-0.000 281 992 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 992 7(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
-0.000 281 992 7(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:

|-0.000 281 992 7| = 0.000 281 992 7


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: 0.000 281 992 7.

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) 0.000 281 992 7 × 2 = 0 + 0.000 563 985 4;
  • 2) 0.000 563 985 4 × 2 = 0 + 0.001 127 970 8;
  • 3) 0.001 127 970 8 × 2 = 0 + 0.002 255 941 6;
  • 4) 0.002 255 941 6 × 2 = 0 + 0.004 511 883 2;
  • 5) 0.004 511 883 2 × 2 = 0 + 0.009 023 766 4;
  • 6) 0.009 023 766 4 × 2 = 0 + 0.018 047 532 8;
  • 7) 0.018 047 532 8 × 2 = 0 + 0.036 095 065 6;
  • 8) 0.036 095 065 6 × 2 = 0 + 0.072 190 131 2;
  • 9) 0.072 190 131 2 × 2 = 0 + 0.144 380 262 4;
  • 10) 0.144 380 262 4 × 2 = 0 + 0.288 760 524 8;
  • 11) 0.288 760 524 8 × 2 = 0 + 0.577 521 049 6;
  • 12) 0.577 521 049 6 × 2 = 1 + 0.155 042 099 2;
  • 13) 0.155 042 099 2 × 2 = 0 + 0.310 084 198 4;
  • 14) 0.310 084 198 4 × 2 = 0 + 0.620 168 396 8;
  • 15) 0.620 168 396 8 × 2 = 1 + 0.240 336 793 6;
  • 16) 0.240 336 793 6 × 2 = 0 + 0.480 673 587 2;
  • 17) 0.480 673 587 2 × 2 = 0 + 0.961 347 174 4;
  • 18) 0.961 347 174 4 × 2 = 1 + 0.922 694 348 8;
  • 19) 0.922 694 348 8 × 2 = 1 + 0.845 388 697 6;
  • 20) 0.845 388 697 6 × 2 = 1 + 0.690 777 395 2;
  • 21) 0.690 777 395 2 × 2 = 1 + 0.381 554 790 4;
  • 22) 0.381 554 790 4 × 2 = 0 + 0.763 109 580 8;
  • 23) 0.763 109 580 8 × 2 = 1 + 0.526 219 161 6;
  • 24) 0.526 219 161 6 × 2 = 1 + 0.052 438 323 2;
  • 25) 0.052 438 323 2 × 2 = 0 + 0.104 876 646 4;
  • 26) 0.104 876 646 4 × 2 = 0 + 0.209 753 292 8;
  • 27) 0.209 753 292 8 × 2 = 0 + 0.419 506 585 6;
  • 28) 0.419 506 585 6 × 2 = 0 + 0.839 013 171 2;
  • 29) 0.839 013 171 2 × 2 = 1 + 0.678 026 342 4;
  • 30) 0.678 026 342 4 × 2 = 1 + 0.356 052 684 8;
  • 31) 0.356 052 684 8 × 2 = 0 + 0.712 105 369 6;
  • 32) 0.712 105 369 6 × 2 = 1 + 0.424 210 739 2;
  • 33) 0.424 210 739 2 × 2 = 0 + 0.848 421 478 4;
  • 34) 0.848 421 478 4 × 2 = 1 + 0.696 842 956 8;
  • 35) 0.696 842 956 8 × 2 = 1 + 0.393 685 913 6;
  • 36) 0.393 685 913 6 × 2 = 0 + 0.787 371 827 2;
  • 37) 0.787 371 827 2 × 2 = 1 + 0.574 743 654 4;
  • 38) 0.574 743 654 4 × 2 = 1 + 0.149 487 308 8;
  • 39) 0.149 487 308 8 × 2 = 0 + 0.298 974 617 6;
  • 40) 0.298 974 617 6 × 2 = 0 + 0.597 949 235 2;
  • 41) 0.597 949 235 2 × 2 = 1 + 0.195 898 470 4;
  • 42) 0.195 898 470 4 × 2 = 0 + 0.391 796 940 8;
  • 43) 0.391 796 940 8 × 2 = 0 + 0.783 593 881 6;
  • 44) 0.783 593 881 6 × 2 = 1 + 0.567 187 763 2;
  • 45) 0.567 187 763 2 × 2 = 1 + 0.134 375 526 4;
  • 46) 0.134 375 526 4 × 2 = 0 + 0.268 751 052 8;
  • 47) 0.268 751 052 8 × 2 = 0 + 0.537 502 105 6;
  • 48) 0.537 502 105 6 × 2 = 1 + 0.075 004 211 2;
  • 49) 0.075 004 211 2 × 2 = 0 + 0.150 008 422 4;
  • 50) 0.150 008 422 4 × 2 = 0 + 0.300 016 844 8;
  • 51) 0.300 016 844 8 × 2 = 0 + 0.600 033 689 6;
  • 52) 0.600 033 689 6 × 2 = 1 + 0.200 067 379 2;
  • 53) 0.200 067 379 2 × 2 = 0 + 0.400 134 758 4;
  • 54) 0.400 134 758 4 × 2 = 0 + 0.800 269 516 8;
  • 55) 0.800 269 516 8 × 2 = 1 + 0.600 539 033 6;
  • 56) 0.600 539 033 6 × 2 = 1 + 0.201 078 067 2;
  • 57) 0.201 078 067 2 × 2 = 0 + 0.402 156 134 4;
  • 58) 0.402 156 134 4 × 2 = 0 + 0.804 312 268 8;
  • 59) 0.804 312 268 8 × 2 = 1 + 0.608 624 537 6;
  • 60) 0.608 624 537 6 × 2 = 1 + 0.217 249 075 2;
  • 61) 0.217 249 075 2 × 2 = 0 + 0.434 498 150 4;
  • 62) 0.434 498 150 4 × 2 = 0 + 0.868 996 300 8;
  • 63) 0.868 996 300 8 × 2 = 1 + 0.737 992 601 6;
  • 64) 0.737 992 601 6 × 2 = 1 + 0.475 985 203 2;

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:


0.000 281 992 7(10) =

0.0000 0000 0001 0010 0111 1011 0000 1101 0110 1100 1001 1001 0001 0011 0011 0011(2)

6. Positive number before normalization:

0.000 281 992 7(10) =

0.0000 0000 0001 0010 0111 1011 0000 1101 0110 1100 1001 1001 0001 0011 0011 0011(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:


0.000 281 992 7(10) =

0.0000 0000 0001 0010 0111 1011 0000 1101 0110 1100 1001 1001 0001 0011 0011 0011(2) =

0.0000 0000 0001 0010 0111 1011 0000 1101 0110 1100 1001 1001 0001 0011 0011 0011(2) × 20 =

1.0010 0111 1011 0000 1101 0110 1100 1001 1001 0001 0011 0011 0011(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 0000 1101 0110 1100 1001 1001 0001 0011 0011 0011


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 0000 1101 0110 1100 1001 1001 0001 0011 0011 0011 =

0010 0111 1011 0000 1101 0110 1100 1001 1001 0001 0011 0011 0011


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 0000 1101 0110 1100 1001 1001 0001 0011 0011 0011


Decimal number -0.000 281 992 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0000 1101 0110 1100 1001 1001 0001 0011 0011 0011

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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