-0.000 282 011 09 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 011 09(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
-0.000 282 011 09(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:

|-0.000 282 011 09| = 0.000 282 011 09


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: 0.000 282 011 09.

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) 0.000 282 011 09 × 2 = 0 + 0.000 564 022 18;
  • 2) 0.000 564 022 18 × 2 = 0 + 0.001 128 044 36;
  • 3) 0.001 128 044 36 × 2 = 0 + 0.002 256 088 72;
  • 4) 0.002 256 088 72 × 2 = 0 + 0.004 512 177 44;
  • 5) 0.004 512 177 44 × 2 = 0 + 0.009 024 354 88;
  • 6) 0.009 024 354 88 × 2 = 0 + 0.018 048 709 76;
  • 7) 0.018 048 709 76 × 2 = 0 + 0.036 097 419 52;
  • 8) 0.036 097 419 52 × 2 = 0 + 0.072 194 839 04;
  • 9) 0.072 194 839 04 × 2 = 0 + 0.144 389 678 08;
  • 10) 0.144 389 678 08 × 2 = 0 + 0.288 779 356 16;
  • 11) 0.288 779 356 16 × 2 = 0 + 0.577 558 712 32;
  • 12) 0.577 558 712 32 × 2 = 1 + 0.155 117 424 64;
  • 13) 0.155 117 424 64 × 2 = 0 + 0.310 234 849 28;
  • 14) 0.310 234 849 28 × 2 = 0 + 0.620 469 698 56;
  • 15) 0.620 469 698 56 × 2 = 1 + 0.240 939 397 12;
  • 16) 0.240 939 397 12 × 2 = 0 + 0.481 878 794 24;
  • 17) 0.481 878 794 24 × 2 = 0 + 0.963 757 588 48;
  • 18) 0.963 757 588 48 × 2 = 1 + 0.927 515 176 96;
  • 19) 0.927 515 176 96 × 2 = 1 + 0.855 030 353 92;
  • 20) 0.855 030 353 92 × 2 = 1 + 0.710 060 707 84;
  • 21) 0.710 060 707 84 × 2 = 1 + 0.420 121 415 68;
  • 22) 0.420 121 415 68 × 2 = 0 + 0.840 242 831 36;
  • 23) 0.840 242 831 36 × 2 = 1 + 0.680 485 662 72;
  • 24) 0.680 485 662 72 × 2 = 1 + 0.360 971 325 44;
  • 25) 0.360 971 325 44 × 2 = 0 + 0.721 942 650 88;
  • 26) 0.721 942 650 88 × 2 = 1 + 0.443 885 301 76;
  • 27) 0.443 885 301 76 × 2 = 0 + 0.887 770 603 52;
  • 28) 0.887 770 603 52 × 2 = 1 + 0.775 541 207 04;
  • 29) 0.775 541 207 04 × 2 = 1 + 0.551 082 414 08;
  • 30) 0.551 082 414 08 × 2 = 1 + 0.102 164 828 16;
  • 31) 0.102 164 828 16 × 2 = 0 + 0.204 329 656 32;
  • 32) 0.204 329 656 32 × 2 = 0 + 0.408 659 312 64;
  • 33) 0.408 659 312 64 × 2 = 0 + 0.817 318 625 28;
  • 34) 0.817 318 625 28 × 2 = 1 + 0.634 637 250 56;
  • 35) 0.634 637 250 56 × 2 = 1 + 0.269 274 501 12;
  • 36) 0.269 274 501 12 × 2 = 0 + 0.538 549 002 24;
  • 37) 0.538 549 002 24 × 2 = 1 + 0.077 098 004 48;
  • 38) 0.077 098 004 48 × 2 = 0 + 0.154 196 008 96;
  • 39) 0.154 196 008 96 × 2 = 0 + 0.308 392 017 92;
  • 40) 0.308 392 017 92 × 2 = 0 + 0.616 784 035 84;
  • 41) 0.616 784 035 84 × 2 = 1 + 0.233 568 071 68;
  • 42) 0.233 568 071 68 × 2 = 0 + 0.467 136 143 36;
  • 43) 0.467 136 143 36 × 2 = 0 + 0.934 272 286 72;
  • 44) 0.934 272 286 72 × 2 = 1 + 0.868 544 573 44;
  • 45) 0.868 544 573 44 × 2 = 1 + 0.737 089 146 88;
  • 46) 0.737 089 146 88 × 2 = 1 + 0.474 178 293 76;
  • 47) 0.474 178 293 76 × 2 = 0 + 0.948 356 587 52;
  • 48) 0.948 356 587 52 × 2 = 1 + 0.896 713 175 04;
  • 49) 0.896 713 175 04 × 2 = 1 + 0.793 426 350 08;
  • 50) 0.793 426 350 08 × 2 = 1 + 0.586 852 700 16;
  • 51) 0.586 852 700 16 × 2 = 1 + 0.173 705 400 32;
  • 52) 0.173 705 400 32 × 2 = 0 + 0.347 410 800 64;
  • 53) 0.347 410 800 64 × 2 = 0 + 0.694 821 601 28;
  • 54) 0.694 821 601 28 × 2 = 1 + 0.389 643 202 56;
  • 55) 0.389 643 202 56 × 2 = 0 + 0.779 286 405 12;
  • 56) 0.779 286 405 12 × 2 = 1 + 0.558 572 810 24;
  • 57) 0.558 572 810 24 × 2 = 1 + 0.117 145 620 48;
  • 58) 0.117 145 620 48 × 2 = 0 + 0.234 291 240 96;
  • 59) 0.234 291 240 96 × 2 = 0 + 0.468 582 481 92;
  • 60) 0.468 582 481 92 × 2 = 0 + 0.937 164 963 84;
  • 61) 0.937 164 963 84 × 2 = 1 + 0.874 329 927 68;
  • 62) 0.874 329 927 68 × 2 = 1 + 0.748 659 855 36;
  • 63) 0.748 659 855 36 × 2 = 1 + 0.497 319 710 72;
  • 64) 0.497 319 710 72 × 2 = 0 + 0.994 639 421 44;

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:


0.000 282 011 09(10) =

0.0000 0000 0001 0010 0111 1011 0101 1100 0110 1000 1001 1101 1110 0101 1000 1110(2)

6. Positive number before normalization:

0.000 282 011 09(10) =

0.0000 0000 0001 0010 0111 1011 0101 1100 0110 1000 1001 1101 1110 0101 1000 1110(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:


0.000 282 011 09(10) =

0.0000 0000 0001 0010 0111 1011 0101 1100 0110 1000 1001 1101 1110 0101 1000 1110(2) =

0.0000 0000 0001 0010 0111 1011 0101 1100 0110 1000 1001 1101 1110 0101 1000 1110(2) × 20 =

1.0010 0111 1011 0101 1100 0110 1000 1001 1101 1110 0101 1000 1110(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 0101 1100 0110 1000 1001 1101 1110 0101 1000 1110


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 0101 1100 0110 1000 1001 1101 1110 0101 1000 1110 =

0010 0111 1011 0101 1100 0110 1000 1001 1101 1110 0101 1000 1110


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 0101 1100 0110 1000 1001 1101 1110 0101 1000 1110


Decimal number -0.000 282 011 09 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 1100 0110 1000 1001 1101 1110 0101 1000 1110

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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