-0.000 282 014 07 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 014 07(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
-0.000 282 014 07(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:

|-0.000 282 014 07| = 0.000 282 014 07


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: 0.000 282 014 07.

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) 0.000 282 014 07 × 2 = 0 + 0.000 564 028 14;
  • 2) 0.000 564 028 14 × 2 = 0 + 0.001 128 056 28;
  • 3) 0.001 128 056 28 × 2 = 0 + 0.002 256 112 56;
  • 4) 0.002 256 112 56 × 2 = 0 + 0.004 512 225 12;
  • 5) 0.004 512 225 12 × 2 = 0 + 0.009 024 450 24;
  • 6) 0.009 024 450 24 × 2 = 0 + 0.018 048 900 48;
  • 7) 0.018 048 900 48 × 2 = 0 + 0.036 097 800 96;
  • 8) 0.036 097 800 96 × 2 = 0 + 0.072 195 601 92;
  • 9) 0.072 195 601 92 × 2 = 0 + 0.144 391 203 84;
  • 10) 0.144 391 203 84 × 2 = 0 + 0.288 782 407 68;
  • 11) 0.288 782 407 68 × 2 = 0 + 0.577 564 815 36;
  • 12) 0.577 564 815 36 × 2 = 1 + 0.155 129 630 72;
  • 13) 0.155 129 630 72 × 2 = 0 + 0.310 259 261 44;
  • 14) 0.310 259 261 44 × 2 = 0 + 0.620 518 522 88;
  • 15) 0.620 518 522 88 × 2 = 1 + 0.241 037 045 76;
  • 16) 0.241 037 045 76 × 2 = 0 + 0.482 074 091 52;
  • 17) 0.482 074 091 52 × 2 = 0 + 0.964 148 183 04;
  • 18) 0.964 148 183 04 × 2 = 1 + 0.928 296 366 08;
  • 19) 0.928 296 366 08 × 2 = 1 + 0.856 592 732 16;
  • 20) 0.856 592 732 16 × 2 = 1 + 0.713 185 464 32;
  • 21) 0.713 185 464 32 × 2 = 1 + 0.426 370 928 64;
  • 22) 0.426 370 928 64 × 2 = 0 + 0.852 741 857 28;
  • 23) 0.852 741 857 28 × 2 = 1 + 0.705 483 714 56;
  • 24) 0.705 483 714 56 × 2 = 1 + 0.410 967 429 12;
  • 25) 0.410 967 429 12 × 2 = 0 + 0.821 934 858 24;
  • 26) 0.821 934 858 24 × 2 = 1 + 0.643 869 716 48;
  • 27) 0.643 869 716 48 × 2 = 1 + 0.287 739 432 96;
  • 28) 0.287 739 432 96 × 2 = 0 + 0.575 478 865 92;
  • 29) 0.575 478 865 92 × 2 = 1 + 0.150 957 731 84;
  • 30) 0.150 957 731 84 × 2 = 0 + 0.301 915 463 68;
  • 31) 0.301 915 463 68 × 2 = 0 + 0.603 830 927 36;
  • 32) 0.603 830 927 36 × 2 = 1 + 0.207 661 854 72;
  • 33) 0.207 661 854 72 × 2 = 0 + 0.415 323 709 44;
  • 34) 0.415 323 709 44 × 2 = 0 + 0.830 647 418 88;
  • 35) 0.830 647 418 88 × 2 = 1 + 0.661 294 837 76;
  • 36) 0.661 294 837 76 × 2 = 1 + 0.322 589 675 52;
  • 37) 0.322 589 675 52 × 2 = 0 + 0.645 179 351 04;
  • 38) 0.645 179 351 04 × 2 = 1 + 0.290 358 702 08;
  • 39) 0.290 358 702 08 × 2 = 0 + 0.580 717 404 16;
  • 40) 0.580 717 404 16 × 2 = 1 + 0.161 434 808 32;
  • 41) 0.161 434 808 32 × 2 = 0 + 0.322 869 616 64;
  • 42) 0.322 869 616 64 × 2 = 0 + 0.645 739 233 28;
  • 43) 0.645 739 233 28 × 2 = 1 + 0.291 478 466 56;
  • 44) 0.291 478 466 56 × 2 = 0 + 0.582 956 933 12;
  • 45) 0.582 956 933 12 × 2 = 1 + 0.165 913 866 24;
  • 46) 0.165 913 866 24 × 2 = 0 + 0.331 827 732 48;
  • 47) 0.331 827 732 48 × 2 = 0 + 0.663 655 464 96;
  • 48) 0.663 655 464 96 × 2 = 1 + 0.327 310 929 92;
  • 49) 0.327 310 929 92 × 2 = 0 + 0.654 621 859 84;
  • 50) 0.654 621 859 84 × 2 = 1 + 0.309 243 719 68;
  • 51) 0.309 243 719 68 × 2 = 0 + 0.618 487 439 36;
  • 52) 0.618 487 439 36 × 2 = 1 + 0.236 974 878 72;
  • 53) 0.236 974 878 72 × 2 = 0 + 0.473 949 757 44;
  • 54) 0.473 949 757 44 × 2 = 0 + 0.947 899 514 88;
  • 55) 0.947 899 514 88 × 2 = 1 + 0.895 799 029 76;
  • 56) 0.895 799 029 76 × 2 = 1 + 0.791 598 059 52;
  • 57) 0.791 598 059 52 × 2 = 1 + 0.583 196 119 04;
  • 58) 0.583 196 119 04 × 2 = 1 + 0.166 392 238 08;
  • 59) 0.166 392 238 08 × 2 = 0 + 0.332 784 476 16;
  • 60) 0.332 784 476 16 × 2 = 0 + 0.665 568 952 32;
  • 61) 0.665 568 952 32 × 2 = 1 + 0.331 137 904 64;
  • 62) 0.331 137 904 64 × 2 = 0 + 0.662 275 809 28;
  • 63) 0.662 275 809 28 × 2 = 1 + 0.324 551 618 56;
  • 64) 0.324 551 618 56 × 2 = 0 + 0.649 103 237 12;

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:


0.000 282 014 07(10) =

0.0000 0000 0001 0010 0111 1011 0110 1001 0011 0101 0010 1001 0101 0011 1100 1010(2)

6. Positive number before normalization:

0.000 282 014 07(10) =

0.0000 0000 0001 0010 0111 1011 0110 1001 0011 0101 0010 1001 0101 0011 1100 1010(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:


0.000 282 014 07(10) =

0.0000 0000 0001 0010 0111 1011 0110 1001 0011 0101 0010 1001 0101 0011 1100 1010(2) =

0.0000 0000 0001 0010 0111 1011 0110 1001 0011 0101 0010 1001 0101 0011 1100 1010(2) × 20 =

1.0010 0111 1011 0110 1001 0011 0101 0010 1001 0101 0011 1100 1010(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 0110 1001 0011 0101 0010 1001 0101 0011 1100 1010


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 0110 1001 0011 0101 0010 1001 0101 0011 1100 1010 =

0010 0111 1011 0110 1001 0011 0101 0010 1001 0101 0011 1100 1010


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 0110 1001 0011 0101 0010 1001 0101 0011 1100 1010


Decimal number -0.000 282 014 07 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0110 1001 0011 0101 0010 1001 0101 0011 1100 1010

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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