-0.000 282 004 39 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 004 39(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
-0.000 282 004 39(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:

|-0.000 282 004 39| = 0.000 282 004 39


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: 0.000 282 004 39.

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) 0.000 282 004 39 × 2 = 0 + 0.000 564 008 78;
  • 2) 0.000 564 008 78 × 2 = 0 + 0.001 128 017 56;
  • 3) 0.001 128 017 56 × 2 = 0 + 0.002 256 035 12;
  • 4) 0.002 256 035 12 × 2 = 0 + 0.004 512 070 24;
  • 5) 0.004 512 070 24 × 2 = 0 + 0.009 024 140 48;
  • 6) 0.009 024 140 48 × 2 = 0 + 0.018 048 280 96;
  • 7) 0.018 048 280 96 × 2 = 0 + 0.036 096 561 92;
  • 8) 0.036 096 561 92 × 2 = 0 + 0.072 193 123 84;
  • 9) 0.072 193 123 84 × 2 = 0 + 0.144 386 247 68;
  • 10) 0.144 386 247 68 × 2 = 0 + 0.288 772 495 36;
  • 11) 0.288 772 495 36 × 2 = 0 + 0.577 544 990 72;
  • 12) 0.577 544 990 72 × 2 = 1 + 0.155 089 981 44;
  • 13) 0.155 089 981 44 × 2 = 0 + 0.310 179 962 88;
  • 14) 0.310 179 962 88 × 2 = 0 + 0.620 359 925 76;
  • 15) 0.620 359 925 76 × 2 = 1 + 0.240 719 851 52;
  • 16) 0.240 719 851 52 × 2 = 0 + 0.481 439 703 04;
  • 17) 0.481 439 703 04 × 2 = 0 + 0.962 879 406 08;
  • 18) 0.962 879 406 08 × 2 = 1 + 0.925 758 812 16;
  • 19) 0.925 758 812 16 × 2 = 1 + 0.851 517 624 32;
  • 20) 0.851 517 624 32 × 2 = 1 + 0.703 035 248 64;
  • 21) 0.703 035 248 64 × 2 = 1 + 0.406 070 497 28;
  • 22) 0.406 070 497 28 × 2 = 0 + 0.812 140 994 56;
  • 23) 0.812 140 994 56 × 2 = 1 + 0.624 281 989 12;
  • 24) 0.624 281 989 12 × 2 = 1 + 0.248 563 978 24;
  • 25) 0.248 563 978 24 × 2 = 0 + 0.497 127 956 48;
  • 26) 0.497 127 956 48 × 2 = 0 + 0.994 255 912 96;
  • 27) 0.994 255 912 96 × 2 = 1 + 0.988 511 825 92;
  • 28) 0.988 511 825 92 × 2 = 1 + 0.977 023 651 84;
  • 29) 0.977 023 651 84 × 2 = 1 + 0.954 047 303 68;
  • 30) 0.954 047 303 68 × 2 = 1 + 0.908 094 607 36;
  • 31) 0.908 094 607 36 × 2 = 1 + 0.816 189 214 72;
  • 32) 0.816 189 214 72 × 2 = 1 + 0.632 378 429 44;
  • 33) 0.632 378 429 44 × 2 = 1 + 0.264 756 858 88;
  • 34) 0.264 756 858 88 × 2 = 0 + 0.529 513 717 76;
  • 35) 0.529 513 717 76 × 2 = 1 + 0.059 027 435 52;
  • 36) 0.059 027 435 52 × 2 = 0 + 0.118 054 871 04;
  • 37) 0.118 054 871 04 × 2 = 0 + 0.236 109 742 08;
  • 38) 0.236 109 742 08 × 2 = 0 + 0.472 219 484 16;
  • 39) 0.472 219 484 16 × 2 = 0 + 0.944 438 968 32;
  • 40) 0.944 438 968 32 × 2 = 1 + 0.888 877 936 64;
  • 41) 0.888 877 936 64 × 2 = 1 + 0.777 755 873 28;
  • 42) 0.777 755 873 28 × 2 = 1 + 0.555 511 746 56;
  • 43) 0.555 511 746 56 × 2 = 1 + 0.111 023 493 12;
  • 44) 0.111 023 493 12 × 2 = 0 + 0.222 046 986 24;
  • 45) 0.222 046 986 24 × 2 = 0 + 0.444 093 972 48;
  • 46) 0.444 093 972 48 × 2 = 0 + 0.888 187 944 96;
  • 47) 0.888 187 944 96 × 2 = 1 + 0.776 375 889 92;
  • 48) 0.776 375 889 92 × 2 = 1 + 0.552 751 779 84;
  • 49) 0.552 751 779 84 × 2 = 1 + 0.105 503 559 68;
  • 50) 0.105 503 559 68 × 2 = 0 + 0.211 007 119 36;
  • 51) 0.211 007 119 36 × 2 = 0 + 0.422 014 238 72;
  • 52) 0.422 014 238 72 × 2 = 0 + 0.844 028 477 44;
  • 53) 0.844 028 477 44 × 2 = 1 + 0.688 056 954 88;
  • 54) 0.688 056 954 88 × 2 = 1 + 0.376 113 909 76;
  • 55) 0.376 113 909 76 × 2 = 0 + 0.752 227 819 52;
  • 56) 0.752 227 819 52 × 2 = 1 + 0.504 455 639 04;
  • 57) 0.504 455 639 04 × 2 = 1 + 0.008 911 278 08;
  • 58) 0.008 911 278 08 × 2 = 0 + 0.017 822 556 16;
  • 59) 0.017 822 556 16 × 2 = 0 + 0.035 645 112 32;
  • 60) 0.035 645 112 32 × 2 = 0 + 0.071 290 224 64;
  • 61) 0.071 290 224 64 × 2 = 0 + 0.142 580 449 28;
  • 62) 0.142 580 449 28 × 2 = 0 + 0.285 160 898 56;
  • 63) 0.285 160 898 56 × 2 = 0 + 0.570 321 797 12;
  • 64) 0.570 321 797 12 × 2 = 1 + 0.140 643 594 24;

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:


0.000 282 004 39(10) =

0.0000 0000 0001 0010 0111 1011 0011 1111 1010 0001 1110 0011 1000 1101 1000 0001(2)

6. Positive number before normalization:

0.000 282 004 39(10) =

0.0000 0000 0001 0010 0111 1011 0011 1111 1010 0001 1110 0011 1000 1101 1000 0001(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:


0.000 282 004 39(10) =

0.0000 0000 0001 0010 0111 1011 0011 1111 1010 0001 1110 0011 1000 1101 1000 0001(2) =

0.0000 0000 0001 0010 0111 1011 0011 1111 1010 0001 1110 0011 1000 1101 1000 0001(2) × 20 =

1.0010 0111 1011 0011 1111 1010 0001 1110 0011 1000 1101 1000 0001(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 0011 1111 1010 0001 1110 0011 1000 1101 1000 0001


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 0011 1111 1010 0001 1110 0011 1000 1101 1000 0001 =

0010 0111 1011 0011 1111 1010 0001 1110 0011 1000 1101 1000 0001


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 0011 1111 1010 0001 1110 0011 1000 1101 1000 0001


Decimal number -0.000 282 004 39 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0011 1111 1010 0001 1110 0011 1000 1101 1000 0001

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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