-0.000 282 012 67 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 012 67(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 67(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 67| = 0.000 282 012 67


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 67.

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 67 × 2 = 0 + 0.000 564 025 34;
  • 2) 0.000 564 025 34 × 2 = 0 + 0.001 128 050 68;
  • 3) 0.001 128 050 68 × 2 = 0 + 0.002 256 101 36;
  • 4) 0.002 256 101 36 × 2 = 0 + 0.004 512 202 72;
  • 5) 0.004 512 202 72 × 2 = 0 + 0.009 024 405 44;
  • 6) 0.009 024 405 44 × 2 = 0 + 0.018 048 810 88;
  • 7) 0.018 048 810 88 × 2 = 0 + 0.036 097 621 76;
  • 8) 0.036 097 621 76 × 2 = 0 + 0.072 195 243 52;
  • 9) 0.072 195 243 52 × 2 = 0 + 0.144 390 487 04;
  • 10) 0.144 390 487 04 × 2 = 0 + 0.288 780 974 08;
  • 11) 0.288 780 974 08 × 2 = 0 + 0.577 561 948 16;
  • 12) 0.577 561 948 16 × 2 = 1 + 0.155 123 896 32;
  • 13) 0.155 123 896 32 × 2 = 0 + 0.310 247 792 64;
  • 14) 0.310 247 792 64 × 2 = 0 + 0.620 495 585 28;
  • 15) 0.620 495 585 28 × 2 = 1 + 0.240 991 170 56;
  • 16) 0.240 991 170 56 × 2 = 0 + 0.481 982 341 12;
  • 17) 0.481 982 341 12 × 2 = 0 + 0.963 964 682 24;
  • 18) 0.963 964 682 24 × 2 = 1 + 0.927 929 364 48;
  • 19) 0.927 929 364 48 × 2 = 1 + 0.855 858 728 96;
  • 20) 0.855 858 728 96 × 2 = 1 + 0.711 717 457 92;
  • 21) 0.711 717 457 92 × 2 = 1 + 0.423 434 915 84;
  • 22) 0.423 434 915 84 × 2 = 0 + 0.846 869 831 68;
  • 23) 0.846 869 831 68 × 2 = 1 + 0.693 739 663 36;
  • 24) 0.693 739 663 36 × 2 = 1 + 0.387 479 326 72;
  • 25) 0.387 479 326 72 × 2 = 0 + 0.774 958 653 44;
  • 26) 0.774 958 653 44 × 2 = 1 + 0.549 917 306 88;
  • 27) 0.549 917 306 88 × 2 = 1 + 0.099 834 613 76;
  • 28) 0.099 834 613 76 × 2 = 0 + 0.199 669 227 52;
  • 29) 0.199 669 227 52 × 2 = 0 + 0.399 338 455 04;
  • 30) 0.399 338 455 04 × 2 = 0 + 0.798 676 910 08;
  • 31) 0.798 676 910 08 × 2 = 1 + 0.597 353 820 16;
  • 32) 0.597 353 820 16 × 2 = 1 + 0.194 707 640 32;
  • 33) 0.194 707 640 32 × 2 = 0 + 0.389 415 280 64;
  • 34) 0.389 415 280 64 × 2 = 0 + 0.778 830 561 28;
  • 35) 0.778 830 561 28 × 2 = 1 + 0.557 661 122 56;
  • 36) 0.557 661 122 56 × 2 = 1 + 0.115 322 245 12;
  • 37) 0.115 322 245 12 × 2 = 0 + 0.230 644 490 24;
  • 38) 0.230 644 490 24 × 2 = 0 + 0.461 288 980 48;
  • 39) 0.461 288 980 48 × 2 = 0 + 0.922 577 960 96;
  • 40) 0.922 577 960 96 × 2 = 1 + 0.845 155 921 92;
  • 41) 0.845 155 921 92 × 2 = 1 + 0.690 311 843 84;
  • 42) 0.690 311 843 84 × 2 = 1 + 0.380 623 687 68;
  • 43) 0.380 623 687 68 × 2 = 0 + 0.761 247 375 36;
  • 44) 0.761 247 375 36 × 2 = 1 + 0.522 494 750 72;
  • 45) 0.522 494 750 72 × 2 = 1 + 0.044 989 501 44;
  • 46) 0.044 989 501 44 × 2 = 0 + 0.089 979 002 88;
  • 47) 0.089 979 002 88 × 2 = 0 + 0.179 958 005 76;
  • 48) 0.179 958 005 76 × 2 = 0 + 0.359 916 011 52;
  • 49) 0.359 916 011 52 × 2 = 0 + 0.719 832 023 04;
  • 50) 0.719 832 023 04 × 2 = 1 + 0.439 664 046 08;
  • 51) 0.439 664 046 08 × 2 = 0 + 0.879 328 092 16;
  • 52) 0.879 328 092 16 × 2 = 1 + 0.758 656 184 32;
  • 53) 0.758 656 184 32 × 2 = 1 + 0.517 312 368 64;
  • 54) 0.517 312 368 64 × 2 = 1 + 0.034 624 737 28;
  • 55) 0.034 624 737 28 × 2 = 0 + 0.069 249 474 56;
  • 56) 0.069 249 474 56 × 2 = 0 + 0.138 498 949 12;
  • 57) 0.138 498 949 12 × 2 = 0 + 0.276 997 898 24;
  • 58) 0.276 997 898 24 × 2 = 0 + 0.553 995 796 48;
  • 59) 0.553 995 796 48 × 2 = 1 + 0.107 991 592 96;
  • 60) 0.107 991 592 96 × 2 = 0 + 0.215 983 185 92;
  • 61) 0.215 983 185 92 × 2 = 0 + 0.431 966 371 84;
  • 62) 0.431 966 371 84 × 2 = 0 + 0.863 932 743 68;
  • 63) 0.863 932 743 68 × 2 = 1 + 0.727 865 487 36;
  • 64) 0.727 865 487 36 × 2 = 1 + 0.455 730 974 72;

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 67(10) =

0.0000 0000 0001 0010 0111 1011 0110 0011 0011 0001 1101 1000 0101 1100 0010 0011(2)

6. Positive number before normalization:

0.000 282 012 67(10) =

0.0000 0000 0001 0010 0111 1011 0110 0011 0011 0001 1101 1000 0101 1100 0010 0011(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 67(10) =

0.0000 0000 0001 0010 0111 1011 0110 0011 0011 0001 1101 1000 0101 1100 0010 0011(2) =

0.0000 0000 0001 0010 0111 1011 0110 0011 0011 0001 1101 1000 0101 1100 0010 0011(2) × 20 =

1.0010 0111 1011 0110 0011 0011 0001 1101 1000 0101 1100 0010 0011(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 0011 0011 0001 1101 1000 0101 1100 0010 0011


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 0011 0011 0001 1101 1000 0101 1100 0010 0011 =

0010 0111 1011 0110 0011 0011 0001 1101 1000 0101 1100 0010 0011


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 0011 0011 0001 1101 1000 0101 1100 0010 0011


Decimal number -0.000 282 012 67 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0110 0011 0011 0001 1101 1000 0101 1100 0010 0011

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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