-0.000 281 980 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 980 4(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
-0.000 281 980 4(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:

|-0.000 281 980 4| = 0.000 281 980 4


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: 0.000 281 980 4.

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) 0.000 281 980 4 × 2 = 0 + 0.000 563 960 8;
  • 2) 0.000 563 960 8 × 2 = 0 + 0.001 127 921 6;
  • 3) 0.001 127 921 6 × 2 = 0 + 0.002 255 843 2;
  • 4) 0.002 255 843 2 × 2 = 0 + 0.004 511 686 4;
  • 5) 0.004 511 686 4 × 2 = 0 + 0.009 023 372 8;
  • 6) 0.009 023 372 8 × 2 = 0 + 0.018 046 745 6;
  • 7) 0.018 046 745 6 × 2 = 0 + 0.036 093 491 2;
  • 8) 0.036 093 491 2 × 2 = 0 + 0.072 186 982 4;
  • 9) 0.072 186 982 4 × 2 = 0 + 0.144 373 964 8;
  • 10) 0.144 373 964 8 × 2 = 0 + 0.288 747 929 6;
  • 11) 0.288 747 929 6 × 2 = 0 + 0.577 495 859 2;
  • 12) 0.577 495 859 2 × 2 = 1 + 0.154 991 718 4;
  • 13) 0.154 991 718 4 × 2 = 0 + 0.309 983 436 8;
  • 14) 0.309 983 436 8 × 2 = 0 + 0.619 966 873 6;
  • 15) 0.619 966 873 6 × 2 = 1 + 0.239 933 747 2;
  • 16) 0.239 933 747 2 × 2 = 0 + 0.479 867 494 4;
  • 17) 0.479 867 494 4 × 2 = 0 + 0.959 734 988 8;
  • 18) 0.959 734 988 8 × 2 = 1 + 0.919 469 977 6;
  • 19) 0.919 469 977 6 × 2 = 1 + 0.838 939 955 2;
  • 20) 0.838 939 955 2 × 2 = 1 + 0.677 879 910 4;
  • 21) 0.677 879 910 4 × 2 = 1 + 0.355 759 820 8;
  • 22) 0.355 759 820 8 × 2 = 0 + 0.711 519 641 6;
  • 23) 0.711 519 641 6 × 2 = 1 + 0.423 039 283 2;
  • 24) 0.423 039 283 2 × 2 = 0 + 0.846 078 566 4;
  • 25) 0.846 078 566 4 × 2 = 1 + 0.692 157 132 8;
  • 26) 0.692 157 132 8 × 2 = 1 + 0.384 314 265 6;
  • 27) 0.384 314 265 6 × 2 = 0 + 0.768 628 531 2;
  • 28) 0.768 628 531 2 × 2 = 1 + 0.537 257 062 4;
  • 29) 0.537 257 062 4 × 2 = 1 + 0.074 514 124 8;
  • 30) 0.074 514 124 8 × 2 = 0 + 0.149 028 249 6;
  • 31) 0.149 028 249 6 × 2 = 0 + 0.298 056 499 2;
  • 32) 0.298 056 499 2 × 2 = 0 + 0.596 112 998 4;
  • 33) 0.596 112 998 4 × 2 = 1 + 0.192 225 996 8;
  • 34) 0.192 225 996 8 × 2 = 0 + 0.384 451 993 6;
  • 35) 0.384 451 993 6 × 2 = 0 + 0.768 903 987 2;
  • 36) 0.768 903 987 2 × 2 = 1 + 0.537 807 974 4;
  • 37) 0.537 807 974 4 × 2 = 1 + 0.075 615 948 8;
  • 38) 0.075 615 948 8 × 2 = 0 + 0.151 231 897 6;
  • 39) 0.151 231 897 6 × 2 = 0 + 0.302 463 795 2;
  • 40) 0.302 463 795 2 × 2 = 0 + 0.604 927 590 4;
  • 41) 0.604 927 590 4 × 2 = 1 + 0.209 855 180 8;
  • 42) 0.209 855 180 8 × 2 = 0 + 0.419 710 361 6;
  • 43) 0.419 710 361 6 × 2 = 0 + 0.839 420 723 2;
  • 44) 0.839 420 723 2 × 2 = 1 + 0.678 841 446 4;
  • 45) 0.678 841 446 4 × 2 = 1 + 0.357 682 892 8;
  • 46) 0.357 682 892 8 × 2 = 0 + 0.715 365 785 6;
  • 47) 0.715 365 785 6 × 2 = 1 + 0.430 731 571 2;
  • 48) 0.430 731 571 2 × 2 = 0 + 0.861 463 142 4;
  • 49) 0.861 463 142 4 × 2 = 1 + 0.722 926 284 8;
  • 50) 0.722 926 284 8 × 2 = 1 + 0.445 852 569 6;
  • 51) 0.445 852 569 6 × 2 = 0 + 0.891 705 139 2;
  • 52) 0.891 705 139 2 × 2 = 1 + 0.783 410 278 4;
  • 53) 0.783 410 278 4 × 2 = 1 + 0.566 820 556 8;
  • 54) 0.566 820 556 8 × 2 = 1 + 0.133 641 113 6;
  • 55) 0.133 641 113 6 × 2 = 0 + 0.267 282 227 2;
  • 56) 0.267 282 227 2 × 2 = 0 + 0.534 564 454 4;
  • 57) 0.534 564 454 4 × 2 = 1 + 0.069 128 908 8;
  • 58) 0.069 128 908 8 × 2 = 0 + 0.138 257 817 6;
  • 59) 0.138 257 817 6 × 2 = 0 + 0.276 515 635 2;
  • 60) 0.276 515 635 2 × 2 = 0 + 0.553 031 270 4;
  • 61) 0.553 031 270 4 × 2 = 1 + 0.106 062 540 8;
  • 62) 0.106 062 540 8 × 2 = 0 + 0.212 125 081 6;
  • 63) 0.212 125 081 6 × 2 = 0 + 0.424 250 163 2;
  • 64) 0.424 250 163 2 × 2 = 0 + 0.848 500 326 4;

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:


0.000 281 980 4(10) =

0.0000 0000 0001 0010 0111 1010 1101 1000 1001 1000 1001 1010 1101 1100 1000 1000(2)

6. Positive number before normalization:

0.000 281 980 4(10) =

0.0000 0000 0001 0010 0111 1010 1101 1000 1001 1000 1001 1010 1101 1100 1000 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:


0.000 281 980 4(10) =

0.0000 0000 0001 0010 0111 1010 1101 1000 1001 1000 1001 1010 1101 1100 1000 1000(2) =

0.0000 0000 0001 0010 0111 1010 1101 1000 1001 1000 1001 1010 1101 1100 1000 1000(2) × 20 =

1.0010 0111 1010 1101 1000 1001 1000 1001 1010 1101 1100 1000 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 1010 1101 1000 1001 1000 1001 1010 1101 1100 1000 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 1010 1101 1000 1001 1000 1001 1010 1101 1100 1000 1000 =

0010 0111 1010 1101 1000 1001 1000 1001 1010 1101 1100 1000 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 1010 1101 1000 1001 1000 1001 1010 1101 1100 1000 1000


Decimal number -0.000 281 980 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1010 1101 1000 1001 1000 1001 1010 1101 1100 1000 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