-0.000 282 005 903 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 903 1(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
-0.000 282 005 903 1(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:

|-0.000 282 005 903 1| = 0.000 282 005 903 1


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: 0.000 282 005 903 1.

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) 0.000 282 005 903 1 × 2 = 0 + 0.000 564 011 806 2;
  • 2) 0.000 564 011 806 2 × 2 = 0 + 0.001 128 023 612 4;
  • 3) 0.001 128 023 612 4 × 2 = 0 + 0.002 256 047 224 8;
  • 4) 0.002 256 047 224 8 × 2 = 0 + 0.004 512 094 449 6;
  • 5) 0.004 512 094 449 6 × 2 = 0 + 0.009 024 188 899 2;
  • 6) 0.009 024 188 899 2 × 2 = 0 + 0.018 048 377 798 4;
  • 7) 0.018 048 377 798 4 × 2 = 0 + 0.036 096 755 596 8;
  • 8) 0.036 096 755 596 8 × 2 = 0 + 0.072 193 511 193 6;
  • 9) 0.072 193 511 193 6 × 2 = 0 + 0.144 387 022 387 2;
  • 10) 0.144 387 022 387 2 × 2 = 0 + 0.288 774 044 774 4;
  • 11) 0.288 774 044 774 4 × 2 = 0 + 0.577 548 089 548 8;
  • 12) 0.577 548 089 548 8 × 2 = 1 + 0.155 096 179 097 6;
  • 13) 0.155 096 179 097 6 × 2 = 0 + 0.310 192 358 195 2;
  • 14) 0.310 192 358 195 2 × 2 = 0 + 0.620 384 716 390 4;
  • 15) 0.620 384 716 390 4 × 2 = 1 + 0.240 769 432 780 8;
  • 16) 0.240 769 432 780 8 × 2 = 0 + 0.481 538 865 561 6;
  • 17) 0.481 538 865 561 6 × 2 = 0 + 0.963 077 731 123 2;
  • 18) 0.963 077 731 123 2 × 2 = 1 + 0.926 155 462 246 4;
  • 19) 0.926 155 462 246 4 × 2 = 1 + 0.852 310 924 492 8;
  • 20) 0.852 310 924 492 8 × 2 = 1 + 0.704 621 848 985 6;
  • 21) 0.704 621 848 985 6 × 2 = 1 + 0.409 243 697 971 2;
  • 22) 0.409 243 697 971 2 × 2 = 0 + 0.818 487 395 942 4;
  • 23) 0.818 487 395 942 4 × 2 = 1 + 0.636 974 791 884 8;
  • 24) 0.636 974 791 884 8 × 2 = 1 + 0.273 949 583 769 6;
  • 25) 0.273 949 583 769 6 × 2 = 0 + 0.547 899 167 539 2;
  • 26) 0.547 899 167 539 2 × 2 = 1 + 0.095 798 335 078 4;
  • 27) 0.095 798 335 078 4 × 2 = 0 + 0.191 596 670 156 8;
  • 28) 0.191 596 670 156 8 × 2 = 0 + 0.383 193 340 313 6;
  • 29) 0.383 193 340 313 6 × 2 = 0 + 0.766 386 680 627 2;
  • 30) 0.766 386 680 627 2 × 2 = 1 + 0.532 773 361 254 4;
  • 31) 0.532 773 361 254 4 × 2 = 1 + 0.065 546 722 508 8;
  • 32) 0.065 546 722 508 8 × 2 = 0 + 0.131 093 445 017 6;
  • 33) 0.131 093 445 017 6 × 2 = 0 + 0.262 186 890 035 2;
  • 34) 0.262 186 890 035 2 × 2 = 0 + 0.524 373 780 070 4;
  • 35) 0.524 373 780 070 4 × 2 = 1 + 0.048 747 560 140 8;
  • 36) 0.048 747 560 140 8 × 2 = 0 + 0.097 495 120 281 6;
  • 37) 0.097 495 120 281 6 × 2 = 0 + 0.194 990 240 563 2;
  • 38) 0.194 990 240 563 2 × 2 = 0 + 0.389 980 481 126 4;
  • 39) 0.389 980 481 126 4 × 2 = 0 + 0.779 960 962 252 8;
  • 40) 0.779 960 962 252 8 × 2 = 1 + 0.559 921 924 505 6;
  • 41) 0.559 921 924 505 6 × 2 = 1 + 0.119 843 849 011 2;
  • 42) 0.119 843 849 011 2 × 2 = 0 + 0.239 687 698 022 4;
  • 43) 0.239 687 698 022 4 × 2 = 0 + 0.479 375 396 044 8;
  • 44) 0.479 375 396 044 8 × 2 = 0 + 0.958 750 792 089 6;
  • 45) 0.958 750 792 089 6 × 2 = 1 + 0.917 501 584 179 2;
  • 46) 0.917 501 584 179 2 × 2 = 1 + 0.835 003 168 358 4;
  • 47) 0.835 003 168 358 4 × 2 = 1 + 0.670 006 336 716 8;
  • 48) 0.670 006 336 716 8 × 2 = 1 + 0.340 012 673 433 6;
  • 49) 0.340 012 673 433 6 × 2 = 0 + 0.680 025 346 867 2;
  • 50) 0.680 025 346 867 2 × 2 = 1 + 0.360 050 693 734 4;
  • 51) 0.360 050 693 734 4 × 2 = 0 + 0.720 101 387 468 8;
  • 52) 0.720 101 387 468 8 × 2 = 1 + 0.440 202 774 937 6;
  • 53) 0.440 202 774 937 6 × 2 = 0 + 0.880 405 549 875 2;
  • 54) 0.880 405 549 875 2 × 2 = 1 + 0.760 811 099 750 4;
  • 55) 0.760 811 099 750 4 × 2 = 1 + 0.521 622 199 500 8;
  • 56) 0.521 622 199 500 8 × 2 = 1 + 0.043 244 399 001 6;
  • 57) 0.043 244 399 001 6 × 2 = 0 + 0.086 488 798 003 2;
  • 58) 0.086 488 798 003 2 × 2 = 0 + 0.172 977 596 006 4;
  • 59) 0.172 977 596 006 4 × 2 = 0 + 0.345 955 192 012 8;
  • 60) 0.345 955 192 012 8 × 2 = 0 + 0.691 910 384 025 6;
  • 61) 0.691 910 384 025 6 × 2 = 1 + 0.383 820 768 051 2;
  • 62) 0.383 820 768 051 2 × 2 = 0 + 0.767 641 536 102 4;
  • 63) 0.767 641 536 102 4 × 2 = 1 + 0.535 283 072 204 8;
  • 64) 0.535 283 072 204 8 × 2 = 1 + 0.070 566 144 409 6;

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:


0.000 282 005 903 1(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0001 1000 1111 0101 0111 0000 1011(2)

6. Positive number before normalization:

0.000 282 005 903 1(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0001 1000 1111 0101 0111 0000 1011(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:


0.000 282 005 903 1(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0001 1000 1111 0101 0111 0000 1011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0001 1000 1111 0101 0111 0000 1011(2) × 20 =

1.0010 0111 1011 0100 0110 0010 0001 1000 1111 0101 0111 0000 1011(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 0010 0001 1000 1111 0101 0111 0000 1011


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 0010 0001 1000 1111 0101 0111 0000 1011 =

0010 0111 1011 0100 0110 0010 0001 1000 1111 0101 0111 0000 1011


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 0010 0001 1000 1111 0101 0111 0000 1011


Decimal number -0.000 282 005 903 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 0001 1000 1111 0101 0111 0000 1011

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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