-0.000 282 005 917 47 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 917 47(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
-0.000 282 005 917 47(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:

|-0.000 282 005 917 47| = 0.000 282 005 917 47


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: 0.000 282 005 917 47.

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) 0.000 282 005 917 47 × 2 = 0 + 0.000 564 011 834 94;
  • 2) 0.000 564 011 834 94 × 2 = 0 + 0.001 128 023 669 88;
  • 3) 0.001 128 023 669 88 × 2 = 0 + 0.002 256 047 339 76;
  • 4) 0.002 256 047 339 76 × 2 = 0 + 0.004 512 094 679 52;
  • 5) 0.004 512 094 679 52 × 2 = 0 + 0.009 024 189 359 04;
  • 6) 0.009 024 189 359 04 × 2 = 0 + 0.018 048 378 718 08;
  • 7) 0.018 048 378 718 08 × 2 = 0 + 0.036 096 757 436 16;
  • 8) 0.036 096 757 436 16 × 2 = 0 + 0.072 193 514 872 32;
  • 9) 0.072 193 514 872 32 × 2 = 0 + 0.144 387 029 744 64;
  • 10) 0.144 387 029 744 64 × 2 = 0 + 0.288 774 059 489 28;
  • 11) 0.288 774 059 489 28 × 2 = 0 + 0.577 548 118 978 56;
  • 12) 0.577 548 118 978 56 × 2 = 1 + 0.155 096 237 957 12;
  • 13) 0.155 096 237 957 12 × 2 = 0 + 0.310 192 475 914 24;
  • 14) 0.310 192 475 914 24 × 2 = 0 + 0.620 384 951 828 48;
  • 15) 0.620 384 951 828 48 × 2 = 1 + 0.240 769 903 656 96;
  • 16) 0.240 769 903 656 96 × 2 = 0 + 0.481 539 807 313 92;
  • 17) 0.481 539 807 313 92 × 2 = 0 + 0.963 079 614 627 84;
  • 18) 0.963 079 614 627 84 × 2 = 1 + 0.926 159 229 255 68;
  • 19) 0.926 159 229 255 68 × 2 = 1 + 0.852 318 458 511 36;
  • 20) 0.852 318 458 511 36 × 2 = 1 + 0.704 636 917 022 72;
  • 21) 0.704 636 917 022 72 × 2 = 1 + 0.409 273 834 045 44;
  • 22) 0.409 273 834 045 44 × 2 = 0 + 0.818 547 668 090 88;
  • 23) 0.818 547 668 090 88 × 2 = 1 + 0.637 095 336 181 76;
  • 24) 0.637 095 336 181 76 × 2 = 1 + 0.274 190 672 363 52;
  • 25) 0.274 190 672 363 52 × 2 = 0 + 0.548 381 344 727 04;
  • 26) 0.548 381 344 727 04 × 2 = 1 + 0.096 762 689 454 08;
  • 27) 0.096 762 689 454 08 × 2 = 0 + 0.193 525 378 908 16;
  • 28) 0.193 525 378 908 16 × 2 = 0 + 0.387 050 757 816 32;
  • 29) 0.387 050 757 816 32 × 2 = 0 + 0.774 101 515 632 64;
  • 30) 0.774 101 515 632 64 × 2 = 1 + 0.548 203 031 265 28;
  • 31) 0.548 203 031 265 28 × 2 = 1 + 0.096 406 062 530 56;
  • 32) 0.096 406 062 530 56 × 2 = 0 + 0.192 812 125 061 12;
  • 33) 0.192 812 125 061 12 × 2 = 0 + 0.385 624 250 122 24;
  • 34) 0.385 624 250 122 24 × 2 = 0 + 0.771 248 500 244 48;
  • 35) 0.771 248 500 244 48 × 2 = 1 + 0.542 497 000 488 96;
  • 36) 0.542 497 000 488 96 × 2 = 1 + 0.084 994 000 977 92;
  • 37) 0.084 994 000 977 92 × 2 = 0 + 0.169 988 001 955 84;
  • 38) 0.169 988 001 955 84 × 2 = 0 + 0.339 976 003 911 68;
  • 39) 0.339 976 003 911 68 × 2 = 0 + 0.679 952 007 823 36;
  • 40) 0.679 952 007 823 36 × 2 = 1 + 0.359 904 015 646 72;
  • 41) 0.359 904 015 646 72 × 2 = 0 + 0.719 808 031 293 44;
  • 42) 0.719 808 031 293 44 × 2 = 1 + 0.439 616 062 586 88;
  • 43) 0.439 616 062 586 88 × 2 = 0 + 0.879 232 125 173 76;
  • 44) 0.879 232 125 173 76 × 2 = 1 + 0.758 464 250 347 52;
  • 45) 0.758 464 250 347 52 × 2 = 1 + 0.516 928 500 695 04;
  • 46) 0.516 928 500 695 04 × 2 = 1 + 0.033 857 001 390 08;
  • 47) 0.033 857 001 390 08 × 2 = 0 + 0.067 714 002 780 16;
  • 48) 0.067 714 002 780 16 × 2 = 0 + 0.135 428 005 560 32;
  • 49) 0.135 428 005 560 32 × 2 = 0 + 0.270 856 011 120 64;
  • 50) 0.270 856 011 120 64 × 2 = 0 + 0.541 712 022 241 28;
  • 51) 0.541 712 022 241 28 × 2 = 1 + 0.083 424 044 482 56;
  • 52) 0.083 424 044 482 56 × 2 = 0 + 0.166 848 088 965 12;
  • 53) 0.166 848 088 965 12 × 2 = 0 + 0.333 696 177 930 24;
  • 54) 0.333 696 177 930 24 × 2 = 0 + 0.667 392 355 860 48;
  • 55) 0.667 392 355 860 48 × 2 = 1 + 0.334 784 711 720 96;
  • 56) 0.334 784 711 720 96 × 2 = 0 + 0.669 569 423 441 92;
  • 57) 0.669 569 423 441 92 × 2 = 1 + 0.339 138 846 883 84;
  • 58) 0.339 138 846 883 84 × 2 = 0 + 0.678 277 693 767 68;
  • 59) 0.678 277 693 767 68 × 2 = 1 + 0.356 555 387 535 36;
  • 60) 0.356 555 387 535 36 × 2 = 0 + 0.713 110 775 070 72;
  • 61) 0.713 110 775 070 72 × 2 = 1 + 0.426 221 550 141 44;
  • 62) 0.426 221 550 141 44 × 2 = 0 + 0.852 443 100 282 88;
  • 63) 0.852 443 100 282 88 × 2 = 1 + 0.704 886 200 565 76;
  • 64) 0.704 886 200 565 76 × 2 = 1 + 0.409 772 401 131 52;

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:


0.000 282 005 917 47(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0001 0101 1100 0010 0010 1010 1011(2)

6. Positive number before normalization:

0.000 282 005 917 47(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0001 0101 1100 0010 0010 1010 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:


0.000 282 005 917 47(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0001 0101 1100 0010 0010 1010 1011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0001 0101 1100 0010 0010 1010 1011(2) × 20 =

1.0010 0111 1011 0100 0110 0011 0001 0101 1100 0010 0010 1010 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 0011 0001 0101 1100 0010 0010 1010 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 0011 0001 0101 1100 0010 0010 1010 1011 =

0010 0111 1011 0100 0110 0011 0001 0101 1100 0010 0010 1010 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 0011 0001 0101 1100 0010 0010 1010 1011


Decimal number -0.000 282 005 917 47 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0011 0001 0101 1100 0010 0010 1010 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