-0.000 282 023 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 023(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
-0.000 282 023(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:

|-0.000 282 023| = 0.000 282 023


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: 0.000 282 023.

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) 0.000 282 023 × 2 = 0 + 0.000 564 046;
  • 2) 0.000 564 046 × 2 = 0 + 0.001 128 092;
  • 3) 0.001 128 092 × 2 = 0 + 0.002 256 184;
  • 4) 0.002 256 184 × 2 = 0 + 0.004 512 368;
  • 5) 0.004 512 368 × 2 = 0 + 0.009 024 736;
  • 6) 0.009 024 736 × 2 = 0 + 0.018 049 472;
  • 7) 0.018 049 472 × 2 = 0 + 0.036 098 944;
  • 8) 0.036 098 944 × 2 = 0 + 0.072 197 888;
  • 9) 0.072 197 888 × 2 = 0 + 0.144 395 776;
  • 10) 0.144 395 776 × 2 = 0 + 0.288 791 552;
  • 11) 0.288 791 552 × 2 = 0 + 0.577 583 104;
  • 12) 0.577 583 104 × 2 = 1 + 0.155 166 208;
  • 13) 0.155 166 208 × 2 = 0 + 0.310 332 416;
  • 14) 0.310 332 416 × 2 = 0 + 0.620 664 832;
  • 15) 0.620 664 832 × 2 = 1 + 0.241 329 664;
  • 16) 0.241 329 664 × 2 = 0 + 0.482 659 328;
  • 17) 0.482 659 328 × 2 = 0 + 0.965 318 656;
  • 18) 0.965 318 656 × 2 = 1 + 0.930 637 312;
  • 19) 0.930 637 312 × 2 = 1 + 0.861 274 624;
  • 20) 0.861 274 624 × 2 = 1 + 0.722 549 248;
  • 21) 0.722 549 248 × 2 = 1 + 0.445 098 496;
  • 22) 0.445 098 496 × 2 = 0 + 0.890 196 992;
  • 23) 0.890 196 992 × 2 = 1 + 0.780 393 984;
  • 24) 0.780 393 984 × 2 = 1 + 0.560 787 968;
  • 25) 0.560 787 968 × 2 = 1 + 0.121 575 936;
  • 26) 0.121 575 936 × 2 = 0 + 0.243 151 872;
  • 27) 0.243 151 872 × 2 = 0 + 0.486 303 744;
  • 28) 0.486 303 744 × 2 = 0 + 0.972 607 488;
  • 29) 0.972 607 488 × 2 = 1 + 0.945 214 976;
  • 30) 0.945 214 976 × 2 = 1 + 0.890 429 952;
  • 31) 0.890 429 952 × 2 = 1 + 0.780 859 904;
  • 32) 0.780 859 904 × 2 = 1 + 0.561 719 808;
  • 33) 0.561 719 808 × 2 = 1 + 0.123 439 616;
  • 34) 0.123 439 616 × 2 = 0 + 0.246 879 232;
  • 35) 0.246 879 232 × 2 = 0 + 0.493 758 464;
  • 36) 0.493 758 464 × 2 = 0 + 0.987 516 928;
  • 37) 0.987 516 928 × 2 = 1 + 0.975 033 856;
  • 38) 0.975 033 856 × 2 = 1 + 0.950 067 712;
  • 39) 0.950 067 712 × 2 = 1 + 0.900 135 424;
  • 40) 0.900 135 424 × 2 = 1 + 0.800 270 848;
  • 41) 0.800 270 848 × 2 = 1 + 0.600 541 696;
  • 42) 0.600 541 696 × 2 = 1 + 0.201 083 392;
  • 43) 0.201 083 392 × 2 = 0 + 0.402 166 784;
  • 44) 0.402 166 784 × 2 = 0 + 0.804 333 568;
  • 45) 0.804 333 568 × 2 = 1 + 0.608 667 136;
  • 46) 0.608 667 136 × 2 = 1 + 0.217 334 272;
  • 47) 0.217 334 272 × 2 = 0 + 0.434 668 544;
  • 48) 0.434 668 544 × 2 = 0 + 0.869 337 088;
  • 49) 0.869 337 088 × 2 = 1 + 0.738 674 176;
  • 50) 0.738 674 176 × 2 = 1 + 0.477 348 352;
  • 51) 0.477 348 352 × 2 = 0 + 0.954 696 704;
  • 52) 0.954 696 704 × 2 = 1 + 0.909 393 408;
  • 53) 0.909 393 408 × 2 = 1 + 0.818 786 816;
  • 54) 0.818 786 816 × 2 = 1 + 0.637 573 632;
  • 55) 0.637 573 632 × 2 = 1 + 0.275 147 264;
  • 56) 0.275 147 264 × 2 = 0 + 0.550 294 528;
  • 57) 0.550 294 528 × 2 = 1 + 0.100 589 056;
  • 58) 0.100 589 056 × 2 = 0 + 0.201 178 112;
  • 59) 0.201 178 112 × 2 = 0 + 0.402 356 224;
  • 60) 0.402 356 224 × 2 = 0 + 0.804 712 448;
  • 61) 0.804 712 448 × 2 = 1 + 0.609 424 896;
  • 62) 0.609 424 896 × 2 = 1 + 0.218 849 792;
  • 63) 0.218 849 792 × 2 = 0 + 0.437 699 584;
  • 64) 0.437 699 584 × 2 = 0 + 0.875 399 168;

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:


0.000 282 023(10) =

0.0000 0000 0001 0010 0111 1011 1000 1111 1000 1111 1100 1100 1101 1110 1000 1100(2)

6. Positive number before normalization:

0.000 282 023(10) =

0.0000 0000 0001 0010 0111 1011 1000 1111 1000 1111 1100 1100 1101 1110 1000 1100(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:


0.000 282 023(10) =

0.0000 0000 0001 0010 0111 1011 1000 1111 1000 1111 1100 1100 1101 1110 1000 1100(2) =

0.0000 0000 0001 0010 0111 1011 1000 1111 1000 1111 1100 1100 1101 1110 1000 1100(2) × 20 =

1.0010 0111 1011 1000 1111 1000 1111 1100 1100 1101 1110 1000 1100(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 1000 1111 1000 1111 1100 1100 1101 1110 1000 1100


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 1000 1111 1000 1111 1100 1100 1101 1110 1000 1100 =

0010 0111 1011 1000 1111 1000 1111 1100 1100 1101 1110 1000 1100


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 1000 1111 1000 1111 1100 1100 1101 1110 1000 1100


Decimal number -0.000 282 023 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 1000 1111 1000 1111 1100 1100 1101 1110 1000 1100

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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