-0.000 282 011 67 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 011 67(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
-0.000 282 011 67(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:

|-0.000 282 011 67| = 0.000 282 011 67


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: 0.000 282 011 67.

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) 0.000 282 011 67 × 2 = 0 + 0.000 564 023 34;
  • 2) 0.000 564 023 34 × 2 = 0 + 0.001 128 046 68;
  • 3) 0.001 128 046 68 × 2 = 0 + 0.002 256 093 36;
  • 4) 0.002 256 093 36 × 2 = 0 + 0.004 512 186 72;
  • 5) 0.004 512 186 72 × 2 = 0 + 0.009 024 373 44;
  • 6) 0.009 024 373 44 × 2 = 0 + 0.018 048 746 88;
  • 7) 0.018 048 746 88 × 2 = 0 + 0.036 097 493 76;
  • 8) 0.036 097 493 76 × 2 = 0 + 0.072 194 987 52;
  • 9) 0.072 194 987 52 × 2 = 0 + 0.144 389 975 04;
  • 10) 0.144 389 975 04 × 2 = 0 + 0.288 779 950 08;
  • 11) 0.288 779 950 08 × 2 = 0 + 0.577 559 900 16;
  • 12) 0.577 559 900 16 × 2 = 1 + 0.155 119 800 32;
  • 13) 0.155 119 800 32 × 2 = 0 + 0.310 239 600 64;
  • 14) 0.310 239 600 64 × 2 = 0 + 0.620 479 201 28;
  • 15) 0.620 479 201 28 × 2 = 1 + 0.240 958 402 56;
  • 16) 0.240 958 402 56 × 2 = 0 + 0.481 916 805 12;
  • 17) 0.481 916 805 12 × 2 = 0 + 0.963 833 610 24;
  • 18) 0.963 833 610 24 × 2 = 1 + 0.927 667 220 48;
  • 19) 0.927 667 220 48 × 2 = 1 + 0.855 334 440 96;
  • 20) 0.855 334 440 96 × 2 = 1 + 0.710 668 881 92;
  • 21) 0.710 668 881 92 × 2 = 1 + 0.421 337 763 84;
  • 22) 0.421 337 763 84 × 2 = 0 + 0.842 675 527 68;
  • 23) 0.842 675 527 68 × 2 = 1 + 0.685 351 055 36;
  • 24) 0.685 351 055 36 × 2 = 1 + 0.370 702 110 72;
  • 25) 0.370 702 110 72 × 2 = 0 + 0.741 404 221 44;
  • 26) 0.741 404 221 44 × 2 = 1 + 0.482 808 442 88;
  • 27) 0.482 808 442 88 × 2 = 0 + 0.965 616 885 76;
  • 28) 0.965 616 885 76 × 2 = 1 + 0.931 233 771 52;
  • 29) 0.931 233 771 52 × 2 = 1 + 0.862 467 543 04;
  • 30) 0.862 467 543 04 × 2 = 1 + 0.724 935 086 08;
  • 31) 0.724 935 086 08 × 2 = 1 + 0.449 870 172 16;
  • 32) 0.449 870 172 16 × 2 = 0 + 0.899 740 344 32;
  • 33) 0.899 740 344 32 × 2 = 1 + 0.799 480 688 64;
  • 34) 0.799 480 688 64 × 2 = 1 + 0.598 961 377 28;
  • 35) 0.598 961 377 28 × 2 = 1 + 0.197 922 754 56;
  • 36) 0.197 922 754 56 × 2 = 0 + 0.395 845 509 12;
  • 37) 0.395 845 509 12 × 2 = 0 + 0.791 691 018 24;
  • 38) 0.791 691 018 24 × 2 = 1 + 0.583 382 036 48;
  • 39) 0.583 382 036 48 × 2 = 1 + 0.166 764 072 96;
  • 40) 0.166 764 072 96 × 2 = 0 + 0.333 528 145 92;
  • 41) 0.333 528 145 92 × 2 = 0 + 0.667 056 291 84;
  • 42) 0.667 056 291 84 × 2 = 1 + 0.334 112 583 68;
  • 43) 0.334 112 583 68 × 2 = 0 + 0.668 225 167 36;
  • 44) 0.668 225 167 36 × 2 = 1 + 0.336 450 334 72;
  • 45) 0.336 450 334 72 × 2 = 0 + 0.672 900 669 44;
  • 46) 0.672 900 669 44 × 2 = 1 + 0.345 801 338 88;
  • 47) 0.345 801 338 88 × 2 = 0 + 0.691 602 677 76;
  • 48) 0.691 602 677 76 × 2 = 1 + 0.383 205 355 52;
  • 49) 0.383 205 355 52 × 2 = 0 + 0.766 410 711 04;
  • 50) 0.766 410 711 04 × 2 = 1 + 0.532 821 422 08;
  • 51) 0.532 821 422 08 × 2 = 1 + 0.065 642 844 16;
  • 52) 0.065 642 844 16 × 2 = 0 + 0.131 285 688 32;
  • 53) 0.131 285 688 32 × 2 = 0 + 0.262 571 376 64;
  • 54) 0.262 571 376 64 × 2 = 0 + 0.525 142 753 28;
  • 55) 0.525 142 753 28 × 2 = 1 + 0.050 285 506 56;
  • 56) 0.050 285 506 56 × 2 = 0 + 0.100 571 013 12;
  • 57) 0.100 571 013 12 × 2 = 0 + 0.201 142 026 24;
  • 58) 0.201 142 026 24 × 2 = 0 + 0.402 284 052 48;
  • 59) 0.402 284 052 48 × 2 = 0 + 0.804 568 104 96;
  • 60) 0.804 568 104 96 × 2 = 1 + 0.609 136 209 92;
  • 61) 0.609 136 209 92 × 2 = 1 + 0.218 272 419 84;
  • 62) 0.218 272 419 84 × 2 = 0 + 0.436 544 839 68;
  • 63) 0.436 544 839 68 × 2 = 0 + 0.873 089 679 36;
  • 64) 0.873 089 679 36 × 2 = 1 + 0.746 179 358 72;

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:


0.000 282 011 67(10) =

0.0000 0000 0001 0010 0111 1011 0101 1110 1110 0110 0101 0101 0110 0010 0001 1001(2)

6. Positive number before normalization:

0.000 282 011 67(10) =

0.0000 0000 0001 0010 0111 1011 0101 1110 1110 0110 0101 0101 0110 0010 0001 1001(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:


0.000 282 011 67(10) =

0.0000 0000 0001 0010 0111 1011 0101 1110 1110 0110 0101 0101 0110 0010 0001 1001(2) =

0.0000 0000 0001 0010 0111 1011 0101 1110 1110 0110 0101 0101 0110 0010 0001 1001(2) × 20 =

1.0010 0111 1011 0101 1110 1110 0110 0101 0101 0110 0010 0001 1001(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 1110 1110 0110 0101 0101 0110 0010 0001 1001


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 1110 1110 0110 0101 0101 0110 0010 0001 1001 =

0010 0111 1011 0101 1110 1110 0110 0101 0101 0110 0010 0001 1001


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 1110 1110 0110 0101 0101 0110 0010 0001 1001


Decimal number -0.000 282 011 67 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 1110 1110 0110 0101 0101 0110 0010 0001 1001

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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