-0.000 282 010 78 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 010 78(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
-0.000 282 010 78(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:

|-0.000 282 010 78| = 0.000 282 010 78


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: 0.000 282 010 78.

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) 0.000 282 010 78 × 2 = 0 + 0.000 564 021 56;
  • 2) 0.000 564 021 56 × 2 = 0 + 0.001 128 043 12;
  • 3) 0.001 128 043 12 × 2 = 0 + 0.002 256 086 24;
  • 4) 0.002 256 086 24 × 2 = 0 + 0.004 512 172 48;
  • 5) 0.004 512 172 48 × 2 = 0 + 0.009 024 344 96;
  • 6) 0.009 024 344 96 × 2 = 0 + 0.018 048 689 92;
  • 7) 0.018 048 689 92 × 2 = 0 + 0.036 097 379 84;
  • 8) 0.036 097 379 84 × 2 = 0 + 0.072 194 759 68;
  • 9) 0.072 194 759 68 × 2 = 0 + 0.144 389 519 36;
  • 10) 0.144 389 519 36 × 2 = 0 + 0.288 779 038 72;
  • 11) 0.288 779 038 72 × 2 = 0 + 0.577 558 077 44;
  • 12) 0.577 558 077 44 × 2 = 1 + 0.155 116 154 88;
  • 13) 0.155 116 154 88 × 2 = 0 + 0.310 232 309 76;
  • 14) 0.310 232 309 76 × 2 = 0 + 0.620 464 619 52;
  • 15) 0.620 464 619 52 × 2 = 1 + 0.240 929 239 04;
  • 16) 0.240 929 239 04 × 2 = 0 + 0.481 858 478 08;
  • 17) 0.481 858 478 08 × 2 = 0 + 0.963 716 956 16;
  • 18) 0.963 716 956 16 × 2 = 1 + 0.927 433 912 32;
  • 19) 0.927 433 912 32 × 2 = 1 + 0.854 867 824 64;
  • 20) 0.854 867 824 64 × 2 = 1 + 0.709 735 649 28;
  • 21) 0.709 735 649 28 × 2 = 1 + 0.419 471 298 56;
  • 22) 0.419 471 298 56 × 2 = 0 + 0.838 942 597 12;
  • 23) 0.838 942 597 12 × 2 = 1 + 0.677 885 194 24;
  • 24) 0.677 885 194 24 × 2 = 1 + 0.355 770 388 48;
  • 25) 0.355 770 388 48 × 2 = 0 + 0.711 540 776 96;
  • 26) 0.711 540 776 96 × 2 = 1 + 0.423 081 553 92;
  • 27) 0.423 081 553 92 × 2 = 0 + 0.846 163 107 84;
  • 28) 0.846 163 107 84 × 2 = 1 + 0.692 326 215 68;
  • 29) 0.692 326 215 68 × 2 = 1 + 0.384 652 431 36;
  • 30) 0.384 652 431 36 × 2 = 0 + 0.769 304 862 72;
  • 31) 0.769 304 862 72 × 2 = 1 + 0.538 609 725 44;
  • 32) 0.538 609 725 44 × 2 = 1 + 0.077 219 450 88;
  • 33) 0.077 219 450 88 × 2 = 0 + 0.154 438 901 76;
  • 34) 0.154 438 901 76 × 2 = 0 + 0.308 877 803 52;
  • 35) 0.308 877 803 52 × 2 = 0 + 0.617 755 607 04;
  • 36) 0.617 755 607 04 × 2 = 1 + 0.235 511 214 08;
  • 37) 0.235 511 214 08 × 2 = 0 + 0.471 022 428 16;
  • 38) 0.471 022 428 16 × 2 = 0 + 0.942 044 856 32;
  • 39) 0.942 044 856 32 × 2 = 1 + 0.884 089 712 64;
  • 40) 0.884 089 712 64 × 2 = 1 + 0.768 179 425 28;
  • 41) 0.768 179 425 28 × 2 = 1 + 0.536 358 850 56;
  • 42) 0.536 358 850 56 × 2 = 1 + 0.072 717 701 12;
  • 43) 0.072 717 701 12 × 2 = 0 + 0.145 435 402 24;
  • 44) 0.145 435 402 24 × 2 = 0 + 0.290 870 804 48;
  • 45) 0.290 870 804 48 × 2 = 0 + 0.581 741 608 96;
  • 46) 0.581 741 608 96 × 2 = 1 + 0.163 483 217 92;
  • 47) 0.163 483 217 92 × 2 = 0 + 0.326 966 435 84;
  • 48) 0.326 966 435 84 × 2 = 0 + 0.653 932 871 68;
  • 49) 0.653 932 871 68 × 2 = 1 + 0.307 865 743 36;
  • 50) 0.307 865 743 36 × 2 = 0 + 0.615 731 486 72;
  • 51) 0.615 731 486 72 × 2 = 1 + 0.231 462 973 44;
  • 52) 0.231 462 973 44 × 2 = 0 + 0.462 925 946 88;
  • 53) 0.462 925 946 88 × 2 = 0 + 0.925 851 893 76;
  • 54) 0.925 851 893 76 × 2 = 1 + 0.851 703 787 52;
  • 55) 0.851 703 787 52 × 2 = 1 + 0.703 407 575 04;
  • 56) 0.703 407 575 04 × 2 = 1 + 0.406 815 150 08;
  • 57) 0.406 815 150 08 × 2 = 0 + 0.813 630 300 16;
  • 58) 0.813 630 300 16 × 2 = 1 + 0.627 260 600 32;
  • 59) 0.627 260 600 32 × 2 = 1 + 0.254 521 200 64;
  • 60) 0.254 521 200 64 × 2 = 0 + 0.509 042 401 28;
  • 61) 0.509 042 401 28 × 2 = 1 + 0.018 084 802 56;
  • 62) 0.018 084 802 56 × 2 = 0 + 0.036 169 605 12;
  • 63) 0.036 169 605 12 × 2 = 0 + 0.072 339 210 24;
  • 64) 0.072 339 210 24 × 2 = 0 + 0.144 678 420 48;

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:


0.000 282 010 78(10) =

0.0000 0000 0001 0010 0111 1011 0101 1011 0001 0011 1100 0100 1010 0111 0110 1000(2)

6. Positive number before normalization:

0.000 282 010 78(10) =

0.0000 0000 0001 0010 0111 1011 0101 1011 0001 0011 1100 0100 1010 0111 0110 1000(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:


0.000 282 010 78(10) =

0.0000 0000 0001 0010 0111 1011 0101 1011 0001 0011 1100 0100 1010 0111 0110 1000(2) =

0.0000 0000 0001 0010 0111 1011 0101 1011 0001 0011 1100 0100 1010 0111 0110 1000(2) × 20 =

1.0010 0111 1011 0101 1011 0001 0011 1100 0100 1010 0111 0110 1000(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 1011 0001 0011 1100 0100 1010 0111 0110 1000


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 1011 0001 0011 1100 0100 1010 0111 0110 1000 =

0010 0111 1011 0101 1011 0001 0011 1100 0100 1010 0111 0110 1000


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 1011 0001 0011 1100 0100 1010 0111 0110 1000


Decimal number -0.000 282 010 78 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 1011 0001 0011 1100 0100 1010 0111 0110 1000

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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