-0.000 282 623 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 623(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
-0.000 282 623(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:

|-0.000 282 623| = 0.000 282 623


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: 0.000 282 623.

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) 0.000 282 623 × 2 = 0 + 0.000 565 246;
  • 2) 0.000 565 246 × 2 = 0 + 0.001 130 492;
  • 3) 0.001 130 492 × 2 = 0 + 0.002 260 984;
  • 4) 0.002 260 984 × 2 = 0 + 0.004 521 968;
  • 5) 0.004 521 968 × 2 = 0 + 0.009 043 936;
  • 6) 0.009 043 936 × 2 = 0 + 0.018 087 872;
  • 7) 0.018 087 872 × 2 = 0 + 0.036 175 744;
  • 8) 0.036 175 744 × 2 = 0 + 0.072 351 488;
  • 9) 0.072 351 488 × 2 = 0 + 0.144 702 976;
  • 10) 0.144 702 976 × 2 = 0 + 0.289 405 952;
  • 11) 0.289 405 952 × 2 = 0 + 0.578 811 904;
  • 12) 0.578 811 904 × 2 = 1 + 0.157 623 808;
  • 13) 0.157 623 808 × 2 = 0 + 0.315 247 616;
  • 14) 0.315 247 616 × 2 = 0 + 0.630 495 232;
  • 15) 0.630 495 232 × 2 = 1 + 0.260 990 464;
  • 16) 0.260 990 464 × 2 = 0 + 0.521 980 928;
  • 17) 0.521 980 928 × 2 = 1 + 0.043 961 856;
  • 18) 0.043 961 856 × 2 = 0 + 0.087 923 712;
  • 19) 0.087 923 712 × 2 = 0 + 0.175 847 424;
  • 20) 0.175 847 424 × 2 = 0 + 0.351 694 848;
  • 21) 0.351 694 848 × 2 = 0 + 0.703 389 696;
  • 22) 0.703 389 696 × 2 = 1 + 0.406 779 392;
  • 23) 0.406 779 392 × 2 = 0 + 0.813 558 784;
  • 24) 0.813 558 784 × 2 = 1 + 0.627 117 568;
  • 25) 0.627 117 568 × 2 = 1 + 0.254 235 136;
  • 26) 0.254 235 136 × 2 = 0 + 0.508 470 272;
  • 27) 0.508 470 272 × 2 = 1 + 0.016 940 544;
  • 28) 0.016 940 544 × 2 = 0 + 0.033 881 088;
  • 29) 0.033 881 088 × 2 = 0 + 0.067 762 176;
  • 30) 0.067 762 176 × 2 = 0 + 0.135 524 352;
  • 31) 0.135 524 352 × 2 = 0 + 0.271 048 704;
  • 32) 0.271 048 704 × 2 = 0 + 0.542 097 408;
  • 33) 0.542 097 408 × 2 = 1 + 0.084 194 816;
  • 34) 0.084 194 816 × 2 = 0 + 0.168 389 632;
  • 35) 0.168 389 632 × 2 = 0 + 0.336 779 264;
  • 36) 0.336 779 264 × 2 = 0 + 0.673 558 528;
  • 37) 0.673 558 528 × 2 = 1 + 0.347 117 056;
  • 38) 0.347 117 056 × 2 = 0 + 0.694 234 112;
  • 39) 0.694 234 112 × 2 = 1 + 0.388 468 224;
  • 40) 0.388 468 224 × 2 = 0 + 0.776 936 448;
  • 41) 0.776 936 448 × 2 = 1 + 0.553 872 896;
  • 42) 0.553 872 896 × 2 = 1 + 0.107 745 792;
  • 43) 0.107 745 792 × 2 = 0 + 0.215 491 584;
  • 44) 0.215 491 584 × 2 = 0 + 0.430 983 168;
  • 45) 0.430 983 168 × 2 = 0 + 0.861 966 336;
  • 46) 0.861 966 336 × 2 = 1 + 0.723 932 672;
  • 47) 0.723 932 672 × 2 = 1 + 0.447 865 344;
  • 48) 0.447 865 344 × 2 = 0 + 0.895 730 688;
  • 49) 0.895 730 688 × 2 = 1 + 0.791 461 376;
  • 50) 0.791 461 376 × 2 = 1 + 0.582 922 752;
  • 51) 0.582 922 752 × 2 = 1 + 0.165 845 504;
  • 52) 0.165 845 504 × 2 = 0 + 0.331 691 008;
  • 53) 0.331 691 008 × 2 = 0 + 0.663 382 016;
  • 54) 0.663 382 016 × 2 = 1 + 0.326 764 032;
  • 55) 0.326 764 032 × 2 = 0 + 0.653 528 064;
  • 56) 0.653 528 064 × 2 = 1 + 0.307 056 128;
  • 57) 0.307 056 128 × 2 = 0 + 0.614 112 256;
  • 58) 0.614 112 256 × 2 = 1 + 0.228 224 512;
  • 59) 0.228 224 512 × 2 = 0 + 0.456 449 024;
  • 60) 0.456 449 024 × 2 = 0 + 0.912 898 048;
  • 61) 0.912 898 048 × 2 = 1 + 0.825 796 096;
  • 62) 0.825 796 096 × 2 = 1 + 0.651 592 192;
  • 63) 0.651 592 192 × 2 = 1 + 0.303 184 384;
  • 64) 0.303 184 384 × 2 = 0 + 0.606 368 768;

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:


0.000 282 623(10) =

0.0000 0000 0001 0010 1000 0101 1010 0000 1000 1010 1100 0110 1110 0101 0100 1110(2)

6. Positive number before normalization:

0.000 282 623(10) =

0.0000 0000 0001 0010 1000 0101 1010 0000 1000 1010 1100 0110 1110 0101 0100 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:


0.000 282 623(10) =

0.0000 0000 0001 0010 1000 0101 1010 0000 1000 1010 1100 0110 1110 0101 0100 1110(2) =

0.0000 0000 0001 0010 1000 0101 1010 0000 1000 1010 1100 0110 1110 0101 0100 1110(2) × 20 =

1.0010 1000 0101 1010 0000 1000 1010 1100 0110 1110 0101 0100 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 1000 0101 1010 0000 1000 1010 1100 0110 1110 0101 0100 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 1000 0101 1010 0000 1000 1010 1100 0110 1110 0101 0100 1110 =

0010 1000 0101 1010 0000 1000 1010 1100 0110 1110 0101 0100 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 1000 0101 1010 0000 1000 1010 1100 0110 1110 0101 0100 1110


Decimal number -0.000 282 623 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0101 1010 0000 1000 1010 1100 0110 1110 0101 0100 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