-0.000 282 005 890 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 890 9(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
-0.000 282 005 890 9(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:

|-0.000 282 005 890 9| = 0.000 282 005 890 9


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: 0.000 282 005 890 9.

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) 0.000 282 005 890 9 × 2 = 0 + 0.000 564 011 781 8;
  • 2) 0.000 564 011 781 8 × 2 = 0 + 0.001 128 023 563 6;
  • 3) 0.001 128 023 563 6 × 2 = 0 + 0.002 256 047 127 2;
  • 4) 0.002 256 047 127 2 × 2 = 0 + 0.004 512 094 254 4;
  • 5) 0.004 512 094 254 4 × 2 = 0 + 0.009 024 188 508 8;
  • 6) 0.009 024 188 508 8 × 2 = 0 + 0.018 048 377 017 6;
  • 7) 0.018 048 377 017 6 × 2 = 0 + 0.036 096 754 035 2;
  • 8) 0.036 096 754 035 2 × 2 = 0 + 0.072 193 508 070 4;
  • 9) 0.072 193 508 070 4 × 2 = 0 + 0.144 387 016 140 8;
  • 10) 0.144 387 016 140 8 × 2 = 0 + 0.288 774 032 281 6;
  • 11) 0.288 774 032 281 6 × 2 = 0 + 0.577 548 064 563 2;
  • 12) 0.577 548 064 563 2 × 2 = 1 + 0.155 096 129 126 4;
  • 13) 0.155 096 129 126 4 × 2 = 0 + 0.310 192 258 252 8;
  • 14) 0.310 192 258 252 8 × 2 = 0 + 0.620 384 516 505 6;
  • 15) 0.620 384 516 505 6 × 2 = 1 + 0.240 769 033 011 2;
  • 16) 0.240 769 033 011 2 × 2 = 0 + 0.481 538 066 022 4;
  • 17) 0.481 538 066 022 4 × 2 = 0 + 0.963 076 132 044 8;
  • 18) 0.963 076 132 044 8 × 2 = 1 + 0.926 152 264 089 6;
  • 19) 0.926 152 264 089 6 × 2 = 1 + 0.852 304 528 179 2;
  • 20) 0.852 304 528 179 2 × 2 = 1 + 0.704 609 056 358 4;
  • 21) 0.704 609 056 358 4 × 2 = 1 + 0.409 218 112 716 8;
  • 22) 0.409 218 112 716 8 × 2 = 0 + 0.818 436 225 433 6;
  • 23) 0.818 436 225 433 6 × 2 = 1 + 0.636 872 450 867 2;
  • 24) 0.636 872 450 867 2 × 2 = 1 + 0.273 744 901 734 4;
  • 25) 0.273 744 901 734 4 × 2 = 0 + 0.547 489 803 468 8;
  • 26) 0.547 489 803 468 8 × 2 = 1 + 0.094 979 606 937 6;
  • 27) 0.094 979 606 937 6 × 2 = 0 + 0.189 959 213 875 2;
  • 28) 0.189 959 213 875 2 × 2 = 0 + 0.379 918 427 750 4;
  • 29) 0.379 918 427 750 4 × 2 = 0 + 0.759 836 855 500 8;
  • 30) 0.759 836 855 500 8 × 2 = 1 + 0.519 673 711 001 6;
  • 31) 0.519 673 711 001 6 × 2 = 1 + 0.039 347 422 003 2;
  • 32) 0.039 347 422 003 2 × 2 = 0 + 0.078 694 844 006 4;
  • 33) 0.078 694 844 006 4 × 2 = 0 + 0.157 389 688 012 8;
  • 34) 0.157 389 688 012 8 × 2 = 0 + 0.314 779 376 025 6;
  • 35) 0.314 779 376 025 6 × 2 = 0 + 0.629 558 752 051 2;
  • 36) 0.629 558 752 051 2 × 2 = 1 + 0.259 117 504 102 4;
  • 37) 0.259 117 504 102 4 × 2 = 0 + 0.518 235 008 204 8;
  • 38) 0.518 235 008 204 8 × 2 = 1 + 0.036 470 016 409 6;
  • 39) 0.036 470 016 409 6 × 2 = 0 + 0.072 940 032 819 2;
  • 40) 0.072 940 032 819 2 × 2 = 0 + 0.145 880 065 638 4;
  • 41) 0.145 880 065 638 4 × 2 = 0 + 0.291 760 131 276 8;
  • 42) 0.291 760 131 276 8 × 2 = 0 + 0.583 520 262 553 6;
  • 43) 0.583 520 262 553 6 × 2 = 1 + 0.167 040 525 107 2;
  • 44) 0.167 040 525 107 2 × 2 = 0 + 0.334 081 050 214 4;
  • 45) 0.334 081 050 214 4 × 2 = 0 + 0.668 162 100 428 8;
  • 46) 0.668 162 100 428 8 × 2 = 1 + 0.336 324 200 857 6;
  • 47) 0.336 324 200 857 6 × 2 = 0 + 0.672 648 401 715 2;
  • 48) 0.672 648 401 715 2 × 2 = 1 + 0.345 296 803 430 4;
  • 49) 0.345 296 803 430 4 × 2 = 0 + 0.690 593 606 860 8;
  • 50) 0.690 593 606 860 8 × 2 = 1 + 0.381 187 213 721 6;
  • 51) 0.381 187 213 721 6 × 2 = 0 + 0.762 374 427 443 2;
  • 52) 0.762 374 427 443 2 × 2 = 1 + 0.524 748 854 886 4;
  • 53) 0.524 748 854 886 4 × 2 = 1 + 0.049 497 709 772 8;
  • 54) 0.049 497 709 772 8 × 2 = 0 + 0.098 995 419 545 6;
  • 55) 0.098 995 419 545 6 × 2 = 0 + 0.197 990 839 091 2;
  • 56) 0.197 990 839 091 2 × 2 = 0 + 0.395 981 678 182 4;
  • 57) 0.395 981 678 182 4 × 2 = 0 + 0.791 963 356 364 8;
  • 58) 0.791 963 356 364 8 × 2 = 1 + 0.583 926 712 729 6;
  • 59) 0.583 926 712 729 6 × 2 = 1 + 0.167 853 425 459 2;
  • 60) 0.167 853 425 459 2 × 2 = 0 + 0.335 706 850 918 4;
  • 61) 0.335 706 850 918 4 × 2 = 0 + 0.671 413 701 836 8;
  • 62) 0.671 413 701 836 8 × 2 = 1 + 0.342 827 403 673 6;
  • 63) 0.342 827 403 673 6 × 2 = 0 + 0.685 654 807 347 2;
  • 64) 0.685 654 807 347 2 × 2 = 1 + 0.371 309 614 694 4;

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:


0.000 282 005 890 9(10) =

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

6. Positive number before normalization:

0.000 282 005 890 9(10) =

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


0.000 282 005 890 9(10) =

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

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

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


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 0001 0100 0010 0101 0101 1000 0110 0101 =

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


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 0001 0100 0010 0101 0101 1000 0110 0101


Decimal number -0.000 282 005 890 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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