-0.000 282 005 930 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 930 3(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
-0.000 282 005 930 3(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:

|-0.000 282 005 930 3| = 0.000 282 005 930 3


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: 0.000 282 005 930 3.

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) 0.000 282 005 930 3 × 2 = 0 + 0.000 564 011 860 6;
  • 2) 0.000 564 011 860 6 × 2 = 0 + 0.001 128 023 721 2;
  • 3) 0.001 128 023 721 2 × 2 = 0 + 0.002 256 047 442 4;
  • 4) 0.002 256 047 442 4 × 2 = 0 + 0.004 512 094 884 8;
  • 5) 0.004 512 094 884 8 × 2 = 0 + 0.009 024 189 769 6;
  • 6) 0.009 024 189 769 6 × 2 = 0 + 0.018 048 379 539 2;
  • 7) 0.018 048 379 539 2 × 2 = 0 + 0.036 096 759 078 4;
  • 8) 0.036 096 759 078 4 × 2 = 0 + 0.072 193 518 156 8;
  • 9) 0.072 193 518 156 8 × 2 = 0 + 0.144 387 036 313 6;
  • 10) 0.144 387 036 313 6 × 2 = 0 + 0.288 774 072 627 2;
  • 11) 0.288 774 072 627 2 × 2 = 0 + 0.577 548 145 254 4;
  • 12) 0.577 548 145 254 4 × 2 = 1 + 0.155 096 290 508 8;
  • 13) 0.155 096 290 508 8 × 2 = 0 + 0.310 192 581 017 6;
  • 14) 0.310 192 581 017 6 × 2 = 0 + 0.620 385 162 035 2;
  • 15) 0.620 385 162 035 2 × 2 = 1 + 0.240 770 324 070 4;
  • 16) 0.240 770 324 070 4 × 2 = 0 + 0.481 540 648 140 8;
  • 17) 0.481 540 648 140 8 × 2 = 0 + 0.963 081 296 281 6;
  • 18) 0.963 081 296 281 6 × 2 = 1 + 0.926 162 592 563 2;
  • 19) 0.926 162 592 563 2 × 2 = 1 + 0.852 325 185 126 4;
  • 20) 0.852 325 185 126 4 × 2 = 1 + 0.704 650 370 252 8;
  • 21) 0.704 650 370 252 8 × 2 = 1 + 0.409 300 740 505 6;
  • 22) 0.409 300 740 505 6 × 2 = 0 + 0.818 601 481 011 2;
  • 23) 0.818 601 481 011 2 × 2 = 1 + 0.637 202 962 022 4;
  • 24) 0.637 202 962 022 4 × 2 = 1 + 0.274 405 924 044 8;
  • 25) 0.274 405 924 044 8 × 2 = 0 + 0.548 811 848 089 6;
  • 26) 0.548 811 848 089 6 × 2 = 1 + 0.097 623 696 179 2;
  • 27) 0.097 623 696 179 2 × 2 = 0 + 0.195 247 392 358 4;
  • 28) 0.195 247 392 358 4 × 2 = 0 + 0.390 494 784 716 8;
  • 29) 0.390 494 784 716 8 × 2 = 0 + 0.780 989 569 433 6;
  • 30) 0.780 989 569 433 6 × 2 = 1 + 0.561 979 138 867 2;
  • 31) 0.561 979 138 867 2 × 2 = 1 + 0.123 958 277 734 4;
  • 32) 0.123 958 277 734 4 × 2 = 0 + 0.247 916 555 468 8;
  • 33) 0.247 916 555 468 8 × 2 = 0 + 0.495 833 110 937 6;
  • 34) 0.495 833 110 937 6 × 2 = 0 + 0.991 666 221 875 2;
  • 35) 0.991 666 221 875 2 × 2 = 1 + 0.983 332 443 750 4;
  • 36) 0.983 332 443 750 4 × 2 = 1 + 0.966 664 887 500 8;
  • 37) 0.966 664 887 500 8 × 2 = 1 + 0.933 329 775 001 6;
  • 38) 0.933 329 775 001 6 × 2 = 1 + 0.866 659 550 003 2;
  • 39) 0.866 659 550 003 2 × 2 = 1 + 0.733 319 100 006 4;
  • 40) 0.733 319 100 006 4 × 2 = 1 + 0.466 638 200 012 8;
  • 41) 0.466 638 200 012 8 × 2 = 0 + 0.933 276 400 025 6;
  • 42) 0.933 276 400 025 6 × 2 = 1 + 0.866 552 800 051 2;
  • 43) 0.866 552 800 051 2 × 2 = 1 + 0.733 105 600 102 4;
  • 44) 0.733 105 600 102 4 × 2 = 1 + 0.466 211 200 204 8;
  • 45) 0.466 211 200 204 8 × 2 = 0 + 0.932 422 400 409 6;
  • 46) 0.932 422 400 409 6 × 2 = 1 + 0.864 844 800 819 2;
  • 47) 0.864 844 800 819 2 × 2 = 1 + 0.729 689 601 638 4;
  • 48) 0.729 689 601 638 4 × 2 = 1 + 0.459 379 203 276 8;
  • 49) 0.459 379 203 276 8 × 2 = 0 + 0.918 758 406 553 6;
  • 50) 0.918 758 406 553 6 × 2 = 1 + 0.837 516 813 107 2;
  • 51) 0.837 516 813 107 2 × 2 = 1 + 0.675 033 626 214 4;
  • 52) 0.675 033 626 214 4 × 2 = 1 + 0.350 067 252 428 8;
  • 53) 0.350 067 252 428 8 × 2 = 0 + 0.700 134 504 857 6;
  • 54) 0.700 134 504 857 6 × 2 = 1 + 0.400 269 009 715 2;
  • 55) 0.400 269 009 715 2 × 2 = 0 + 0.800 538 019 430 4;
  • 56) 0.800 538 019 430 4 × 2 = 1 + 0.601 076 038 860 8;
  • 57) 0.601 076 038 860 8 × 2 = 1 + 0.202 152 077 721 6;
  • 58) 0.202 152 077 721 6 × 2 = 0 + 0.404 304 155 443 2;
  • 59) 0.404 304 155 443 2 × 2 = 0 + 0.808 608 310 886 4;
  • 60) 0.808 608 310 886 4 × 2 = 1 + 0.617 216 621 772 8;
  • 61) 0.617 216 621 772 8 × 2 = 1 + 0.234 433 243 545 6;
  • 62) 0.234 433 243 545 6 × 2 = 0 + 0.468 866 487 091 2;
  • 63) 0.468 866 487 091 2 × 2 = 0 + 0.937 732 974 182 4;
  • 64) 0.937 732 974 182 4 × 2 = 1 + 0.875 465 948 364 8;

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:


0.000 282 005 930 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 1111 0111 0111 0111 0101 1001 1001(2)

6. Positive number before normalization:

0.000 282 005 930 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 1111 0111 0111 0111 0101 1001 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:


0.000 282 005 930 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 1111 0111 0111 0111 0101 1001 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 1111 0111 0111 0111 0101 1001 1001(2) × 20 =

1.0010 0111 1011 0100 0110 0011 1111 0111 0111 0111 0101 1001 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 0110 0011 1111 0111 0111 0111 0101 1001 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 0110 0011 1111 0111 0111 0111 0101 1001 1001 =

0010 0111 1011 0100 0110 0011 1111 0111 0111 0111 0101 1001 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 0110 0011 1111 0111 0111 0111 0101 1001 1001


Decimal number -0.000 282 005 930 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0011 1111 0111 0111 0111 0101 1001 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