-0.000 282 005 37 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 37(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
-0.000 282 005 37(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:

|-0.000 282 005 37| = 0.000 282 005 37


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: 0.000 282 005 37.

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) 0.000 282 005 37 × 2 = 0 + 0.000 564 010 74;
  • 2) 0.000 564 010 74 × 2 = 0 + 0.001 128 021 48;
  • 3) 0.001 128 021 48 × 2 = 0 + 0.002 256 042 96;
  • 4) 0.002 256 042 96 × 2 = 0 + 0.004 512 085 92;
  • 5) 0.004 512 085 92 × 2 = 0 + 0.009 024 171 84;
  • 6) 0.009 024 171 84 × 2 = 0 + 0.018 048 343 68;
  • 7) 0.018 048 343 68 × 2 = 0 + 0.036 096 687 36;
  • 8) 0.036 096 687 36 × 2 = 0 + 0.072 193 374 72;
  • 9) 0.072 193 374 72 × 2 = 0 + 0.144 386 749 44;
  • 10) 0.144 386 749 44 × 2 = 0 + 0.288 773 498 88;
  • 11) 0.288 773 498 88 × 2 = 0 + 0.577 546 997 76;
  • 12) 0.577 546 997 76 × 2 = 1 + 0.155 093 995 52;
  • 13) 0.155 093 995 52 × 2 = 0 + 0.310 187 991 04;
  • 14) 0.310 187 991 04 × 2 = 0 + 0.620 375 982 08;
  • 15) 0.620 375 982 08 × 2 = 1 + 0.240 751 964 16;
  • 16) 0.240 751 964 16 × 2 = 0 + 0.481 503 928 32;
  • 17) 0.481 503 928 32 × 2 = 0 + 0.963 007 856 64;
  • 18) 0.963 007 856 64 × 2 = 1 + 0.926 015 713 28;
  • 19) 0.926 015 713 28 × 2 = 1 + 0.852 031 426 56;
  • 20) 0.852 031 426 56 × 2 = 1 + 0.704 062 853 12;
  • 21) 0.704 062 853 12 × 2 = 1 + 0.408 125 706 24;
  • 22) 0.408 125 706 24 × 2 = 0 + 0.816 251 412 48;
  • 23) 0.816 251 412 48 × 2 = 1 + 0.632 502 824 96;
  • 24) 0.632 502 824 96 × 2 = 1 + 0.265 005 649 92;
  • 25) 0.265 005 649 92 × 2 = 0 + 0.530 011 299 84;
  • 26) 0.530 011 299 84 × 2 = 1 + 0.060 022 599 68;
  • 27) 0.060 022 599 68 × 2 = 0 + 0.120 045 199 36;
  • 28) 0.120 045 199 36 × 2 = 0 + 0.240 090 398 72;
  • 29) 0.240 090 398 72 × 2 = 0 + 0.480 180 797 44;
  • 30) 0.480 180 797 44 × 2 = 0 + 0.960 361 594 88;
  • 31) 0.960 361 594 88 × 2 = 1 + 0.920 723 189 76;
  • 32) 0.920 723 189 76 × 2 = 1 + 0.841 446 379 52;
  • 33) 0.841 446 379 52 × 2 = 1 + 0.682 892 759 04;
  • 34) 0.682 892 759 04 × 2 = 1 + 0.365 785 518 08;
  • 35) 0.365 785 518 08 × 2 = 0 + 0.731 571 036 16;
  • 36) 0.731 571 036 16 × 2 = 1 + 0.463 142 072 32;
  • 37) 0.463 142 072 32 × 2 = 0 + 0.926 284 144 64;
  • 38) 0.926 284 144 64 × 2 = 1 + 0.852 568 289 28;
  • 39) 0.852 568 289 28 × 2 = 1 + 0.705 136 578 56;
  • 40) 0.705 136 578 56 × 2 = 1 + 0.410 273 157 12;
  • 41) 0.410 273 157 12 × 2 = 0 + 0.820 546 314 24;
  • 42) 0.820 546 314 24 × 2 = 1 + 0.641 092 628 48;
  • 43) 0.641 092 628 48 × 2 = 1 + 0.282 185 256 96;
  • 44) 0.282 185 256 96 × 2 = 0 + 0.564 370 513 92;
  • 45) 0.564 370 513 92 × 2 = 1 + 0.128 741 027 84;
  • 46) 0.128 741 027 84 × 2 = 0 + 0.257 482 055 68;
  • 47) 0.257 482 055 68 × 2 = 0 + 0.514 964 111 36;
  • 48) 0.514 964 111 36 × 2 = 1 + 0.029 928 222 72;
  • 49) 0.029 928 222 72 × 2 = 0 + 0.059 856 445 44;
  • 50) 0.059 856 445 44 × 2 = 0 + 0.119 712 890 88;
  • 51) 0.119 712 890 88 × 2 = 0 + 0.239 425 781 76;
  • 52) 0.239 425 781 76 × 2 = 0 + 0.478 851 563 52;
  • 53) 0.478 851 563 52 × 2 = 0 + 0.957 703 127 04;
  • 54) 0.957 703 127 04 × 2 = 1 + 0.915 406 254 08;
  • 55) 0.915 406 254 08 × 2 = 1 + 0.830 812 508 16;
  • 56) 0.830 812 508 16 × 2 = 1 + 0.661 625 016 32;
  • 57) 0.661 625 016 32 × 2 = 1 + 0.323 250 032 64;
  • 58) 0.323 250 032 64 × 2 = 0 + 0.646 500 065 28;
  • 59) 0.646 500 065 28 × 2 = 1 + 0.293 000 130 56;
  • 60) 0.293 000 130 56 × 2 = 0 + 0.586 000 261 12;
  • 61) 0.586 000 261 12 × 2 = 1 + 0.172 000 522 24;
  • 62) 0.172 000 522 24 × 2 = 0 + 0.344 001 044 48;
  • 63) 0.344 001 044 48 × 2 = 0 + 0.688 002 088 96;
  • 64) 0.688 002 088 96 × 2 = 1 + 0.376 004 177 92;

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:


0.000 282 005 37(10) =

0.0000 0000 0001 0010 0111 1011 0100 0011 1101 0111 0110 1001 0000 0111 1010 1001(2)

6. Positive number before normalization:

0.000 282 005 37(10) =

0.0000 0000 0001 0010 0111 1011 0100 0011 1101 0111 0110 1001 0000 0111 1010 1001(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:


0.000 282 005 37(10) =

0.0000 0000 0001 0010 0111 1011 0100 0011 1101 0111 0110 1001 0000 0111 1010 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0011 1101 0111 0110 1001 0000 0111 1010 1001(2) × 20 =

1.0010 0111 1011 0100 0011 1101 0111 0110 1001 0000 0111 1010 1001(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 0100 0011 1101 0111 0110 1001 0000 0111 1010 1001


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 0100 0011 1101 0111 0110 1001 0000 0111 1010 1001 =

0010 0111 1011 0100 0011 1101 0111 0110 1001 0000 0111 1010 1001


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 0100 0011 1101 0111 0110 1001 0000 0111 1010 1001


Decimal number -0.000 282 005 37 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0011 1101 0111 0110 1001 0000 0111 1010 1001

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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