-0.000 282 005 04 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 04(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 04(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 04| = 0.000 282 005 04


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 04.

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 04 × 2 = 0 + 0.000 564 010 08;
  • 2) 0.000 564 010 08 × 2 = 0 + 0.001 128 020 16;
  • 3) 0.001 128 020 16 × 2 = 0 + 0.002 256 040 32;
  • 4) 0.002 256 040 32 × 2 = 0 + 0.004 512 080 64;
  • 5) 0.004 512 080 64 × 2 = 0 + 0.009 024 161 28;
  • 6) 0.009 024 161 28 × 2 = 0 + 0.018 048 322 56;
  • 7) 0.018 048 322 56 × 2 = 0 + 0.036 096 645 12;
  • 8) 0.036 096 645 12 × 2 = 0 + 0.072 193 290 24;
  • 9) 0.072 193 290 24 × 2 = 0 + 0.144 386 580 48;
  • 10) 0.144 386 580 48 × 2 = 0 + 0.288 773 160 96;
  • 11) 0.288 773 160 96 × 2 = 0 + 0.577 546 321 92;
  • 12) 0.577 546 321 92 × 2 = 1 + 0.155 092 643 84;
  • 13) 0.155 092 643 84 × 2 = 0 + 0.310 185 287 68;
  • 14) 0.310 185 287 68 × 2 = 0 + 0.620 370 575 36;
  • 15) 0.620 370 575 36 × 2 = 1 + 0.240 741 150 72;
  • 16) 0.240 741 150 72 × 2 = 0 + 0.481 482 301 44;
  • 17) 0.481 482 301 44 × 2 = 0 + 0.962 964 602 88;
  • 18) 0.962 964 602 88 × 2 = 1 + 0.925 929 205 76;
  • 19) 0.925 929 205 76 × 2 = 1 + 0.851 858 411 52;
  • 20) 0.851 858 411 52 × 2 = 1 + 0.703 716 823 04;
  • 21) 0.703 716 823 04 × 2 = 1 + 0.407 433 646 08;
  • 22) 0.407 433 646 08 × 2 = 0 + 0.814 867 292 16;
  • 23) 0.814 867 292 16 × 2 = 1 + 0.629 734 584 32;
  • 24) 0.629 734 584 32 × 2 = 1 + 0.259 469 168 64;
  • 25) 0.259 469 168 64 × 2 = 0 + 0.518 938 337 28;
  • 26) 0.518 938 337 28 × 2 = 1 + 0.037 876 674 56;
  • 27) 0.037 876 674 56 × 2 = 0 + 0.075 753 349 12;
  • 28) 0.075 753 349 12 × 2 = 0 + 0.151 506 698 24;
  • 29) 0.151 506 698 24 × 2 = 0 + 0.303 013 396 48;
  • 30) 0.303 013 396 48 × 2 = 0 + 0.606 026 792 96;
  • 31) 0.606 026 792 96 × 2 = 1 + 0.212 053 585 92;
  • 32) 0.212 053 585 92 × 2 = 0 + 0.424 107 171 84;
  • 33) 0.424 107 171 84 × 2 = 0 + 0.848 214 343 68;
  • 34) 0.848 214 343 68 × 2 = 1 + 0.696 428 687 36;
  • 35) 0.696 428 687 36 × 2 = 1 + 0.392 857 374 72;
  • 36) 0.392 857 374 72 × 2 = 0 + 0.785 714 749 44;
  • 37) 0.785 714 749 44 × 2 = 1 + 0.571 429 498 88;
  • 38) 0.571 429 498 88 × 2 = 1 + 0.142 858 997 76;
  • 39) 0.142 858 997 76 × 2 = 0 + 0.285 717 995 52;
  • 40) 0.285 717 995 52 × 2 = 0 + 0.571 435 991 04;
  • 41) 0.571 435 991 04 × 2 = 1 + 0.142 871 982 08;
  • 42) 0.142 871 982 08 × 2 = 0 + 0.285 743 964 16;
  • 43) 0.285 743 964 16 × 2 = 0 + 0.571 487 928 32;
  • 44) 0.571 487 928 32 × 2 = 1 + 0.142 975 856 64;
  • 45) 0.142 975 856 64 × 2 = 0 + 0.285 951 713 28;
  • 46) 0.285 951 713 28 × 2 = 0 + 0.571 903 426 56;
  • 47) 0.571 903 426 56 × 2 = 1 + 0.143 806 853 12;
  • 48) 0.143 806 853 12 × 2 = 0 + 0.287 613 706 24;
  • 49) 0.287 613 706 24 × 2 = 0 + 0.575 227 412 48;
  • 50) 0.575 227 412 48 × 2 = 1 + 0.150 454 824 96;
  • 51) 0.150 454 824 96 × 2 = 0 + 0.300 909 649 92;
  • 52) 0.300 909 649 92 × 2 = 0 + 0.601 819 299 84;
  • 53) 0.601 819 299 84 × 2 = 1 + 0.203 638 599 68;
  • 54) 0.203 638 599 68 × 2 = 0 + 0.407 277 199 36;
  • 55) 0.407 277 199 36 × 2 = 0 + 0.814 554 398 72;
  • 56) 0.814 554 398 72 × 2 = 1 + 0.629 108 797 44;
  • 57) 0.629 108 797 44 × 2 = 1 + 0.258 217 594 88;
  • 58) 0.258 217 594 88 × 2 = 0 + 0.516 435 189 76;
  • 59) 0.516 435 189 76 × 2 = 1 + 0.032 870 379 52;
  • 60) 0.032 870 379 52 × 2 = 0 + 0.065 740 759 04;
  • 61) 0.065 740 759 04 × 2 = 0 + 0.131 481 518 08;
  • 62) 0.131 481 518 08 × 2 = 0 + 0.262 963 036 16;
  • 63) 0.262 963 036 16 × 2 = 0 + 0.525 926 072 32;
  • 64) 0.525 926 072 32 × 2 = 1 + 0.051 852 144 64;

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 04(10) =

0.0000 0000 0001 0010 0111 1011 0100 0010 0110 1100 1001 0010 0100 1001 1010 0001(2)

6. Positive number before normalization:

0.000 282 005 04(10) =

0.0000 0000 0001 0010 0111 1011 0100 0010 0110 1100 1001 0010 0100 1001 1010 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:


0.000 282 005 04(10) =

0.0000 0000 0001 0010 0111 1011 0100 0010 0110 1100 1001 0010 0100 1001 1010 0001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0010 0110 1100 1001 0010 0100 1001 1010 0001(2) × 20 =

1.0010 0111 1011 0100 0010 0110 1100 1001 0010 0100 1001 1010 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 0100 0010 0110 1100 1001 0010 0100 1001 1010 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 0100 0010 0110 1100 1001 0010 0100 1001 1010 0001 =

0010 0111 1011 0100 0010 0110 1100 1001 0010 0100 1001 1010 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 0100 0010 0110 1100 1001 0010 0100 1001 1010 0001


Decimal number -0.000 282 005 04 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0010 0110 1100 1001 0010 0100 1001 1010 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