-0.000 282 012 31 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 012 31(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
-0.000 282 012 31(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:

|-0.000 282 012 31| = 0.000 282 012 31


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: 0.000 282 012 31.

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) 0.000 282 012 31 × 2 = 0 + 0.000 564 024 62;
  • 2) 0.000 564 024 62 × 2 = 0 + 0.001 128 049 24;
  • 3) 0.001 128 049 24 × 2 = 0 + 0.002 256 098 48;
  • 4) 0.002 256 098 48 × 2 = 0 + 0.004 512 196 96;
  • 5) 0.004 512 196 96 × 2 = 0 + 0.009 024 393 92;
  • 6) 0.009 024 393 92 × 2 = 0 + 0.018 048 787 84;
  • 7) 0.018 048 787 84 × 2 = 0 + 0.036 097 575 68;
  • 8) 0.036 097 575 68 × 2 = 0 + 0.072 195 151 36;
  • 9) 0.072 195 151 36 × 2 = 0 + 0.144 390 302 72;
  • 10) 0.144 390 302 72 × 2 = 0 + 0.288 780 605 44;
  • 11) 0.288 780 605 44 × 2 = 0 + 0.577 561 210 88;
  • 12) 0.577 561 210 88 × 2 = 1 + 0.155 122 421 76;
  • 13) 0.155 122 421 76 × 2 = 0 + 0.310 244 843 52;
  • 14) 0.310 244 843 52 × 2 = 0 + 0.620 489 687 04;
  • 15) 0.620 489 687 04 × 2 = 1 + 0.240 979 374 08;
  • 16) 0.240 979 374 08 × 2 = 0 + 0.481 958 748 16;
  • 17) 0.481 958 748 16 × 2 = 0 + 0.963 917 496 32;
  • 18) 0.963 917 496 32 × 2 = 1 + 0.927 834 992 64;
  • 19) 0.927 834 992 64 × 2 = 1 + 0.855 669 985 28;
  • 20) 0.855 669 985 28 × 2 = 1 + 0.711 339 970 56;
  • 21) 0.711 339 970 56 × 2 = 1 + 0.422 679 941 12;
  • 22) 0.422 679 941 12 × 2 = 0 + 0.845 359 882 24;
  • 23) 0.845 359 882 24 × 2 = 1 + 0.690 719 764 48;
  • 24) 0.690 719 764 48 × 2 = 1 + 0.381 439 528 96;
  • 25) 0.381 439 528 96 × 2 = 0 + 0.762 879 057 92;
  • 26) 0.762 879 057 92 × 2 = 1 + 0.525 758 115 84;
  • 27) 0.525 758 115 84 × 2 = 1 + 0.051 516 231 68;
  • 28) 0.051 516 231 68 × 2 = 0 + 0.103 032 463 36;
  • 29) 0.103 032 463 36 × 2 = 0 + 0.206 064 926 72;
  • 30) 0.206 064 926 72 × 2 = 0 + 0.412 129 853 44;
  • 31) 0.412 129 853 44 × 2 = 0 + 0.824 259 706 88;
  • 32) 0.824 259 706 88 × 2 = 1 + 0.648 519 413 76;
  • 33) 0.648 519 413 76 × 2 = 1 + 0.297 038 827 52;
  • 34) 0.297 038 827 52 × 2 = 0 + 0.594 077 655 04;
  • 35) 0.594 077 655 04 × 2 = 1 + 0.188 155 310 08;
  • 36) 0.188 155 310 08 × 2 = 0 + 0.376 310 620 16;
  • 37) 0.376 310 620 16 × 2 = 0 + 0.752 621 240 32;
  • 38) 0.752 621 240 32 × 2 = 1 + 0.505 242 480 64;
  • 39) 0.505 242 480 64 × 2 = 1 + 0.010 484 961 28;
  • 40) 0.010 484 961 28 × 2 = 0 + 0.020 969 922 56;
  • 41) 0.020 969 922 56 × 2 = 0 + 0.041 939 845 12;
  • 42) 0.041 939 845 12 × 2 = 0 + 0.083 879 690 24;
  • 43) 0.083 879 690 24 × 2 = 0 + 0.167 759 380 48;
  • 44) 0.167 759 380 48 × 2 = 0 + 0.335 518 760 96;
  • 45) 0.335 518 760 96 × 2 = 0 + 0.671 037 521 92;
  • 46) 0.671 037 521 92 × 2 = 1 + 0.342 075 043 84;
  • 47) 0.342 075 043 84 × 2 = 0 + 0.684 150 087 68;
  • 48) 0.684 150 087 68 × 2 = 1 + 0.368 300 175 36;
  • 49) 0.368 300 175 36 × 2 = 0 + 0.736 600 350 72;
  • 50) 0.736 600 350 72 × 2 = 1 + 0.473 200 701 44;
  • 51) 0.473 200 701 44 × 2 = 0 + 0.946 401 402 88;
  • 52) 0.946 401 402 88 × 2 = 1 + 0.892 802 805 76;
  • 53) 0.892 802 805 76 × 2 = 1 + 0.785 605 611 52;
  • 54) 0.785 605 611 52 × 2 = 1 + 0.571 211 223 04;
  • 55) 0.571 211 223 04 × 2 = 1 + 0.142 422 446 08;
  • 56) 0.142 422 446 08 × 2 = 0 + 0.284 844 892 16;
  • 57) 0.284 844 892 16 × 2 = 0 + 0.569 689 784 32;
  • 58) 0.569 689 784 32 × 2 = 1 + 0.139 379 568 64;
  • 59) 0.139 379 568 64 × 2 = 0 + 0.278 759 137 28;
  • 60) 0.278 759 137 28 × 2 = 0 + 0.557 518 274 56;
  • 61) 0.557 518 274 56 × 2 = 1 + 0.115 036 549 12;
  • 62) 0.115 036 549 12 × 2 = 0 + 0.230 073 098 24;
  • 63) 0.230 073 098 24 × 2 = 0 + 0.460 146 196 48;
  • 64) 0.460 146 196 48 × 2 = 0 + 0.920 292 392 96;

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:


0.000 282 012 31(10) =

0.0000 0000 0001 0010 0111 1011 0110 0001 1010 0110 0000 0101 0101 1110 0100 1000(2)

6. Positive number before normalization:

0.000 282 012 31(10) =

0.0000 0000 0001 0010 0111 1011 0110 0001 1010 0110 0000 0101 0101 1110 0100 1000(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:


0.000 282 012 31(10) =

0.0000 0000 0001 0010 0111 1011 0110 0001 1010 0110 0000 0101 0101 1110 0100 1000(2) =

0.0000 0000 0001 0010 0111 1011 0110 0001 1010 0110 0000 0101 0101 1110 0100 1000(2) × 20 =

1.0010 0111 1011 0110 0001 1010 0110 0000 0101 0101 1110 0100 1000(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 0001 1010 0110 0000 0101 0101 1110 0100 1000


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 0001 1010 0110 0000 0101 0101 1110 0100 1000 =

0010 0111 1011 0110 0001 1010 0110 0000 0101 0101 1110 0100 1000


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 0001 1010 0110 0000 0101 0101 1110 0100 1000


Decimal number -0.000 282 012 31 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0110 0001 1010 0110 0000 0101 0101 1110 0100 1000

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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