-0.000 282 005 913 38 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 913 38(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
-0.000 282 005 913 38(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:

|-0.000 282 005 913 38| = 0.000 282 005 913 38


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: 0.000 282 005 913 38.

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) 0.000 282 005 913 38 × 2 = 0 + 0.000 564 011 826 76;
  • 2) 0.000 564 011 826 76 × 2 = 0 + 0.001 128 023 653 52;
  • 3) 0.001 128 023 653 52 × 2 = 0 + 0.002 256 047 307 04;
  • 4) 0.002 256 047 307 04 × 2 = 0 + 0.004 512 094 614 08;
  • 5) 0.004 512 094 614 08 × 2 = 0 + 0.009 024 189 228 16;
  • 6) 0.009 024 189 228 16 × 2 = 0 + 0.018 048 378 456 32;
  • 7) 0.018 048 378 456 32 × 2 = 0 + 0.036 096 756 912 64;
  • 8) 0.036 096 756 912 64 × 2 = 0 + 0.072 193 513 825 28;
  • 9) 0.072 193 513 825 28 × 2 = 0 + 0.144 387 027 650 56;
  • 10) 0.144 387 027 650 56 × 2 = 0 + 0.288 774 055 301 12;
  • 11) 0.288 774 055 301 12 × 2 = 0 + 0.577 548 110 602 24;
  • 12) 0.577 548 110 602 24 × 2 = 1 + 0.155 096 221 204 48;
  • 13) 0.155 096 221 204 48 × 2 = 0 + 0.310 192 442 408 96;
  • 14) 0.310 192 442 408 96 × 2 = 0 + 0.620 384 884 817 92;
  • 15) 0.620 384 884 817 92 × 2 = 1 + 0.240 769 769 635 84;
  • 16) 0.240 769 769 635 84 × 2 = 0 + 0.481 539 539 271 68;
  • 17) 0.481 539 539 271 68 × 2 = 0 + 0.963 079 078 543 36;
  • 18) 0.963 079 078 543 36 × 2 = 1 + 0.926 158 157 086 72;
  • 19) 0.926 158 157 086 72 × 2 = 1 + 0.852 316 314 173 44;
  • 20) 0.852 316 314 173 44 × 2 = 1 + 0.704 632 628 346 88;
  • 21) 0.704 632 628 346 88 × 2 = 1 + 0.409 265 256 693 76;
  • 22) 0.409 265 256 693 76 × 2 = 0 + 0.818 530 513 387 52;
  • 23) 0.818 530 513 387 52 × 2 = 1 + 0.637 061 026 775 04;
  • 24) 0.637 061 026 775 04 × 2 = 1 + 0.274 122 053 550 08;
  • 25) 0.274 122 053 550 08 × 2 = 0 + 0.548 244 107 100 16;
  • 26) 0.548 244 107 100 16 × 2 = 1 + 0.096 488 214 200 32;
  • 27) 0.096 488 214 200 32 × 2 = 0 + 0.192 976 428 400 64;
  • 28) 0.192 976 428 400 64 × 2 = 0 + 0.385 952 856 801 28;
  • 29) 0.385 952 856 801 28 × 2 = 0 + 0.771 905 713 602 56;
  • 30) 0.771 905 713 602 56 × 2 = 1 + 0.543 811 427 205 12;
  • 31) 0.543 811 427 205 12 × 2 = 1 + 0.087 622 854 410 24;
  • 32) 0.087 622 854 410 24 × 2 = 0 + 0.175 245 708 820 48;
  • 33) 0.175 245 708 820 48 × 2 = 0 + 0.350 491 417 640 96;
  • 34) 0.350 491 417 640 96 × 2 = 0 + 0.700 982 835 281 92;
  • 35) 0.700 982 835 281 92 × 2 = 1 + 0.401 965 670 563 84;
  • 36) 0.401 965 670 563 84 × 2 = 0 + 0.803 931 341 127 68;
  • 37) 0.803 931 341 127 68 × 2 = 1 + 0.607 862 682 255 36;
  • 38) 0.607 862 682 255 36 × 2 = 1 + 0.215 725 364 510 72;
  • 39) 0.215 725 364 510 72 × 2 = 0 + 0.431 450 729 021 44;
  • 40) 0.431 450 729 021 44 × 2 = 0 + 0.862 901 458 042 88;
  • 41) 0.862 901 458 042 88 × 2 = 1 + 0.725 802 916 085 76;
  • 42) 0.725 802 916 085 76 × 2 = 1 + 0.451 605 832 171 52;
  • 43) 0.451 605 832 171 52 × 2 = 0 + 0.903 211 664 343 04;
  • 44) 0.903 211 664 343 04 × 2 = 1 + 0.806 423 328 686 08;
  • 45) 0.806 423 328 686 08 × 2 = 1 + 0.612 846 657 372 16;
  • 46) 0.612 846 657 372 16 × 2 = 1 + 0.225 693 314 744 32;
  • 47) 0.225 693 314 744 32 × 2 = 0 + 0.451 386 629 488 64;
  • 48) 0.451 386 629 488 64 × 2 = 0 + 0.902 773 258 977 28;
  • 49) 0.902 773 258 977 28 × 2 = 1 + 0.805 546 517 954 56;
  • 50) 0.805 546 517 954 56 × 2 = 1 + 0.611 093 035 909 12;
  • 51) 0.611 093 035 909 12 × 2 = 1 + 0.222 186 071 818 24;
  • 52) 0.222 186 071 818 24 × 2 = 0 + 0.444 372 143 636 48;
  • 53) 0.444 372 143 636 48 × 2 = 0 + 0.888 744 287 272 96;
  • 54) 0.888 744 287 272 96 × 2 = 1 + 0.777 488 574 545 92;
  • 55) 0.777 488 574 545 92 × 2 = 1 + 0.554 977 149 091 84;
  • 56) 0.554 977 149 091 84 × 2 = 1 + 0.109 954 298 183 68;
  • 57) 0.109 954 298 183 68 × 2 = 0 + 0.219 908 596 367 36;
  • 58) 0.219 908 596 367 36 × 2 = 0 + 0.439 817 192 734 72;
  • 59) 0.439 817 192 734 72 × 2 = 0 + 0.879 634 385 469 44;
  • 60) 0.879 634 385 469 44 × 2 = 1 + 0.759 268 770 938 88;
  • 61) 0.759 268 770 938 88 × 2 = 1 + 0.518 537 541 877 76;
  • 62) 0.518 537 541 877 76 × 2 = 1 + 0.037 075 083 755 52;
  • 63) 0.037 075 083 755 52 × 2 = 0 + 0.074 150 167 511 04;
  • 64) 0.074 150 167 511 04 × 2 = 0 + 0.148 300 335 022 08;

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:


0.000 282 005 913 38(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1101 1100 1110 0111 0001 1100(2)

6. Positive number before normalization:

0.000 282 005 913 38(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1101 1100 1110 0111 0001 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:


0.000 282 005 913 38(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1101 1100 1110 0111 0001 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1101 1100 1110 0111 0001 1100(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1100 1101 1100 1110 0111 0001 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 0100 0110 0010 1100 1101 1100 1110 0111 0001 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 0100 0110 0010 1100 1101 1100 1110 0111 0001 1100 =

0010 0111 1011 0100 0110 0010 1100 1101 1100 1110 0111 0001 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 0100 0110 0010 1100 1101 1100 1110 0111 0001 1100


Decimal number -0.000 282 005 913 38 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1100 1101 1100 1110 0111 0001 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