-0.000 282 004 62 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 004 62(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
-0.000 282 004 62(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:

|-0.000 282 004 62| = 0.000 282 004 62


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: 0.000 282 004 62.

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) 0.000 282 004 62 × 2 = 0 + 0.000 564 009 24;
  • 2) 0.000 564 009 24 × 2 = 0 + 0.001 128 018 48;
  • 3) 0.001 128 018 48 × 2 = 0 + 0.002 256 036 96;
  • 4) 0.002 256 036 96 × 2 = 0 + 0.004 512 073 92;
  • 5) 0.004 512 073 92 × 2 = 0 + 0.009 024 147 84;
  • 6) 0.009 024 147 84 × 2 = 0 + 0.018 048 295 68;
  • 7) 0.018 048 295 68 × 2 = 0 + 0.036 096 591 36;
  • 8) 0.036 096 591 36 × 2 = 0 + 0.072 193 182 72;
  • 9) 0.072 193 182 72 × 2 = 0 + 0.144 386 365 44;
  • 10) 0.144 386 365 44 × 2 = 0 + 0.288 772 730 88;
  • 11) 0.288 772 730 88 × 2 = 0 + 0.577 545 461 76;
  • 12) 0.577 545 461 76 × 2 = 1 + 0.155 090 923 52;
  • 13) 0.155 090 923 52 × 2 = 0 + 0.310 181 847 04;
  • 14) 0.310 181 847 04 × 2 = 0 + 0.620 363 694 08;
  • 15) 0.620 363 694 08 × 2 = 1 + 0.240 727 388 16;
  • 16) 0.240 727 388 16 × 2 = 0 + 0.481 454 776 32;
  • 17) 0.481 454 776 32 × 2 = 0 + 0.962 909 552 64;
  • 18) 0.962 909 552 64 × 2 = 1 + 0.925 819 105 28;
  • 19) 0.925 819 105 28 × 2 = 1 + 0.851 638 210 56;
  • 20) 0.851 638 210 56 × 2 = 1 + 0.703 276 421 12;
  • 21) 0.703 276 421 12 × 2 = 1 + 0.406 552 842 24;
  • 22) 0.406 552 842 24 × 2 = 0 + 0.813 105 684 48;
  • 23) 0.813 105 684 48 × 2 = 1 + 0.626 211 368 96;
  • 24) 0.626 211 368 96 × 2 = 1 + 0.252 422 737 92;
  • 25) 0.252 422 737 92 × 2 = 0 + 0.504 845 475 84;
  • 26) 0.504 845 475 84 × 2 = 1 + 0.009 690 951 68;
  • 27) 0.009 690 951 68 × 2 = 0 + 0.019 381 903 36;
  • 28) 0.019 381 903 36 × 2 = 0 + 0.038 763 806 72;
  • 29) 0.038 763 806 72 × 2 = 0 + 0.077 527 613 44;
  • 30) 0.077 527 613 44 × 2 = 0 + 0.155 055 226 88;
  • 31) 0.155 055 226 88 × 2 = 0 + 0.310 110 453 76;
  • 32) 0.310 110 453 76 × 2 = 0 + 0.620 220 907 52;
  • 33) 0.620 220 907 52 × 2 = 1 + 0.240 441 815 04;
  • 34) 0.240 441 815 04 × 2 = 0 + 0.480 883 630 08;
  • 35) 0.480 883 630 08 × 2 = 0 + 0.961 767 260 16;
  • 36) 0.961 767 260 16 × 2 = 1 + 0.923 534 520 32;
  • 37) 0.923 534 520 32 × 2 = 1 + 0.847 069 040 64;
  • 38) 0.847 069 040 64 × 2 = 1 + 0.694 138 081 28;
  • 39) 0.694 138 081 28 × 2 = 1 + 0.388 276 162 56;
  • 40) 0.388 276 162 56 × 2 = 0 + 0.776 552 325 12;
  • 41) 0.776 552 325 12 × 2 = 1 + 0.553 104 650 24;
  • 42) 0.553 104 650 24 × 2 = 1 + 0.106 209 300 48;
  • 43) 0.106 209 300 48 × 2 = 0 + 0.212 418 600 96;
  • 44) 0.212 418 600 96 × 2 = 0 + 0.424 837 201 92;
  • 45) 0.424 837 201 92 × 2 = 0 + 0.849 674 403 84;
  • 46) 0.849 674 403 84 × 2 = 1 + 0.699 348 807 68;
  • 47) 0.699 348 807 68 × 2 = 1 + 0.398 697 615 36;
  • 48) 0.398 697 615 36 × 2 = 0 + 0.797 395 230 72;
  • 49) 0.797 395 230 72 × 2 = 1 + 0.594 790 461 44;
  • 50) 0.594 790 461 44 × 2 = 1 + 0.189 580 922 88;
  • 51) 0.189 580 922 88 × 2 = 0 + 0.379 161 845 76;
  • 52) 0.379 161 845 76 × 2 = 0 + 0.758 323 691 52;
  • 53) 0.758 323 691 52 × 2 = 1 + 0.516 647 383 04;
  • 54) 0.516 647 383 04 × 2 = 1 + 0.033 294 766 08;
  • 55) 0.033 294 766 08 × 2 = 0 + 0.066 589 532 16;
  • 56) 0.066 589 532 16 × 2 = 0 + 0.133 179 064 32;
  • 57) 0.133 179 064 32 × 2 = 0 + 0.266 358 128 64;
  • 58) 0.266 358 128 64 × 2 = 0 + 0.532 716 257 28;
  • 59) 0.532 716 257 28 × 2 = 1 + 0.065 432 514 56;
  • 60) 0.065 432 514 56 × 2 = 0 + 0.130 865 029 12;
  • 61) 0.130 865 029 12 × 2 = 0 + 0.261 730 058 24;
  • 62) 0.261 730 058 24 × 2 = 0 + 0.523 460 116 48;
  • 63) 0.523 460 116 48 × 2 = 1 + 0.046 920 232 96;
  • 64) 0.046 920 232 96 × 2 = 0 + 0.093 840 465 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:


0.000 282 004 62(10) =

0.0000 0000 0001 0010 0111 1011 0100 0000 1001 1110 1100 0110 1100 1100 0010 0010(2)

6. Positive number before normalization:

0.000 282 004 62(10) =

0.0000 0000 0001 0010 0111 1011 0100 0000 1001 1110 1100 0110 1100 1100 0010 0010(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:


0.000 282 004 62(10) =

0.0000 0000 0001 0010 0111 1011 0100 0000 1001 1110 1100 0110 1100 1100 0010 0010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0000 1001 1110 1100 0110 1100 1100 0010 0010(2) × 20 =

1.0010 0111 1011 0100 0000 1001 1110 1100 0110 1100 1100 0010 0010(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 0000 1001 1110 1100 0110 1100 1100 0010 0010


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 0000 1001 1110 1100 0110 1100 1100 0010 0010 =

0010 0111 1011 0100 0000 1001 1110 1100 0110 1100 1100 0010 0010


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 0000 1001 1110 1100 0110 1100 1100 0010 0010


Decimal number -0.000 282 004 62 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0000 1001 1110 1100 0110 1100 1100 0010 0010

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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