-0.000 282 005 872 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 872 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
-0.000 282 005 872 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:

|-0.000 282 005 872 3| = 0.000 282 005 872 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: 0.000 282 005 872 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) 0.000 282 005 872 3 × 2 = 0 + 0.000 564 011 744 6;
  • 2) 0.000 564 011 744 6 × 2 = 0 + 0.001 128 023 489 2;
  • 3) 0.001 128 023 489 2 × 2 = 0 + 0.002 256 046 978 4;
  • 4) 0.002 256 046 978 4 × 2 = 0 + 0.004 512 093 956 8;
  • 5) 0.004 512 093 956 8 × 2 = 0 + 0.009 024 187 913 6;
  • 6) 0.009 024 187 913 6 × 2 = 0 + 0.018 048 375 827 2;
  • 7) 0.018 048 375 827 2 × 2 = 0 + 0.036 096 751 654 4;
  • 8) 0.036 096 751 654 4 × 2 = 0 + 0.072 193 503 308 8;
  • 9) 0.072 193 503 308 8 × 2 = 0 + 0.144 387 006 617 6;
  • 10) 0.144 387 006 617 6 × 2 = 0 + 0.288 774 013 235 2;
  • 11) 0.288 774 013 235 2 × 2 = 0 + 0.577 548 026 470 4;
  • 12) 0.577 548 026 470 4 × 2 = 1 + 0.155 096 052 940 8;
  • 13) 0.155 096 052 940 8 × 2 = 0 + 0.310 192 105 881 6;
  • 14) 0.310 192 105 881 6 × 2 = 0 + 0.620 384 211 763 2;
  • 15) 0.620 384 211 763 2 × 2 = 1 + 0.240 768 423 526 4;
  • 16) 0.240 768 423 526 4 × 2 = 0 + 0.481 536 847 052 8;
  • 17) 0.481 536 847 052 8 × 2 = 0 + 0.963 073 694 105 6;
  • 18) 0.963 073 694 105 6 × 2 = 1 + 0.926 147 388 211 2;
  • 19) 0.926 147 388 211 2 × 2 = 1 + 0.852 294 776 422 4;
  • 20) 0.852 294 776 422 4 × 2 = 1 + 0.704 589 552 844 8;
  • 21) 0.704 589 552 844 8 × 2 = 1 + 0.409 179 105 689 6;
  • 22) 0.409 179 105 689 6 × 2 = 0 + 0.818 358 211 379 2;
  • 23) 0.818 358 211 379 2 × 2 = 1 + 0.636 716 422 758 4;
  • 24) 0.636 716 422 758 4 × 2 = 1 + 0.273 432 845 516 8;
  • 25) 0.273 432 845 516 8 × 2 = 0 + 0.546 865 691 033 6;
  • 26) 0.546 865 691 033 6 × 2 = 1 + 0.093 731 382 067 2;
  • 27) 0.093 731 382 067 2 × 2 = 0 + 0.187 462 764 134 4;
  • 28) 0.187 462 764 134 4 × 2 = 0 + 0.374 925 528 268 8;
  • 29) 0.374 925 528 268 8 × 2 = 0 + 0.749 851 056 537 6;
  • 30) 0.749 851 056 537 6 × 2 = 1 + 0.499 702 113 075 2;
  • 31) 0.499 702 113 075 2 × 2 = 0 + 0.999 404 226 150 4;
  • 32) 0.999 404 226 150 4 × 2 = 1 + 0.998 808 452 300 8;
  • 33) 0.998 808 452 300 8 × 2 = 1 + 0.997 616 904 601 6;
  • 34) 0.997 616 904 601 6 × 2 = 1 + 0.995 233 809 203 2;
  • 35) 0.995 233 809 203 2 × 2 = 1 + 0.990 467 618 406 4;
  • 36) 0.990 467 618 406 4 × 2 = 1 + 0.980 935 236 812 8;
  • 37) 0.980 935 236 812 8 × 2 = 1 + 0.961 870 473 625 6;
  • 38) 0.961 870 473 625 6 × 2 = 1 + 0.923 740 947 251 2;
  • 39) 0.923 740 947 251 2 × 2 = 1 + 0.847 481 894 502 4;
  • 40) 0.847 481 894 502 4 × 2 = 1 + 0.694 963 789 004 8;
  • 41) 0.694 963 789 004 8 × 2 = 1 + 0.389 927 578 009 6;
  • 42) 0.389 927 578 009 6 × 2 = 0 + 0.779 855 156 019 2;
  • 43) 0.779 855 156 019 2 × 2 = 1 + 0.559 710 312 038 4;
  • 44) 0.559 710 312 038 4 × 2 = 1 + 0.119 420 624 076 8;
  • 45) 0.119 420 624 076 8 × 2 = 0 + 0.238 841 248 153 6;
  • 46) 0.238 841 248 153 6 × 2 = 0 + 0.477 682 496 307 2;
  • 47) 0.477 682 496 307 2 × 2 = 0 + 0.955 364 992 614 4;
  • 48) 0.955 364 992 614 4 × 2 = 1 + 0.910 729 985 228 8;
  • 49) 0.910 729 985 228 8 × 2 = 1 + 0.821 459 970 457 6;
  • 50) 0.821 459 970 457 6 × 2 = 1 + 0.642 919 940 915 2;
  • 51) 0.642 919 940 915 2 × 2 = 1 + 0.285 839 881 830 4;
  • 52) 0.285 839 881 830 4 × 2 = 0 + 0.571 679 763 660 8;
  • 53) 0.571 679 763 660 8 × 2 = 1 + 0.143 359 527 321 6;
  • 54) 0.143 359 527 321 6 × 2 = 0 + 0.286 719 054 643 2;
  • 55) 0.286 719 054 643 2 × 2 = 0 + 0.573 438 109 286 4;
  • 56) 0.573 438 109 286 4 × 2 = 1 + 0.146 876 218 572 8;
  • 57) 0.146 876 218 572 8 × 2 = 0 + 0.293 752 437 145 6;
  • 58) 0.293 752 437 145 6 × 2 = 0 + 0.587 504 874 291 2;
  • 59) 0.587 504 874 291 2 × 2 = 1 + 0.175 009 748 582 4;
  • 60) 0.175 009 748 582 4 × 2 = 0 + 0.350 019 497 164 8;
  • 61) 0.350 019 497 164 8 × 2 = 0 + 0.700 038 994 329 6;
  • 62) 0.700 038 994 329 6 × 2 = 1 + 0.400 077 988 659 2;
  • 63) 0.400 077 988 659 2 × 2 = 0 + 0.800 155 977 318 4;
  • 64) 0.800 155 977 318 4 × 2 = 1 + 0.600 311 954 636 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:


0.000 282 005 872 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1111 1011 0001 1110 1001 0010 0101(2)

6. Positive number before normalization:

0.000 282 005 872 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1111 1011 0001 1110 1001 0010 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:


0.000 282 005 872 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1111 1011 0001 1110 1001 0010 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1111 1011 0001 1110 1001 0010 0101(2) × 20 =

1.0010 0111 1011 0100 0101 1111 1111 1011 0001 1110 1001 0010 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 0101 1111 1111 1011 0001 1110 1001 0010 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 0101 1111 1111 1011 0001 1110 1001 0010 0101 =

0010 0111 1011 0100 0101 1111 1111 1011 0001 1110 1001 0010 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 0101 1111 1111 1011 0001 1110 1001 0010 0101


Decimal number -0.000 282 005 872 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 1111 1111 1011 0001 1110 1001 0010 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