 6;
  • 34) 0.851 401 113 6 × 2 = 1 + 0.702 802 227 2;
  • 35) 0.702 802 227 2 × 2 = 1 + 0.405 604 454 4;
  • 36) 0.405 604 454 4 × 2 = 0 + 0.811 208 908 8;
  • 37) 0.811 208 908 8 × 2 = 1 + 0.622 417 817 6;
  • 38) 0.622 417 817 6 × 2 = 1 + 0.244 835 635 2;
  • 39) 0.244 835 635 2 × 2 = 0 + 0.489 671 270 4;
  • 40) 0.489 671 270 4 × 2 = 0 + 0.979 342 540 8;
  • 41) 0.979 342 540 8 × 2 = 1 + 0.958 685 081 6;
  • 42) 0.958 685 081 6 × 2 = 1 + 0.917 370 163 2;
  • 43) 0.917 370 163 2 × 2 = 1 + 0.834 740 326 4;
  • 44) 0.834 740 326 4 × 2 = 1 + 0.669 480 652 8;
  • 45) 0.669 480 652 8 × 2 = 1 + 0.338 961 305 6;
  • 46) 0.338 961 305 6 × 2 = 0 + 0.677 922 611 2;
  • 47) 0.677 922 611 2 × 2 = 1 + 0.355 845 222 4;
  • 48) 0.355 845 222 4 × 2 = 0 + 0.711 690 444 8;
  • 49) 0.711 690 444 8 × 2 = 1 + 0.423 380 889 6;
  • 50) 0.423 380 889 6 × 2 = 0 + 0.846 761 779 2;
  • 51) 0.846 761 779 2 × 2 = 1 + 0.693 523 558 4;
  • 52) 0.693 523 558 4 × 2 = 1 + 0.387 047 116 8;
  • 53) 0.387 047 116 8 × 2 = 0 + 0.774 094 233 6;
  • 54) 0.774 094 233 6 × 2 = 1 + 0.548 188 467 2;
  • 55) 0.548 188 467 2 × 2 = 1 + 0.096 376 934 4;
  • 56) 0.096 376 934 4 × 2 = 0 + 0.192 753 868 8;
  • 57) 0.192 753 868 8 × 2 = 0 + 0.385 507 737 6;
  • 58) 0.385 507 737 6 × 2 = 0 + 0.771 015 475 2;
  • 59) 0.771 015 475 2 × 2 = 1 + 0.542 030 950 4;
  • 60) 0.542 030 950 4 × 2 = 1 + 0.084 061 900 8;
  • 61) 0.084 061 900 8 × 2 = 0 + 0.168 123 801 6;
  • 62) 0.168 123 801 6 × 2 = 0 + 0.336 247 603 2;
  • 63) 0.336 247 603 2 × 2 = 0 + 0.672 495 206 4;
  • 64) 0.672 495 206 4 × 2 = 1 + 0.344 990 412 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:


0.000 282 008 3(10) =

0.0000 0000 0001 0010 0111 1011 0101 0000 0110 1100 1111 1010 1011 0110 0011 0001(2)

6. Positive number before normalization:

0.000 282 008 3(10) =

0.0000 0000 0001 0010 0111 1011 0101 0000 0110 1100 1111 1010 1011 0110 0011 0001(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:


0.000 282 008 3(10) =

0.0000 0000 0001 0010 0111 1011 0101 0000 0110 1100 1111 1010 1011 0110 0011 0001(2) =

0.0000 0000 0001 0010 0111 1011 0101 0000 0110 1100 1111 1010 1011 0110 0011 0001(2) × 20 =

1.0010 0111 1011 0101 0000 0110 1100 1111 1010 1011 0110 0011 0001(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 0000 0110 1100 1111 1010 1011 0110 0011 0001


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 0000 0110 1100 1111 1010 1011 0110 0011 0001 =

0010 0111 1011 0101 0000 0110 1100 1111 1010 1011 0110 0011 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 0011

Mantissa (52 bits) =
0010 0111 1011 0101 0000 0110 1100 1111 1010 1011 0110 0011 0001


Decimal number -0.000 282 008 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 0000 0110 1100 1111 1010 1011 0110 0011 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