-0.000 282 004 01 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 004 01(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
-0.000 282 004 01(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:

|-0.000 282 004 01| = 0.000 282 004 01


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: 0.000 282 004 01.

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) 0.000 282 004 01 × 2 = 0 + 0.000 564 008 02;
  • 2) 0.000 564 008 02 × 2 = 0 + 0.001 128 016 04;
  • 3) 0.001 128 016 04 × 2 = 0 + 0.002 256 032 08;
  • 4) 0.002 256 032 08 × 2 = 0 + 0.004 512 064 16;
  • 5) 0.004 512 064 16 × 2 = 0 + 0.009 024 128 32;
  • 6) 0.009 024 128 32 × 2 = 0 + 0.018 048 256 64;
  • 7) 0.018 048 256 64 × 2 = 0 + 0.036 096 513 28;
  • 8) 0.036 096 513 28 × 2 = 0 + 0.072 193 026 56;
  • 9) 0.072 193 026 56 × 2 = 0 + 0.144 386 053 12;
  • 10) 0.144 386 053 12 × 2 = 0 + 0.288 772 106 24;
  • 11) 0.288 772 106 24 × 2 = 0 + 0.577 544 212 48;
  • 12) 0.577 544 212 48 × 2 = 1 + 0.155 088 424 96;
  • 13) 0.155 088 424 96 × 2 = 0 + 0.310 176 849 92;
  • 14) 0.310 176 849 92 × 2 = 0 + 0.620 353 699 84;
  • 15) 0.620 353 699 84 × 2 = 1 + 0.240 707 399 68;
  • 16) 0.240 707 399 68 × 2 = 0 + 0.481 414 799 36;
  • 17) 0.481 414 799 36 × 2 = 0 + 0.962 829 598 72;
  • 18) 0.962 829 598 72 × 2 = 1 + 0.925 659 197 44;
  • 19) 0.925 659 197 44 × 2 = 1 + 0.851 318 394 88;
  • 20) 0.851 318 394 88 × 2 = 1 + 0.702 636 789 76;
  • 21) 0.702 636 789 76 × 2 = 1 + 0.405 273 579 52;
  • 22) 0.405 273 579 52 × 2 = 0 + 0.810 547 159 04;
  • 23) 0.810 547 159 04 × 2 = 1 + 0.621 094 318 08;
  • 24) 0.621 094 318 08 × 2 = 1 + 0.242 188 636 16;
  • 25) 0.242 188 636 16 × 2 = 0 + 0.484 377 272 32;
  • 26) 0.484 377 272 32 × 2 = 0 + 0.968 754 544 64;
  • 27) 0.968 754 544 64 × 2 = 1 + 0.937 509 089 28;
  • 28) 0.937 509 089 28 × 2 = 1 + 0.875 018 178 56;
  • 29) 0.875 018 178 56 × 2 = 1 + 0.750 036 357 12;
  • 30) 0.750 036 357 12 × 2 = 1 + 0.500 072 714 24;
  • 31) 0.500 072 714 24 × 2 = 1 + 0.000 145 428 48;
  • 32) 0.000 145 428 48 × 2 = 0 + 0.000 290 856 96;
  • 33) 0.000 290 856 96 × 2 = 0 + 0.000 581 713 92;
  • 34) 0.000 581 713 92 × 2 = 0 + 0.001 163 427 84;
  • 35) 0.001 163 427 84 × 2 = 0 + 0.002 326 855 68;
  • 36) 0.002 326 855 68 × 2 = 0 + 0.004 653 711 36;
  • 37) 0.004 653 711 36 × 2 = 0 + 0.009 307 422 72;
  • 38) 0.009 307 422 72 × 2 = 0 + 0.018 614 845 44;
  • 39) 0.018 614 845 44 × 2 = 0 + 0.037 229 690 88;
  • 40) 0.037 229 690 88 × 2 = 0 + 0.074 459 381 76;
  • 41) 0.074 459 381 76 × 2 = 0 + 0.148 918 763 52;
  • 42) 0.148 918 763 52 × 2 = 0 + 0.297 837 527 04;
  • 43) 0.297 837 527 04 × 2 = 0 + 0.595 675 054 08;
  • 44) 0.595 675 054 08 × 2 = 1 + 0.191 350 108 16;
  • 45) 0.191 350 108 16 × 2 = 0 + 0.382 700 216 32;
  • 46) 0.382 700 216 32 × 2 = 0 + 0.765 400 432 64;
  • 47) 0.765 400 432 64 × 2 = 1 + 0.530 800 865 28;
  • 48) 0.530 800 865 28 × 2 = 1 + 0.061 601 730 56;
  • 49) 0.061 601 730 56 × 2 = 0 + 0.123 203 461 12;
  • 50) 0.123 203 461 12 × 2 = 0 + 0.246 406 922 24;
  • 51) 0.246 406 922 24 × 2 = 0 + 0.492 813 844 48;
  • 52) 0.492 813 844 48 × 2 = 0 + 0.985 627 688 96;
  • 53) 0.985 627 688 96 × 2 = 1 + 0.971 255 377 92;
  • 54) 0.971 255 377 92 × 2 = 1 + 0.942 510 755 84;
  • 55) 0.942 510 755 84 × 2 = 1 + 0.885 021 511 68;
  • 56) 0.885 021 511 68 × 2 = 1 + 0.770 043 023 36;
  • 57) 0.770 043 023 36 × 2 = 1 + 0.540 086 046 72;
  • 58) 0.540 086 046 72 × 2 = 1 + 0.080 172 093 44;
  • 59) 0.080 172 093 44 × 2 = 0 + 0.160 344 186 88;
  • 60) 0.160 344 186 88 × 2 = 0 + 0.320 688 373 76;
  • 61) 0.320 688 373 76 × 2 = 0 + 0.641 376 747 52;
  • 62) 0.641 376 747 52 × 2 = 1 + 0.282 753 495 04;
  • 63) 0.282 753 495 04 × 2 = 0 + 0.565 506 990 08;
  • 64) 0.565 506 990 08 × 2 = 1 + 0.131 013 980 16;

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:


0.000 282 004 01(10) =

0.0000 0000 0001 0010 0111 1011 0011 1110 0000 0000 0001 0011 0000 1111 1100 0101(2)

6. Positive number before normalization:

0.000 282 004 01(10) =

0.0000 0000 0001 0010 0111 1011 0011 1110 0000 0000 0001 0011 0000 1111 1100 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:


0.000 282 004 01(10) =

0.0000 0000 0001 0010 0111 1011 0011 1110 0000 0000 0001 0011 0000 1111 1100 0101(2) =

0.0000 0000 0001 0010 0111 1011 0011 1110 0000 0000 0001 0011 0000 1111 1100 0101(2) × 20 =

1.0010 0111 1011 0011 1110 0000 0000 0001 0011 0000 1111 1100 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 0011 1110 0000 0000 0001 0011 0000 1111 1100 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 0011 1110 0000 0000 0001 0011 0000 1111 1100 0101 =

0010 0111 1011 0011 1110 0000 0000 0001 0011 0000 1111 1100 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 0011 1110 0000 0000 0001 0011 0000 1111 1100 0101


Decimal number -0.000 282 004 01 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0011 1110 0000 0000 0001 0011 0000 1111 1100 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