-0.000 282 336 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 336(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
-0.000 282 336(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:

|-0.000 282 336| = 0.000 282 336


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: 0.000 282 336.

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) 0.000 282 336 × 2 = 0 + 0.000 564 672;
  • 2) 0.000 564 672 × 2 = 0 + 0.001 129 344;
  • 3) 0.001 129 344 × 2 = 0 + 0.002 258 688;
  • 4) 0.002 258 688 × 2 = 0 + 0.004 517 376;
  • 5) 0.004 517 376 × 2 = 0 + 0.009 034 752;
  • 6) 0.009 034 752 × 2 = 0 + 0.018 069 504;
  • 7) 0.018 069 504 × 2 = 0 + 0.036 139 008;
  • 8) 0.036 139 008 × 2 = 0 + 0.072 278 016;
  • 9) 0.072 278 016 × 2 = 0 + 0.144 556 032;
  • 10) 0.144 556 032 × 2 = 0 + 0.289 112 064;
  • 11) 0.289 112 064 × 2 = 0 + 0.578 224 128;
  • 12) 0.578 224 128 × 2 = 1 + 0.156 448 256;
  • 13) 0.156 448 256 × 2 = 0 + 0.312 896 512;
  • 14) 0.312 896 512 × 2 = 0 + 0.625 793 024;
  • 15) 0.625 793 024 × 2 = 1 + 0.251 586 048;
  • 16) 0.251 586 048 × 2 = 0 + 0.503 172 096;
  • 17) 0.503 172 096 × 2 = 1 + 0.006 344 192;
  • 18) 0.006 344 192 × 2 = 0 + 0.012 688 384;
  • 19) 0.012 688 384 × 2 = 0 + 0.025 376 768;
  • 20) 0.025 376 768 × 2 = 0 + 0.050 753 536;
  • 21) 0.050 753 536 × 2 = 0 + 0.101 507 072;
  • 22) 0.101 507 072 × 2 = 0 + 0.203 014 144;
  • 23) 0.203 014 144 × 2 = 0 + 0.406 028 288;
  • 24) 0.406 028 288 × 2 = 0 + 0.812 056 576;
  • 25) 0.812 056 576 × 2 = 1 + 0.624 113 152;
  • 26) 0.624 113 152 × 2 = 1 + 0.248 226 304;
  • 27) 0.248 226 304 × 2 = 0 + 0.496 452 608;
  • 28) 0.496 452 608 × 2 = 0 + 0.992 905 216;
  • 29) 0.992 905 216 × 2 = 1 + 0.985 810 432;
  • 30) 0.985 810 432 × 2 = 1 + 0.971 620 864;
  • 31) 0.971 620 864 × 2 = 1 + 0.943 241 728;
  • 32) 0.943 241 728 × 2 = 1 + 0.886 483 456;
  • 33) 0.886 483 456 × 2 = 1 + 0.772 966 912;
  • 34) 0.772 966 912 × 2 = 1 + 0.545 933 824;
  • 35) 0.545 933 824 × 2 = 1 + 0.091 867 648;
  • 36) 0.091 867 648 × 2 = 0 + 0.183 735 296;
  • 37) 0.183 735 296 × 2 = 0 + 0.367 470 592;
  • 38) 0.367 470 592 × 2 = 0 + 0.734 941 184;
  • 39) 0.734 941 184 × 2 = 1 + 0.469 882 368;
  • 40) 0.469 882 368 × 2 = 0 + 0.939 764 736;
  • 41) 0.939 764 736 × 2 = 1 + 0.879 529 472;
  • 42) 0.879 529 472 × 2 = 1 + 0.759 058 944;
  • 43) 0.759 058 944 × 2 = 1 + 0.518 117 888;
  • 44) 0.518 117 888 × 2 = 1 + 0.036 235 776;
  • 45) 0.036 235 776 × 2 = 0 + 0.072 471 552;
  • 46) 0.072 471 552 × 2 = 0 + 0.144 943 104;
  • 47) 0.144 943 104 × 2 = 0 + 0.289 886 208;
  • 48) 0.289 886 208 × 2 = 0 + 0.579 772 416;
  • 49) 0.579 772 416 × 2 = 1 + 0.159 544 832;
  • 50) 0.159 544 832 × 2 = 0 + 0.319 089 664;
  • 51) 0.319 089 664 × 2 = 0 + 0.638 179 328;
  • 52) 0.638 179 328 × 2 = 1 + 0.276 358 656;
  • 53) 0.276 358 656 × 2 = 0 + 0.552 717 312;
  • 54) 0.552 717 312 × 2 = 1 + 0.105 434 624;
  • 55) 0.105 434 624 × 2 = 0 + 0.210 869 248;
  • 56) 0.210 869 248 × 2 = 0 + 0.421 738 496;
  • 57) 0.421 738 496 × 2 = 0 + 0.843 476 992;
  • 58) 0.843 476 992 × 2 = 1 + 0.686 953 984;
  • 59) 0.686 953 984 × 2 = 1 + 0.373 907 968;
  • 60) 0.373 907 968 × 2 = 0 + 0.747 815 936;
  • 61) 0.747 815 936 × 2 = 1 + 0.495 631 872;
  • 62) 0.495 631 872 × 2 = 0 + 0.991 263 744;
  • 63) 0.991 263 744 × 2 = 1 + 0.982 527 488;
  • 64) 0.982 527 488 × 2 = 1 + 0.965 054 976;

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:


0.000 282 336(10) =

0.0000 0000 0001 0010 1000 0000 1100 1111 1110 0010 1111 0000 1001 0100 0110 1011(2)

6. Positive number before normalization:

0.000 282 336(10) =

0.0000 0000 0001 0010 1000 0000 1100 1111 1110 0010 1111 0000 1001 0100 0110 1011(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:


0.000 282 336(10) =

0.0000 0000 0001 0010 1000 0000 1100 1111 1110 0010 1111 0000 1001 0100 0110 1011(2) =

0.0000 0000 0001 0010 1000 0000 1100 1111 1110 0010 1111 0000 1001 0100 0110 1011(2) × 20 =

1.0010 1000 0000 1100 1111 1110 0010 1111 0000 1001 0100 0110 1011(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 0000 1100 1111 1110 0010 1111 0000 1001 0100 0110 1011


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 0000 1100 1111 1110 0010 1111 0000 1001 0100 0110 1011 =

0010 1000 0000 1100 1111 1110 0010 1111 0000 1001 0100 0110 1011


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 0000 1100 1111 1110 0010 1111 0000 1001 0100 0110 1011


Decimal number -0.000 282 336 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0000 1100 1111 1110 0010 1111 0000 1001 0100 0110 1011

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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