-0.000 282 062 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 062(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
-0.000 282 062(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:

|-0.000 282 062| = 0.000 282 062


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: 0.000 282 062.

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) 0.000 282 062 × 2 = 0 + 0.000 564 124;
  • 2) 0.000 564 124 × 2 = 0 + 0.001 128 248;
  • 3) 0.001 128 248 × 2 = 0 + 0.002 256 496;
  • 4) 0.002 256 496 × 2 = 0 + 0.004 512 992;
  • 5) 0.004 512 992 × 2 = 0 + 0.009 025 984;
  • 6) 0.009 025 984 × 2 = 0 + 0.018 051 968;
  • 7) 0.018 051 968 × 2 = 0 + 0.036 103 936;
  • 8) 0.036 103 936 × 2 = 0 + 0.072 207 872;
  • 9) 0.072 207 872 × 2 = 0 + 0.144 415 744;
  • 10) 0.144 415 744 × 2 = 0 + 0.288 831 488;
  • 11) 0.288 831 488 × 2 = 0 + 0.577 662 976;
  • 12) 0.577 662 976 × 2 = 1 + 0.155 325 952;
  • 13) 0.155 325 952 × 2 = 0 + 0.310 651 904;
  • 14) 0.310 651 904 × 2 = 0 + 0.621 303 808;
  • 15) 0.621 303 808 × 2 = 1 + 0.242 607 616;
  • 16) 0.242 607 616 × 2 = 0 + 0.485 215 232;
  • 17) 0.485 215 232 × 2 = 0 + 0.970 430 464;
  • 18) 0.970 430 464 × 2 = 1 + 0.940 860 928;
  • 19) 0.940 860 928 × 2 = 1 + 0.881 721 856;
  • 20) 0.881 721 856 × 2 = 1 + 0.763 443 712;
  • 21) 0.763 443 712 × 2 = 1 + 0.526 887 424;
  • 22) 0.526 887 424 × 2 = 1 + 0.053 774 848;
  • 23) 0.053 774 848 × 2 = 0 + 0.107 549 696;
  • 24) 0.107 549 696 × 2 = 0 + 0.215 099 392;
  • 25) 0.215 099 392 × 2 = 0 + 0.430 198 784;
  • 26) 0.430 198 784 × 2 = 0 + 0.860 397 568;
  • 27) 0.860 397 568 × 2 = 1 + 0.720 795 136;
  • 28) 0.720 795 136 × 2 = 1 + 0.441 590 272;
  • 29) 0.441 590 272 × 2 = 0 + 0.883 180 544;
  • 30) 0.883 180 544 × 2 = 1 + 0.766 361 088;
  • 31) 0.766 361 088 × 2 = 1 + 0.532 722 176;
  • 32) 0.532 722 176 × 2 = 1 + 0.065 444 352;
  • 33) 0.065 444 352 × 2 = 0 + 0.130 888 704;
  • 34) 0.130 888 704 × 2 = 0 + 0.261 777 408;
  • 35) 0.261 777 408 × 2 = 0 + 0.523 554 816;
  • 36) 0.523 554 816 × 2 = 1 + 0.047 109 632;
  • 37) 0.047 109 632 × 2 = 0 + 0.094 219 264;
  • 38) 0.094 219 264 × 2 = 0 + 0.188 438 528;
  • 39) 0.188 438 528 × 2 = 0 + 0.376 877 056;
  • 40) 0.376 877 056 × 2 = 0 + 0.753 754 112;
  • 41) 0.753 754 112 × 2 = 1 + 0.507 508 224;
  • 42) 0.507 508 224 × 2 = 1 + 0.015 016 448;
  • 43) 0.015 016 448 × 2 = 0 + 0.030 032 896;
  • 44) 0.030 032 896 × 2 = 0 + 0.060 065 792;
  • 45) 0.060 065 792 × 2 = 0 + 0.120 131 584;
  • 46) 0.120 131 584 × 2 = 0 + 0.240 263 168;
  • 47) 0.240 263 168 × 2 = 0 + 0.480 526 336;
  • 48) 0.480 526 336 × 2 = 0 + 0.961 052 672;
  • 49) 0.961 052 672 × 2 = 1 + 0.922 105 344;
  • 50) 0.922 105 344 × 2 = 1 + 0.844 210 688;
  • 51) 0.844 210 688 × 2 = 1 + 0.688 421 376;
  • 52) 0.688 421 376 × 2 = 1 + 0.376 842 752;
  • 53) 0.376 842 752 × 2 = 0 + 0.753 685 504;
  • 54) 0.753 685 504 × 2 = 1 + 0.507 371 008;
  • 55) 0.507 371 008 × 2 = 1 + 0.014 742 016;
  • 56) 0.014 742 016 × 2 = 0 + 0.029 484 032;
  • 57) 0.029 484 032 × 2 = 0 + 0.058 968 064;
  • 58) 0.058 968 064 × 2 = 0 + 0.117 936 128;
  • 59) 0.117 936 128 × 2 = 0 + 0.235 872 256;
  • 60) 0.235 872 256 × 2 = 0 + 0.471 744 512;
  • 61) 0.471 744 512 × 2 = 0 + 0.943 489 024;
  • 62) 0.943 489 024 × 2 = 1 + 0.886 978 048;
  • 63) 0.886 978 048 × 2 = 1 + 0.773 956 096;
  • 64) 0.773 956 096 × 2 = 1 + 0.547 912 192;

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:


0.000 282 062(10) =

0.0000 0000 0001 0010 0111 1100 0011 0111 0001 0000 1100 0000 1111 0110 0000 0111(2)

6. Positive number before normalization:

0.000 282 062(10) =

0.0000 0000 0001 0010 0111 1100 0011 0111 0001 0000 1100 0000 1111 0110 0000 0111(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:


0.000 282 062(10) =

0.0000 0000 0001 0010 0111 1100 0011 0111 0001 0000 1100 0000 1111 0110 0000 0111(2) =

0.0000 0000 0001 0010 0111 1100 0011 0111 0001 0000 1100 0000 1111 0110 0000 0111(2) × 20 =

1.0010 0111 1100 0011 0111 0001 0000 1100 0000 1111 0110 0000 0111(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 1100 0011 0111 0001 0000 1100 0000 1111 0110 0000 0111


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 1100 0011 0111 0001 0000 1100 0000 1111 0110 0000 0111 =

0010 0111 1100 0011 0111 0001 0000 1100 0000 1111 0110 0000 0111


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 1100 0011 0111 0001 0000 1100 0000 1111 0110 0000 0111


Decimal number -0.000 282 062 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1100 0011 0111 0001 0000 1100 0000 1111 0110 0000 0111

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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