-0.000 282 003 47 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 003 47(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
-0.000 282 003 47(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:

|-0.000 282 003 47| = 0.000 282 003 47


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: 0.000 282 003 47.

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) 0.000 282 003 47 × 2 = 0 + 0.000 564 006 94;
  • 2) 0.000 564 006 94 × 2 = 0 + 0.001 128 013 88;
  • 3) 0.001 128 013 88 × 2 = 0 + 0.002 256 027 76;
  • 4) 0.002 256 027 76 × 2 = 0 + 0.004 512 055 52;
  • 5) 0.004 512 055 52 × 2 = 0 + 0.009 024 111 04;
  • 6) 0.009 024 111 04 × 2 = 0 + 0.018 048 222 08;
  • 7) 0.018 048 222 08 × 2 = 0 + 0.036 096 444 16;
  • 8) 0.036 096 444 16 × 2 = 0 + 0.072 192 888 32;
  • 9) 0.072 192 888 32 × 2 = 0 + 0.144 385 776 64;
  • 10) 0.144 385 776 64 × 2 = 0 + 0.288 771 553 28;
  • 11) 0.288 771 553 28 × 2 = 0 + 0.577 543 106 56;
  • 12) 0.577 543 106 56 × 2 = 1 + 0.155 086 213 12;
  • 13) 0.155 086 213 12 × 2 = 0 + 0.310 172 426 24;
  • 14) 0.310 172 426 24 × 2 = 0 + 0.620 344 852 48;
  • 15) 0.620 344 852 48 × 2 = 1 + 0.240 689 704 96;
  • 16) 0.240 689 704 96 × 2 = 0 + 0.481 379 409 92;
  • 17) 0.481 379 409 92 × 2 = 0 + 0.962 758 819 84;
  • 18) 0.962 758 819 84 × 2 = 1 + 0.925 517 639 68;
  • 19) 0.925 517 639 68 × 2 = 1 + 0.851 035 279 36;
  • 20) 0.851 035 279 36 × 2 = 1 + 0.702 070 558 72;
  • 21) 0.702 070 558 72 × 2 = 1 + 0.404 141 117 44;
  • 22) 0.404 141 117 44 × 2 = 0 + 0.808 282 234 88;
  • 23) 0.808 282 234 88 × 2 = 1 + 0.616 564 469 76;
  • 24) 0.616 564 469 76 × 2 = 1 + 0.233 128 939 52;
  • 25) 0.233 128 939 52 × 2 = 0 + 0.466 257 879 04;
  • 26) 0.466 257 879 04 × 2 = 0 + 0.932 515 758 08;
  • 27) 0.932 515 758 08 × 2 = 1 + 0.865 031 516 16;
  • 28) 0.865 031 516 16 × 2 = 1 + 0.730 063 032 32;
  • 29) 0.730 063 032 32 × 2 = 1 + 0.460 126 064 64;
  • 30) 0.460 126 064 64 × 2 = 0 + 0.920 252 129 28;
  • 31) 0.920 252 129 28 × 2 = 1 + 0.840 504 258 56;
  • 32) 0.840 504 258 56 × 2 = 1 + 0.681 008 517 12;
  • 33) 0.681 008 517 12 × 2 = 1 + 0.362 017 034 24;
  • 34) 0.362 017 034 24 × 2 = 0 + 0.724 034 068 48;
  • 35) 0.724 034 068 48 × 2 = 1 + 0.448 068 136 96;
  • 36) 0.448 068 136 96 × 2 = 0 + 0.896 136 273 92;
  • 37) 0.896 136 273 92 × 2 = 1 + 0.792 272 547 84;
  • 38) 0.792 272 547 84 × 2 = 1 + 0.584 545 095 68;
  • 39) 0.584 545 095 68 × 2 = 1 + 0.169 090 191 36;
  • 40) 0.169 090 191 36 × 2 = 0 + 0.338 180 382 72;
  • 41) 0.338 180 382 72 × 2 = 0 + 0.676 360 765 44;
  • 42) 0.676 360 765 44 × 2 = 1 + 0.352 721 530 88;
  • 43) 0.352 721 530 88 × 2 = 0 + 0.705 443 061 76;
  • 44) 0.705 443 061 76 × 2 = 1 + 0.410 886 123 52;
  • 45) 0.410 886 123 52 × 2 = 0 + 0.821 772 247 04;
  • 46) 0.821 772 247 04 × 2 = 1 + 0.643 544 494 08;
  • 47) 0.643 544 494 08 × 2 = 1 + 0.287 088 988 16;
  • 48) 0.287 088 988 16 × 2 = 0 + 0.574 177 976 32;
  • 49) 0.574 177 976 32 × 2 = 1 + 0.148 355 952 64;
  • 50) 0.148 355 952 64 × 2 = 0 + 0.296 711 905 28;
  • 51) 0.296 711 905 28 × 2 = 0 + 0.593 423 810 56;
  • 52) 0.593 423 810 56 × 2 = 1 + 0.186 847 621 12;
  • 53) 0.186 847 621 12 × 2 = 0 + 0.373 695 242 24;
  • 54) 0.373 695 242 24 × 2 = 0 + 0.747 390 484 48;
  • 55) 0.747 390 484 48 × 2 = 1 + 0.494 780 968 96;
  • 56) 0.494 780 968 96 × 2 = 0 + 0.989 561 937 92;
  • 57) 0.989 561 937 92 × 2 = 1 + 0.979 123 875 84;
  • 58) 0.979 123 875 84 × 2 = 1 + 0.958 247 751 68;
  • 59) 0.958 247 751 68 × 2 = 1 + 0.916 495 503 36;
  • 60) 0.916 495 503 36 × 2 = 1 + 0.832 991 006 72;
  • 61) 0.832 991 006 72 × 2 = 1 + 0.665 982 013 44;
  • 62) 0.665 982 013 44 × 2 = 1 + 0.331 964 026 88;
  • 63) 0.331 964 026 88 × 2 = 0 + 0.663 928 053 76;
  • 64) 0.663 928 053 76 × 2 = 1 + 0.327 856 107 52;

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:


0.000 282 003 47(10) =

0.0000 0000 0001 0010 0111 1011 0011 1011 1010 1110 0101 0110 1001 0010 1111 1101(2)

6. Positive number before normalization:

0.000 282 003 47(10) =

0.0000 0000 0001 0010 0111 1011 0011 1011 1010 1110 0101 0110 1001 0010 1111 1101(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:


0.000 282 003 47(10) =

0.0000 0000 0001 0010 0111 1011 0011 1011 1010 1110 0101 0110 1001 0010 1111 1101(2) =

0.0000 0000 0001 0010 0111 1011 0011 1011 1010 1110 0101 0110 1001 0010 1111 1101(2) × 20 =

1.0010 0111 1011 0011 1011 1010 1110 0101 0110 1001 0010 1111 1101(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 1011 1010 1110 0101 0110 1001 0010 1111 1101


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 1011 1010 1110 0101 0110 1001 0010 1111 1101 =

0010 0111 1011 0011 1011 1010 1110 0101 0110 1001 0010 1111 1101


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 1011 1010 1110 0101 0110 1001 0010 1111 1101


Decimal number -0.000 282 003 47 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0011 1011 1010 1110 0101 0110 1001 0010 1111 1101

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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