-0.000 282 003 85 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 003 85(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
-0.000 282 003 85(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:

|-0.000 282 003 85| = 0.000 282 003 85


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: 0.000 282 003 85.

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) 0.000 282 003 85 × 2 = 0 + 0.000 564 007 7;
  • 2) 0.000 564 007 7 × 2 = 0 + 0.001 128 015 4;
  • 3) 0.001 128 015 4 × 2 = 0 + 0.002 256 030 8;
  • 4) 0.002 256 030 8 × 2 = 0 + 0.004 512 061 6;
  • 5) 0.004 512 061 6 × 2 = 0 + 0.009 024 123 2;
  • 6) 0.009 024 123 2 × 2 = 0 + 0.018 048 246 4;
  • 7) 0.018 048 246 4 × 2 = 0 + 0.036 096 492 8;
  • 8) 0.036 096 492 8 × 2 = 0 + 0.072 192 985 6;
  • 9) 0.072 192 985 6 × 2 = 0 + 0.144 385 971 2;
  • 10) 0.144 385 971 2 × 2 = 0 + 0.288 771 942 4;
  • 11) 0.288 771 942 4 × 2 = 0 + 0.577 543 884 8;
  • 12) 0.577 543 884 8 × 2 = 1 + 0.155 087 769 6;
  • 13) 0.155 087 769 6 × 2 = 0 + 0.310 175 539 2;
  • 14) 0.310 175 539 2 × 2 = 0 + 0.620 351 078 4;
  • 15) 0.620 351 078 4 × 2 = 1 + 0.240 702 156 8;
  • 16) 0.240 702 156 8 × 2 = 0 + 0.481 404 313 6;
  • 17) 0.481 404 313 6 × 2 = 0 + 0.962 808 627 2;
  • 18) 0.962 808 627 2 × 2 = 1 + 0.925 617 254 4;
  • 19) 0.925 617 254 4 × 2 = 1 + 0.851 234 508 8;
  • 20) 0.851 234 508 8 × 2 = 1 + 0.702 469 017 6;
  • 21) 0.702 469 017 6 × 2 = 1 + 0.404 938 035 2;
  • 22) 0.404 938 035 2 × 2 = 0 + 0.809 876 070 4;
  • 23) 0.809 876 070 4 × 2 = 1 + 0.619 752 140 8;
  • 24) 0.619 752 140 8 × 2 = 1 + 0.239 504 281 6;
  • 25) 0.239 504 281 6 × 2 = 0 + 0.479 008 563 2;
  • 26) 0.479 008 563 2 × 2 = 0 + 0.958 017 126 4;
  • 27) 0.958 017 126 4 × 2 = 1 + 0.916 034 252 8;
  • 28) 0.916 034 252 8 × 2 = 1 + 0.832 068 505 6;
  • 29) 0.832 068 505 6 × 2 = 1 + 0.664 137 011 2;
  • 30) 0.664 137 011 2 × 2 = 1 + 0.328 274 022 4;
  • 31) 0.328 274 022 4 × 2 = 0 + 0.656 548 044 8;
  • 32) 0.656 548 044 8 × 2 = 1 + 0.313 096 089 6;
  • 33) 0.313 096 089 6 × 2 = 0 + 0.626 192 179 2;
  • 34) 0.626 192 179 2 × 2 = 1 + 0.252 384 358 4;
  • 35) 0.252 384 358 4 × 2 = 0 + 0.504 768 716 8;
  • 36) 0.504 768 716 8 × 2 = 1 + 0.009 537 433 6;
  • 37) 0.009 537 433 6 × 2 = 0 + 0.019 074 867 2;
  • 38) 0.019 074 867 2 × 2 = 0 + 0.038 149 734 4;
  • 39) 0.038 149 734 4 × 2 = 0 + 0.076 299 468 8;
  • 40) 0.076 299 468 8 × 2 = 0 + 0.152 598 937 6;
  • 41) 0.152 598 937 6 × 2 = 0 + 0.305 197 875 2;
  • 42) 0.305 197 875 2 × 2 = 0 + 0.610 395 750 4;
  • 43) 0.610 395 750 4 × 2 = 1 + 0.220 791 500 8;
  • 44) 0.220 791 500 8 × 2 = 0 + 0.441 583 001 6;
  • 45) 0.441 583 001 6 × 2 = 0 + 0.883 166 003 2;
  • 46) 0.883 166 003 2 × 2 = 1 + 0.766 332 006 4;
  • 47) 0.766 332 006 4 × 2 = 1 + 0.532 664 012 8;
  • 48) 0.532 664 012 8 × 2 = 1 + 0.065 328 025 6;
  • 49) 0.065 328 025 6 × 2 = 0 + 0.130 656 051 2;
  • 50) 0.130 656 051 2 × 2 = 0 + 0.261 312 102 4;
  • 51) 0.261 312 102 4 × 2 = 0 + 0.522 624 204 8;
  • 52) 0.522 624 204 8 × 2 = 1 + 0.045 248 409 6;
  • 53) 0.045 248 409 6 × 2 = 0 + 0.090 496 819 2;
  • 54) 0.090 496 819 2 × 2 = 0 + 0.180 993 638 4;
  • 55) 0.180 993 638 4 × 2 = 0 + 0.361 987 276 8;
  • 56) 0.361 987 276 8 × 2 = 0 + 0.723 974 553 6;
  • 57) 0.723 974 553 6 × 2 = 1 + 0.447 949 107 2;
  • 58) 0.447 949 107 2 × 2 = 0 + 0.895 898 214 4;
  • 59) 0.895 898 214 4 × 2 = 1 + 0.791 796 428 8;
  • 60) 0.791 796 428 8 × 2 = 1 + 0.583 592 857 6;
  • 61) 0.583 592 857 6 × 2 = 1 + 0.167 185 715 2;
  • 62) 0.167 185 715 2 × 2 = 0 + 0.334 371 430 4;
  • 63) 0.334 371 430 4 × 2 = 0 + 0.668 742 860 8;
  • 64) 0.668 742 860 8 × 2 = 1 + 0.337 485 721 6;

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:


0.000 282 003 85(10) =

0.0000 0000 0001 0010 0111 1011 0011 1101 0101 0000 0010 0111 0001 0000 1011 1001(2)

6. Positive number before normalization:

0.000 282 003 85(10) =

0.0000 0000 0001 0010 0111 1011 0011 1101 0101 0000 0010 0111 0001 0000 1011 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:


0.000 282 003 85(10) =

0.0000 0000 0001 0010 0111 1011 0011 1101 0101 0000 0010 0111 0001 0000 1011 1001(2) =

0.0000 0000 0001 0010 0111 1011 0011 1101 0101 0000 0010 0111 0001 0000 1011 1001(2) × 20 =

1.0010 0111 1011 0011 1101 0101 0000 0010 0111 0001 0000 1011 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 0011 1101 0101 0000 0010 0111 0001 0000 1011 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 0011 1101 0101 0000 0010 0111 0001 0000 1011 1001 =

0010 0111 1011 0011 1101 0101 0000 0010 0111 0001 0000 1011 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 0011 1101 0101 0000 0010 0111 0001 0000 1011 1001


Decimal number -0.000 282 003 85 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0011 1101 0101 0000 0010 0111 0001 0000 1011 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