-0.000 282 033 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 033 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
-0.000 282 033 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:

|-0.000 282 033 3| = 0.000 282 033 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: 0.000 282 033 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) 0.000 282 033 3 × 2 = 0 + 0.000 564 066 6;
  • 2) 0.000 564 066 6 × 2 = 0 + 0.001 128 133 2;
  • 3) 0.001 128 133 2 × 2 = 0 + 0.002 256 266 4;
  • 4) 0.002 256 266 4 × 2 = 0 + 0.004 512 532 8;
  • 5) 0.004 512 532 8 × 2 = 0 + 0.009 025 065 6;
  • 6) 0.009 025 065 6 × 2 = 0 + 0.018 050 131 2;
  • 7) 0.018 050 131 2 × 2 = 0 + 0.036 100 262 4;
  • 8) 0.036 100 262 4 × 2 = 0 + 0.072 200 524 8;
  • 9) 0.072 200 524 8 × 2 = 0 + 0.144 401 049 6;
  • 10) 0.144 401 049 6 × 2 = 0 + 0.288 802 099 2;
  • 11) 0.288 802 099 2 × 2 = 0 + 0.577 604 198 4;
  • 12) 0.577 604 198 4 × 2 = 1 + 0.155 208 396 8;
  • 13) 0.155 208 396 8 × 2 = 0 + 0.310 416 793 6;
  • 14) 0.310 416 793 6 × 2 = 0 + 0.620 833 587 2;
  • 15) 0.620 833 587 2 × 2 = 1 + 0.241 667 174 4;
  • 16) 0.241 667 174 4 × 2 = 0 + 0.483 334 348 8;
  • 17) 0.483 334 348 8 × 2 = 0 + 0.966 668 697 6;
  • 18) 0.966 668 697 6 × 2 = 1 + 0.933 337 395 2;
  • 19) 0.933 337 395 2 × 2 = 1 + 0.866 674 790 4;
  • 20) 0.866 674 790 4 × 2 = 1 + 0.733 349 580 8;
  • 21) 0.733 349 580 8 × 2 = 1 + 0.466 699 161 6;
  • 22) 0.466 699 161 6 × 2 = 0 + 0.933 398 323 2;
  • 23) 0.933 398 323 2 × 2 = 1 + 0.866 796 646 4;
  • 24) 0.866 796 646 4 × 2 = 1 + 0.733 593 292 8;
  • 25) 0.733 593 292 8 × 2 = 1 + 0.467 186 585 6;
  • 26) 0.467 186 585 6 × 2 = 0 + 0.934 373 171 2;
  • 27) 0.934 373 171 2 × 2 = 1 + 0.868 746 342 4;
  • 28) 0.868 746 342 4 × 2 = 1 + 0.737 492 684 8;
  • 29) 0.737 492 684 8 × 2 = 1 + 0.474 985 369 6;
  • 30) 0.474 985 369 6 × 2 = 0 + 0.949 970 739 2;
  • 31) 0.949 970 739 2 × 2 = 1 + 0.899 941 478 4;
  • 32) 0.899 941 478 4 × 2 = 1 + 0.799 882 956 8;
  • 33) 0.799 882 956 8 × 2 = 1 + 0.599 765 913 6;
  • 34) 0.599 765 913 6 × 2 = 1 + 0.199 531 827 2;
  • 35) 0.199 531 827 2 × 2 = 0 + 0.399 063 654 4;
  • 36) 0.399 063 654 4 × 2 = 0 + 0.798 127 308 8;
  • 37) 0.798 127 308 8 × 2 = 1 + 0.596 254 617 6;
  • 38) 0.596 254 617 6 × 2 = 1 + 0.192 509 235 2;
  • 39) 0.192 509 235 2 × 2 = 0 + 0.385 018 470 4;
  • 40) 0.385 018 470 4 × 2 = 0 + 0.770 036 940 8;
  • 41) 0.770 036 940 8 × 2 = 1 + 0.540 073 881 6;
  • 42) 0.540 073 881 6 × 2 = 1 + 0.080 147 763 2;
  • 43) 0.080 147 763 2 × 2 = 0 + 0.160 295 526 4;
  • 44) 0.160 295 526 4 × 2 = 0 + 0.320 591 052 8;
  • 45) 0.320 591 052 8 × 2 = 0 + 0.641 182 105 6;
  • 46) 0.641 182 105 6 × 2 = 1 + 0.282 364 211 2;
  • 47) 0.282 364 211 2 × 2 = 0 + 0.564 728 422 4;
  • 48) 0.564 728 422 4 × 2 = 1 + 0.129 456 844 8;
  • 49) 0.129 456 844 8 × 2 = 0 + 0.258 913 689 6;
  • 50) 0.258 913 689 6 × 2 = 0 + 0.517 827 379 2;
  • 51) 0.517 827 379 2 × 2 = 1 + 0.035 654 758 4;
  • 52) 0.035 654 758 4 × 2 = 0 + 0.071 309 516 8;
  • 53) 0.071 309 516 8 × 2 = 0 + 0.142 619 033 6;
  • 54) 0.142 619 033 6 × 2 = 0 + 0.285 238 067 2;
  • 55) 0.285 238 067 2 × 2 = 0 + 0.570 476 134 4;
  • 56) 0.570 476 134 4 × 2 = 1 + 0.140 952 268 8;
  • 57) 0.140 952 268 8 × 2 = 0 + 0.281 904 537 6;
  • 58) 0.281 904 537 6 × 2 = 0 + 0.563 809 075 2;
  • 59) 0.563 809 075 2 × 2 = 1 + 0.127 618 150 4;
  • 60) 0.127 618 150 4 × 2 = 0 + 0.255 236 300 8;
  • 61) 0.255 236 300 8 × 2 = 0 + 0.510 472 601 6;
  • 62) 0.510 472 601 6 × 2 = 1 + 0.020 945 203 2;
  • 63) 0.020 945 203 2 × 2 = 0 + 0.041 890 406 4;
  • 64) 0.041 890 406 4 × 2 = 0 + 0.083 780 812 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:


0.000 282 033 3(10) =

0.0000 0000 0001 0010 0111 1011 1011 1011 1100 1100 1100 0101 0010 0001 0010 0100(2)

6. Positive number before normalization:

0.000 282 033 3(10) =

0.0000 0000 0001 0010 0111 1011 1011 1011 1100 1100 1100 0101 0010 0001 0010 0100(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:


0.000 282 033 3(10) =

0.0000 0000 0001 0010 0111 1011 1011 1011 1100 1100 1100 0101 0010 0001 0010 0100(2) =

0.0000 0000 0001 0010 0111 1011 1011 1011 1100 1100 1100 0101 0010 0001 0010 0100(2) × 20 =

1.0010 0111 1011 1011 1011 1100 1100 1100 0101 0010 0001 0010 0100(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 1011 1011 1100 1100 1100 0101 0010 0001 0010 0100


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 1011 1011 1100 1100 1100 0101 0010 0001 0010 0100 =

0010 0111 1011 1011 1011 1100 1100 1100 0101 0010 0001 0010 0100


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 1011 1011 1100 1100 1100 0101 0010 0001 0010 0100


Decimal number -0.000 282 033 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 1011 1011 1100 1100 1100 0101 0010 0001 0010 0100

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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