-0.000 281 997 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 997 2(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
-0.000 281 997 2(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:

|-0.000 281 997 2| = 0.000 281 997 2


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: 0.000 281 997 2.

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) 0.000 281 997 2 × 2 = 0 + 0.000 563 994 4;
  • 2) 0.000 563 994 4 × 2 = 0 + 0.001 127 988 8;
  • 3) 0.001 127 988 8 × 2 = 0 + 0.002 255 977 6;
  • 4) 0.002 255 977 6 × 2 = 0 + 0.004 511 955 2;
  • 5) 0.004 511 955 2 × 2 = 0 + 0.009 023 910 4;
  • 6) 0.009 023 910 4 × 2 = 0 + 0.018 047 820 8;
  • 7) 0.018 047 820 8 × 2 = 0 + 0.036 095 641 6;
  • 8) 0.036 095 641 6 × 2 = 0 + 0.072 191 283 2;
  • 9) 0.072 191 283 2 × 2 = 0 + 0.144 382 566 4;
  • 10) 0.144 382 566 4 × 2 = 0 + 0.288 765 132 8;
  • 11) 0.288 765 132 8 × 2 = 0 + 0.577 530 265 6;
  • 12) 0.577 530 265 6 × 2 = 1 + 0.155 060 531 2;
  • 13) 0.155 060 531 2 × 2 = 0 + 0.310 121 062 4;
  • 14) 0.310 121 062 4 × 2 = 0 + 0.620 242 124 8;
  • 15) 0.620 242 124 8 × 2 = 1 + 0.240 484 249 6;
  • 16) 0.240 484 249 6 × 2 = 0 + 0.480 968 499 2;
  • 17) 0.480 968 499 2 × 2 = 0 + 0.961 936 998 4;
  • 18) 0.961 936 998 4 × 2 = 1 + 0.923 873 996 8;
  • 19) 0.923 873 996 8 × 2 = 1 + 0.847 747 993 6;
  • 20) 0.847 747 993 6 × 2 = 1 + 0.695 495 987 2;
  • 21) 0.695 495 987 2 × 2 = 1 + 0.390 991 974 4;
  • 22) 0.390 991 974 4 × 2 = 0 + 0.781 983 948 8;
  • 23) 0.781 983 948 8 × 2 = 1 + 0.563 967 897 6;
  • 24) 0.563 967 897 6 × 2 = 1 + 0.127 935 795 2;
  • 25) 0.127 935 795 2 × 2 = 0 + 0.255 871 590 4;
  • 26) 0.255 871 590 4 × 2 = 0 + 0.511 743 180 8;
  • 27) 0.511 743 180 8 × 2 = 1 + 0.023 486 361 6;
  • 28) 0.023 486 361 6 × 2 = 0 + 0.046 972 723 2;
  • 29) 0.046 972 723 2 × 2 = 0 + 0.093 945 446 4;
  • 30) 0.093 945 446 4 × 2 = 0 + 0.187 890 892 8;
  • 31) 0.187 890 892 8 × 2 = 0 + 0.375 781 785 6;
  • 32) 0.375 781 785 6 × 2 = 0 + 0.751 563 571 2;
  • 33) 0.751 563 571 2 × 2 = 1 + 0.503 127 142 4;
  • 34) 0.503 127 142 4 × 2 = 1 + 0.006 254 284 8;
  • 35) 0.006 254 284 8 × 2 = 0 + 0.012 508 569 6;
  • 36) 0.012 508 569 6 × 2 = 0 + 0.025 017 139 2;
  • 37) 0.025 017 139 2 × 2 = 0 + 0.050 034 278 4;
  • 38) 0.050 034 278 4 × 2 = 0 + 0.100 068 556 8;
  • 39) 0.100 068 556 8 × 2 = 0 + 0.200 137 113 6;
  • 40) 0.200 137 113 6 × 2 = 0 + 0.400 274 227 2;
  • 41) 0.400 274 227 2 × 2 = 0 + 0.800 548 454 4;
  • 42) 0.800 548 454 4 × 2 = 1 + 0.601 096 908 8;
  • 43) 0.601 096 908 8 × 2 = 1 + 0.202 193 817 6;
  • 44) 0.202 193 817 6 × 2 = 0 + 0.404 387 635 2;
  • 45) 0.404 387 635 2 × 2 = 0 + 0.808 775 270 4;
  • 46) 0.808 775 270 4 × 2 = 1 + 0.617 550 540 8;
  • 47) 0.617 550 540 8 × 2 = 1 + 0.235 101 081 6;
  • 48) 0.235 101 081 6 × 2 = 0 + 0.470 202 163 2;
  • 49) 0.470 202 163 2 × 2 = 0 + 0.940 404 326 4;
  • 50) 0.940 404 326 4 × 2 = 1 + 0.880 808 652 8;
  • 51) 0.880 808 652 8 × 2 = 1 + 0.761 617 305 6;
  • 52) 0.761 617 305 6 × 2 = 1 + 0.523 234 611 2;
  • 53) 0.523 234 611 2 × 2 = 1 + 0.046 469 222 4;
  • 54) 0.046 469 222 4 × 2 = 0 + 0.092 938 444 8;
  • 55) 0.092 938 444 8 × 2 = 0 + 0.185 876 889 6;
  • 56) 0.185 876 889 6 × 2 = 0 + 0.371 753 779 2;
  • 57) 0.371 753 779 2 × 2 = 0 + 0.743 507 558 4;
  • 58) 0.743 507 558 4 × 2 = 1 + 0.487 015 116 8;
  • 59) 0.487 015 116 8 × 2 = 0 + 0.974 030 233 6;
  • 60) 0.974 030 233 6 × 2 = 1 + 0.948 060 467 2;
  • 61) 0.948 060 467 2 × 2 = 1 + 0.896 120 934 4;
  • 62) 0.896 120 934 4 × 2 = 1 + 0.792 241 868 8;
  • 63) 0.792 241 868 8 × 2 = 1 + 0.584 483 737 6;
  • 64) 0.584 483 737 6 × 2 = 1 + 0.168 967 475 2;

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:


0.000 281 997 2(10) =

0.0000 0000 0001 0010 0111 1011 0010 0000 1100 0000 0110 0110 0111 1000 0101 1111(2)

6. Positive number before normalization:

0.000 281 997 2(10) =

0.0000 0000 0001 0010 0111 1011 0010 0000 1100 0000 0110 0110 0111 1000 0101 1111(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:


0.000 281 997 2(10) =

0.0000 0000 0001 0010 0111 1011 0010 0000 1100 0000 0110 0110 0111 1000 0101 1111(2) =

0.0000 0000 0001 0010 0111 1011 0010 0000 1100 0000 0110 0110 0111 1000 0101 1111(2) × 20 =

1.0010 0111 1011 0010 0000 1100 0000 0110 0110 0111 1000 0101 1111(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 0010 0000 1100 0000 0110 0110 0111 1000 0101 1111


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 0010 0000 1100 0000 0110 0110 0111 1000 0101 1111 =

0010 0111 1011 0010 0000 1100 0000 0110 0110 0111 1000 0101 1111


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 0010 0000 1100 0000 0110 0110 0111 1000 0101 1111


Decimal number -0.000 281 997 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0010 0000 1100 0000 0110 0110 0111 1000 0101 1111

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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