-0.381 966 011 116 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.381 966 011 116(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
-0.381 966 011 116(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:

|-0.381 966 011 116| = 0.381 966 011 116


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: 0.381 966 011 116.

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) 0.381 966 011 116 × 2 = 0 + 0.763 932 022 232;
  • 2) 0.763 932 022 232 × 2 = 1 + 0.527 864 044 464;
  • 3) 0.527 864 044 464 × 2 = 1 + 0.055 728 088 928;
  • 4) 0.055 728 088 928 × 2 = 0 + 0.111 456 177 856;
  • 5) 0.111 456 177 856 × 2 = 0 + 0.222 912 355 712;
  • 6) 0.222 912 355 712 × 2 = 0 + 0.445 824 711 424;
  • 7) 0.445 824 711 424 × 2 = 0 + 0.891 649 422 848;
  • 8) 0.891 649 422 848 × 2 = 1 + 0.783 298 845 696;
  • 9) 0.783 298 845 696 × 2 = 1 + 0.566 597 691 392;
  • 10) 0.566 597 691 392 × 2 = 1 + 0.133 195 382 784;
  • 11) 0.133 195 382 784 × 2 = 0 + 0.266 390 765 568;
  • 12) 0.266 390 765 568 × 2 = 0 + 0.532 781 531 136;
  • 13) 0.532 781 531 136 × 2 = 1 + 0.065 563 062 272;
  • 14) 0.065 563 062 272 × 2 = 0 + 0.131 126 124 544;
  • 15) 0.131 126 124 544 × 2 = 0 + 0.262 252 249 088;
  • 16) 0.262 252 249 088 × 2 = 0 + 0.524 504 498 176;
  • 17) 0.524 504 498 176 × 2 = 1 + 0.049 008 996 352;
  • 18) 0.049 008 996 352 × 2 = 0 + 0.098 017 992 704;
  • 19) 0.098 017 992 704 × 2 = 0 + 0.196 035 985 408;
  • 20) 0.196 035 985 408 × 2 = 0 + 0.392 071 970 816;
  • 21) 0.392 071 970 816 × 2 = 0 + 0.784 143 941 632;
  • 22) 0.784 143 941 632 × 2 = 1 + 0.568 287 883 264;
  • 23) 0.568 287 883 264 × 2 = 1 + 0.136 575 766 528;
  • 24) 0.136 575 766 528 × 2 = 0 + 0.273 151 533 056;
  • 25) 0.273 151 533 056 × 2 = 0 + 0.546 303 066 112;
  • 26) 0.546 303 066 112 × 2 = 1 + 0.092 606 132 224;
  • 27) 0.092 606 132 224 × 2 = 0 + 0.185 212 264 448;
  • 28) 0.185 212 264 448 × 2 = 0 + 0.370 424 528 896;
  • 29) 0.370 424 528 896 × 2 = 0 + 0.740 849 057 792;
  • 30) 0.740 849 057 792 × 2 = 1 + 0.481 698 115 584;
  • 31) 0.481 698 115 584 × 2 = 0 + 0.963 396 231 168;
  • 32) 0.963 396 231 168 × 2 = 1 + 0.926 792 462 336;
  • 33) 0.926 792 462 336 × 2 = 1 + 0.853 584 924 672;
  • 34) 0.853 584 924 672 × 2 = 1 + 0.707 169 849 344;
  • 35) 0.707 169 849 344 × 2 = 1 + 0.414 339 698 688;
  • 36) 0.414 339 698 688 × 2 = 0 + 0.828 679 397 376;
  • 37) 0.828 679 397 376 × 2 = 1 + 0.657 358 794 752;
  • 38) 0.657 358 794 752 × 2 = 1 + 0.314 717 589 504;
  • 39) 0.314 717 589 504 × 2 = 0 + 0.629 435 179 008;
  • 40) 0.629 435 179 008 × 2 = 1 + 0.258 870 358 016;
  • 41) 0.258 870 358 016 × 2 = 0 + 0.517 740 716 032;
  • 42) 0.517 740 716 032 × 2 = 1 + 0.035 481 432 064;
  • 43) 0.035 481 432 064 × 2 = 0 + 0.070 962 864 128;
  • 44) 0.070 962 864 128 × 2 = 0 + 0.141 925 728 256;
  • 45) 0.141 925 728 256 × 2 = 0 + 0.283 851 456 512;
  • 46) 0.283 851 456 512 × 2 = 0 + 0.567 702 913 024;
  • 47) 0.567 702 913 024 × 2 = 1 + 0.135 405 826 048;
  • 48) 0.135 405 826 048 × 2 = 0 + 0.270 811 652 096;
  • 49) 0.270 811 652 096 × 2 = 0 + 0.541 623 304 192;
  • 50) 0.541 623 304 192 × 2 = 1 + 0.083 246 608 384;
  • 51) 0.083 246 608 384 × 2 = 0 + 0.166 493 216 768;
  • 52) 0.166 493 216 768 × 2 = 0 + 0.332 986 433 536;
  • 53) 0.332 986 433 536 × 2 = 0 + 0.665 972 867 072;
  • 54) 0.665 972 867 072 × 2 = 1 + 0.331 945 734 144;

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:


0.381 966 011 116(10) =

0.0110 0001 1100 1000 1000 0110 0100 0101 1110 1101 0100 0010 0100 01(2)

6. Positive number before normalization:

0.381 966 011 116(10) =

0.0110 0001 1100 1000 1000 0110 0100 0101 1110 1101 0100 0010 0100 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.381 966 011 116(10) =

0.0110 0001 1100 1000 1000 0110 0100 0101 1110 1101 0100 0010 0100 01(2) =

0.0110 0001 1100 1000 1000 0110 0100 0101 1110 1101 0100 0010 0100 01(2) × 20 =

1.1000 0111 0010 0010 0001 1001 0001 0111 1011 0101 0000 1001 0001(2) × 2-2


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): -2

Mantissa (not normalized):
1.1000 0111 0010 0010 0001 1001 0001 0111 1011 0101 0000 1001 0001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-2 + 2(11-1) - 1 =

(-2 + 1 023)(10) =

1 021(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;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1021(10) =

011 1111 1101(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. 1000 0111 0010 0010 0001 1001 0001 0111 1011 0101 0000 1001 0001 =

1000 0111 0010 0010 0001 1001 0001 0111 1011 0101 0000 1001 0001


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 1101

Mantissa (52 bits) =
1000 0111 0010 0010 0001 1001 0001 0111 1011 0101 0000 1001 0001


Decimal number -0.381 966 011 116 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1101 - 1000 0111 0010 0010 0001 1001 0001 0111 1011 0101 0000 1001 0001

» Convert decimal -0.381 966 011 131 to 64 bit double precision IEEE 754 binary floating point representation standard

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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