-0.000 282 034 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 034 5(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
-0.000 282 034 5(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:

|-0.000 282 034 5| = 0.000 282 034 5


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: 0.000 282 034 5.

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) 0.000 282 034 5 × 2 = 0 + 0.000 564 069;
  • 2) 0.000 564 069 × 2 = 0 + 0.001 128 138;
  • 3) 0.001 128 138 × 2 = 0 + 0.002 256 276;
  • 4) 0.002 256 276 × 2 = 0 + 0.004 512 552;
  • 5) 0.004 512 552 × 2 = 0 + 0.009 025 104;
  • 6) 0.009 025 104 × 2 = 0 + 0.018 050 208;
  • 7) 0.018 050 208 × 2 = 0 + 0.036 100 416;
  • 8) 0.036 100 416 × 2 = 0 + 0.072 200 832;
  • 9) 0.072 200 832 × 2 = 0 + 0.144 401 664;
  • 10) 0.144 401 664 × 2 = 0 + 0.288 803 328;
  • 11) 0.288 803 328 × 2 = 0 + 0.577 606 656;
  • 12) 0.577 606 656 × 2 = 1 + 0.155 213 312;
  • 13) 0.155 213 312 × 2 = 0 + 0.310 426 624;
  • 14) 0.310 426 624 × 2 = 0 + 0.620 853 248;
  • 15) 0.620 853 248 × 2 = 1 + 0.241 706 496;
  • 16) 0.241 706 496 × 2 = 0 + 0.483 412 992;
  • 17) 0.483 412 992 × 2 = 0 + 0.966 825 984;
  • 18) 0.966 825 984 × 2 = 1 + 0.933 651 968;
  • 19) 0.933 651 968 × 2 = 1 + 0.867 303 936;
  • 20) 0.867 303 936 × 2 = 1 + 0.734 607 872;
  • 21) 0.734 607 872 × 2 = 1 + 0.469 215 744;
  • 22) 0.469 215 744 × 2 = 0 + 0.938 431 488;
  • 23) 0.938 431 488 × 2 = 1 + 0.876 862 976;
  • 24) 0.876 862 976 × 2 = 1 + 0.753 725 952;
  • 25) 0.753 725 952 × 2 = 1 + 0.507 451 904;
  • 26) 0.507 451 904 × 2 = 1 + 0.014 903 808;
  • 27) 0.014 903 808 × 2 = 0 + 0.029 807 616;
  • 28) 0.029 807 616 × 2 = 0 + 0.059 615 232;
  • 29) 0.059 615 232 × 2 = 0 + 0.119 230 464;
  • 30) 0.119 230 464 × 2 = 0 + 0.238 460 928;
  • 31) 0.238 460 928 × 2 = 0 + 0.476 921 856;
  • 32) 0.476 921 856 × 2 = 0 + 0.953 843 712;
  • 33) 0.953 843 712 × 2 = 1 + 0.907 687 424;
  • 34) 0.907 687 424 × 2 = 1 + 0.815 374 848;
  • 35) 0.815 374 848 × 2 = 1 + 0.630 749 696;
  • 36) 0.630 749 696 × 2 = 1 + 0.261 499 392;
  • 37) 0.261 499 392 × 2 = 0 + 0.522 998 784;
  • 38) 0.522 998 784 × 2 = 1 + 0.045 997 568;
  • 39) 0.045 997 568 × 2 = 0 + 0.091 995 136;
  • 40) 0.091 995 136 × 2 = 0 + 0.183 990 272;
  • 41) 0.183 990 272 × 2 = 0 + 0.367 980 544;
  • 42) 0.367 980 544 × 2 = 0 + 0.735 961 088;
  • 43) 0.735 961 088 × 2 = 1 + 0.471 922 176;
  • 44) 0.471 922 176 × 2 = 0 + 0.943 844 352;
  • 45) 0.943 844 352 × 2 = 1 + 0.887 688 704;
  • 46) 0.887 688 704 × 2 = 1 + 0.775 377 408;
  • 47) 0.775 377 408 × 2 = 1 + 0.550 754 816;
  • 48) 0.550 754 816 × 2 = 1 + 0.101 509 632;
  • 49) 0.101 509 632 × 2 = 0 + 0.203 019 264;
  • 50) 0.203 019 264 × 2 = 0 + 0.406 038 528;
  • 51) 0.406 038 528 × 2 = 0 + 0.812 077 056;
  • 52) 0.812 077 056 × 2 = 1 + 0.624 154 112;
  • 53) 0.624 154 112 × 2 = 1 + 0.248 308 224;
  • 54) 0.248 308 224 × 2 = 0 + 0.496 616 448;
  • 55) 0.496 616 448 × 2 = 0 + 0.993 232 896;
  • 56) 0.993 232 896 × 2 = 1 + 0.986 465 792;
  • 57) 0.986 465 792 × 2 = 1 + 0.972 931 584;
  • 58) 0.972 931 584 × 2 = 1 + 0.945 863 168;
  • 59) 0.945 863 168 × 2 = 1 + 0.891 726 336;
  • 60) 0.891 726 336 × 2 = 1 + 0.783 452 672;
  • 61) 0.783 452 672 × 2 = 1 + 0.566 905 344;
  • 62) 0.566 905 344 × 2 = 1 + 0.133 810 688;
  • 63) 0.133 810 688 × 2 = 0 + 0.267 621 376;
  • 64) 0.267 621 376 × 2 = 0 + 0.535 242 752;

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:


0.000 282 034 5(10) =

0.0000 0000 0001 0010 0111 1011 1100 0000 1111 0100 0010 1111 0001 1001 1111 1100(2)

6. Positive number before normalization:

0.000 282 034 5(10) =

0.0000 0000 0001 0010 0111 1011 1100 0000 1111 0100 0010 1111 0001 1001 1111 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:


0.000 282 034 5(10) =

0.0000 0000 0001 0010 0111 1011 1100 0000 1111 0100 0010 1111 0001 1001 1111 1100(2) =

0.0000 0000 0001 0010 0111 1011 1100 0000 1111 0100 0010 1111 0001 1001 1111 1100(2) × 20 =

1.0010 0111 1011 1100 0000 1111 0100 0010 1111 0001 1001 1111 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 1100 0000 1111 0100 0010 1111 0001 1001 1111 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 1100 0000 1111 0100 0010 1111 0001 1001 1111 1100 =

0010 0111 1011 1100 0000 1111 0100 0010 1111 0001 1001 1111 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 1100 0000 1111 0100 0010 1111 0001 1001 1111 1100


Decimal number -0.000 282 034 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 1100 0000 1111 0100 0010 1111 0001 1001 1111 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