-0.000 282 006 46 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 46(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
-0.000 282 006 46(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:

|-0.000 282 006 46| = 0.000 282 006 46


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: 0.000 282 006 46.

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) 0.000 282 006 46 × 2 = 0 + 0.000 564 012 92;
  • 2) 0.000 564 012 92 × 2 = 0 + 0.001 128 025 84;
  • 3) 0.001 128 025 84 × 2 = 0 + 0.002 256 051 68;
  • 4) 0.002 256 051 68 × 2 = 0 + 0.004 512 103 36;
  • 5) 0.004 512 103 36 × 2 = 0 + 0.009 024 206 72;
  • 6) 0.009 024 206 72 × 2 = 0 + 0.018 048 413 44;
  • 7) 0.018 048 413 44 × 2 = 0 + 0.036 096 826 88;
  • 8) 0.036 096 826 88 × 2 = 0 + 0.072 193 653 76;
  • 9) 0.072 193 653 76 × 2 = 0 + 0.144 387 307 52;
  • 10) 0.144 387 307 52 × 2 = 0 + 0.288 774 615 04;
  • 11) 0.288 774 615 04 × 2 = 0 + 0.577 549 230 08;
  • 12) 0.577 549 230 08 × 2 = 1 + 0.155 098 460 16;
  • 13) 0.155 098 460 16 × 2 = 0 + 0.310 196 920 32;
  • 14) 0.310 196 920 32 × 2 = 0 + 0.620 393 840 64;
  • 15) 0.620 393 840 64 × 2 = 1 + 0.240 787 681 28;
  • 16) 0.240 787 681 28 × 2 = 0 + 0.481 575 362 56;
  • 17) 0.481 575 362 56 × 2 = 0 + 0.963 150 725 12;
  • 18) 0.963 150 725 12 × 2 = 1 + 0.926 301 450 24;
  • 19) 0.926 301 450 24 × 2 = 1 + 0.852 602 900 48;
  • 20) 0.852 602 900 48 × 2 = 1 + 0.705 205 800 96;
  • 21) 0.705 205 800 96 × 2 = 1 + 0.410 411 601 92;
  • 22) 0.410 411 601 92 × 2 = 0 + 0.820 823 203 84;
  • 23) 0.820 823 203 84 × 2 = 1 + 0.641 646 407 68;
  • 24) 0.641 646 407 68 × 2 = 1 + 0.283 292 815 36;
  • 25) 0.283 292 815 36 × 2 = 0 + 0.566 585 630 72;
  • 26) 0.566 585 630 72 × 2 = 1 + 0.133 171 261 44;
  • 27) 0.133 171 261 44 × 2 = 0 + 0.266 342 522 88;
  • 28) 0.266 342 522 88 × 2 = 0 + 0.532 685 045 76;
  • 29) 0.532 685 045 76 × 2 = 1 + 0.065 370 091 52;
  • 30) 0.065 370 091 52 × 2 = 0 + 0.130 740 183 04;
  • 31) 0.130 740 183 04 × 2 = 0 + 0.261 480 366 08;
  • 32) 0.261 480 366 08 × 2 = 0 + 0.522 960 732 16;
  • 33) 0.522 960 732 16 × 2 = 1 + 0.045 921 464 32;
  • 34) 0.045 921 464 32 × 2 = 0 + 0.091 842 928 64;
  • 35) 0.091 842 928 64 × 2 = 0 + 0.183 685 857 28;
  • 36) 0.183 685 857 28 × 2 = 0 + 0.367 371 714 56;
  • 37) 0.367 371 714 56 × 2 = 0 + 0.734 743 429 12;
  • 38) 0.734 743 429 12 × 2 = 1 + 0.469 486 858 24;
  • 39) 0.469 486 858 24 × 2 = 0 + 0.938 973 716 48;
  • 40) 0.938 973 716 48 × 2 = 1 + 0.877 947 432 96;
  • 41) 0.877 947 432 96 × 2 = 1 + 0.755 894 865 92;
  • 42) 0.755 894 865 92 × 2 = 1 + 0.511 789 731 84;
  • 43) 0.511 789 731 84 × 2 = 1 + 0.023 579 463 68;
  • 44) 0.023 579 463 68 × 2 = 0 + 0.047 158 927 36;
  • 45) 0.047 158 927 36 × 2 = 0 + 0.094 317 854 72;
  • 46) 0.094 317 854 72 × 2 = 0 + 0.188 635 709 44;
  • 47) 0.188 635 709 44 × 2 = 0 + 0.377 271 418 88;
  • 48) 0.377 271 418 88 × 2 = 0 + 0.754 542 837 76;
  • 49) 0.754 542 837 76 × 2 = 1 + 0.509 085 675 52;
  • 50) 0.509 085 675 52 × 2 = 1 + 0.018 171 351 04;
  • 51) 0.018 171 351 04 × 2 = 0 + 0.036 342 702 08;
  • 52) 0.036 342 702 08 × 2 = 0 + 0.072 685 404 16;
  • 53) 0.072 685 404 16 × 2 = 0 + 0.145 370 808 32;
  • 54) 0.145 370 808 32 × 2 = 0 + 0.290 741 616 64;
  • 55) 0.290 741 616 64 × 2 = 0 + 0.581 483 233 28;
  • 56) 0.581 483 233 28 × 2 = 1 + 0.162 966 466 56;
  • 57) 0.162 966 466 56 × 2 = 0 + 0.325 932 933 12;
  • 58) 0.325 932 933 12 × 2 = 0 + 0.651 865 866 24;
  • 59) 0.651 865 866 24 × 2 = 1 + 0.303 731 732 48;
  • 60) 0.303 731 732 48 × 2 = 0 + 0.607 463 464 96;
  • 61) 0.607 463 464 96 × 2 = 1 + 0.214 926 929 92;
  • 62) 0.214 926 929 92 × 2 = 0 + 0.429 853 859 84;
  • 63) 0.429 853 859 84 × 2 = 0 + 0.859 707 719 68;
  • 64) 0.859 707 719 68 × 2 = 1 + 0.719 415 439 36;

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:


0.000 282 006 46(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 0101 1110 0000 1100 0001 0010 1001(2)

6. Positive number before normalization:

0.000 282 006 46(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 0101 1110 0000 1100 0001 0010 1001(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:


0.000 282 006 46(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 0101 1110 0000 1100 0001 0010 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 0101 1110 0000 1100 0001 0010 1001(2) × 20 =

1.0010 0111 1011 0100 1000 1000 0101 1110 0000 1100 0001 0010 1001(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 1000 1000 0101 1110 0000 1100 0001 0010 1001


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 1000 1000 0101 1110 0000 1100 0001 0010 1001 =

0010 0111 1011 0100 1000 1000 0101 1110 0000 1100 0001 0010 1001


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 1000 1000 0101 1110 0000 1100 0001 0010 1001


Decimal number -0.000 282 006 46 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1000 1000 0101 1110 0000 1100 0001 0010 1001

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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