-0.000 282 005 69 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 69(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 69(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 69| = 0.000 282 005 69


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 69.

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 69 × 2 = 0 + 0.000 564 011 38;
  • 2) 0.000 564 011 38 × 2 = 0 + 0.001 128 022 76;
  • 3) 0.001 128 022 76 × 2 = 0 + 0.002 256 045 52;
  • 4) 0.002 256 045 52 × 2 = 0 + 0.004 512 091 04;
  • 5) 0.004 512 091 04 × 2 = 0 + 0.009 024 182 08;
  • 6) 0.009 024 182 08 × 2 = 0 + 0.018 048 364 16;
  • 7) 0.018 048 364 16 × 2 = 0 + 0.036 096 728 32;
  • 8) 0.036 096 728 32 × 2 = 0 + 0.072 193 456 64;
  • 9) 0.072 193 456 64 × 2 = 0 + 0.144 386 913 28;
  • 10) 0.144 386 913 28 × 2 = 0 + 0.288 773 826 56;
  • 11) 0.288 773 826 56 × 2 = 0 + 0.577 547 653 12;
  • 12) 0.577 547 653 12 × 2 = 1 + 0.155 095 306 24;
  • 13) 0.155 095 306 24 × 2 = 0 + 0.310 190 612 48;
  • 14) 0.310 190 612 48 × 2 = 0 + 0.620 381 224 96;
  • 15) 0.620 381 224 96 × 2 = 1 + 0.240 762 449 92;
  • 16) 0.240 762 449 92 × 2 = 0 + 0.481 524 899 84;
  • 17) 0.481 524 899 84 × 2 = 0 + 0.963 049 799 68;
  • 18) 0.963 049 799 68 × 2 = 1 + 0.926 099 599 36;
  • 19) 0.926 099 599 36 × 2 = 1 + 0.852 199 198 72;
  • 20) 0.852 199 198 72 × 2 = 1 + 0.704 398 397 44;
  • 21) 0.704 398 397 44 × 2 = 1 + 0.408 796 794 88;
  • 22) 0.408 796 794 88 × 2 = 0 + 0.817 593 589 76;
  • 23) 0.817 593 589 76 × 2 = 1 + 0.635 187 179 52;
  • 24) 0.635 187 179 52 × 2 = 1 + 0.270 374 359 04;
  • 25) 0.270 374 359 04 × 2 = 0 + 0.540 748 718 08;
  • 26) 0.540 748 718 08 × 2 = 1 + 0.081 497 436 16;
  • 27) 0.081 497 436 16 × 2 = 0 + 0.162 994 872 32;
  • 28) 0.162 994 872 32 × 2 = 0 + 0.325 989 744 64;
  • 29) 0.325 989 744 64 × 2 = 0 + 0.651 979 489 28;
  • 30) 0.651 979 489 28 × 2 = 1 + 0.303 958 978 56;
  • 31) 0.303 958 978 56 × 2 = 0 + 0.607 917 957 12;
  • 32) 0.607 917 957 12 × 2 = 1 + 0.215 835 914 24;
  • 33) 0.215 835 914 24 × 2 = 0 + 0.431 671 828 48;
  • 34) 0.431 671 828 48 × 2 = 0 + 0.863 343 656 96;
  • 35) 0.863 343 656 96 × 2 = 1 + 0.726 687 313 92;
  • 36) 0.726 687 313 92 × 2 = 1 + 0.453 374 627 84;
  • 37) 0.453 374 627 84 × 2 = 0 + 0.906 749 255 68;
  • 38) 0.906 749 255 68 × 2 = 1 + 0.813 498 511 36;
  • 39) 0.813 498 511 36 × 2 = 1 + 0.626 997 022 72;
  • 40) 0.626 997 022 72 × 2 = 1 + 0.253 994 045 44;
  • 41) 0.253 994 045 44 × 2 = 0 + 0.507 988 090 88;
  • 42) 0.507 988 090 88 × 2 = 1 + 0.015 976 181 76;
  • 43) 0.015 976 181 76 × 2 = 0 + 0.031 952 363 52;
  • 44) 0.031 952 363 52 × 2 = 0 + 0.063 904 727 04;
  • 45) 0.063 904 727 04 × 2 = 0 + 0.127 809 454 08;
  • 46) 0.127 809 454 08 × 2 = 0 + 0.255 618 908 16;
  • 47) 0.255 618 908 16 × 2 = 0 + 0.511 237 816 32;
  • 48) 0.511 237 816 32 × 2 = 1 + 0.022 475 632 64;
  • 49) 0.022 475 632 64 × 2 = 0 + 0.044 951 265 28;
  • 50) 0.044 951 265 28 × 2 = 0 + 0.089 902 530 56;
  • 51) 0.089 902 530 56 × 2 = 0 + 0.179 805 061 12;
  • 52) 0.179 805 061 12 × 2 = 0 + 0.359 610 122 24;
  • 53) 0.359 610 122 24 × 2 = 0 + 0.719 220 244 48;
  • 54) 0.719 220 244 48 × 2 = 1 + 0.438 440 488 96;
  • 55) 0.438 440 488 96 × 2 = 0 + 0.876 880 977 92;
  • 56) 0.876 880 977 92 × 2 = 1 + 0.753 761 955 84;
  • 57) 0.753 761 955 84 × 2 = 1 + 0.507 523 911 68;
  • 58) 0.507 523 911 68 × 2 = 1 + 0.015 047 823 36;
  • 59) 0.015 047 823 36 × 2 = 0 + 0.030 095 646 72;
  • 60) 0.030 095 646 72 × 2 = 0 + 0.060 191 293 44;
  • 61) 0.060 191 293 44 × 2 = 0 + 0.120 382 586 88;
  • 62) 0.120 382 586 88 × 2 = 0 + 0.240 765 173 76;
  • 63) 0.240 765 173 76 × 2 = 0 + 0.481 530 347 52;
  • 64) 0.481 530 347 52 × 2 = 0 + 0.963 060 695 04;

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 69(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0011 0111 0100 0001 0000 0101 1100 0000(2)

6. Positive number before normalization:

0.000 282 005 69(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0011 0111 0100 0001 0000 0101 1100 0000(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 69(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0011 0111 0100 0001 0000 0101 1100 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0011 0111 0100 0001 0000 0101 1100 0000(2) × 20 =

1.0010 0111 1011 0100 0101 0011 0111 0100 0001 0000 0101 1100 0000(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 0101 0011 0111 0100 0001 0000 0101 1100 0000


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 0101 0011 0111 0100 0001 0000 0101 1100 0000 =

0010 0111 1011 0100 0101 0011 0111 0100 0001 0000 0101 1100 0000


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 0101 0011 0111 0100 0001 0000 0101 1100 0000


Decimal number -0.000 282 005 69 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 0011 0111 0100 0001 0000 0101 1100 0000

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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