-0.381 966 011 254 56 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.381 966 011 254 56(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
-0.381 966 011 254 56(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:

|-0.381 966 011 254 56| = 0.381 966 011 254 56


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: 0.381 966 011 254 56.

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) 0.381 966 011 254 56 × 2 = 0 + 0.763 932 022 509 12;
  • 2) 0.763 932 022 509 12 × 2 = 1 + 0.527 864 045 018 24;
  • 3) 0.527 864 045 018 24 × 2 = 1 + 0.055 728 090 036 48;
  • 4) 0.055 728 090 036 48 × 2 = 0 + 0.111 456 180 072 96;
  • 5) 0.111 456 180 072 96 × 2 = 0 + 0.222 912 360 145 92;
  • 6) 0.222 912 360 145 92 × 2 = 0 + 0.445 824 720 291 84;
  • 7) 0.445 824 720 291 84 × 2 = 0 + 0.891 649 440 583 68;
  • 8) 0.891 649 440 583 68 × 2 = 1 + 0.783 298 881 167 36;
  • 9) 0.783 298 881 167 36 × 2 = 1 + 0.566 597 762 334 72;
  • 10) 0.566 597 762 334 72 × 2 = 1 + 0.133 195 524 669 44;
  • 11) 0.133 195 524 669 44 × 2 = 0 + 0.266 391 049 338 88;
  • 12) 0.266 391 049 338 88 × 2 = 0 + 0.532 782 098 677 76;
  • 13) 0.532 782 098 677 76 × 2 = 1 + 0.065 564 197 355 52;
  • 14) 0.065 564 197 355 52 × 2 = 0 + 0.131 128 394 711 04;
  • 15) 0.131 128 394 711 04 × 2 = 0 + 0.262 256 789 422 08;
  • 16) 0.262 256 789 422 08 × 2 = 0 + 0.524 513 578 844 16;
  • 17) 0.524 513 578 844 16 × 2 = 1 + 0.049 027 157 688 32;
  • 18) 0.049 027 157 688 32 × 2 = 0 + 0.098 054 315 376 64;
  • 19) 0.098 054 315 376 64 × 2 = 0 + 0.196 108 630 753 28;
  • 20) 0.196 108 630 753 28 × 2 = 0 + 0.392 217 261 506 56;
  • 21) 0.392 217 261 506 56 × 2 = 0 + 0.784 434 523 013 12;
  • 22) 0.784 434 523 013 12 × 2 = 1 + 0.568 869 046 026 24;
  • 23) 0.568 869 046 026 24 × 2 = 1 + 0.137 738 092 052 48;
  • 24) 0.137 738 092 052 48 × 2 = 0 + 0.275 476 184 104 96;
  • 25) 0.275 476 184 104 96 × 2 = 0 + 0.550 952 368 209 92;
  • 26) 0.550 952 368 209 92 × 2 = 1 + 0.101 904 736 419 84;
  • 27) 0.101 904 736 419 84 × 2 = 0 + 0.203 809 472 839 68;
  • 28) 0.203 809 472 839 68 × 2 = 0 + 0.407 618 945 679 36;
  • 29) 0.407 618 945 679 36 × 2 = 0 + 0.815 237 891 358 72;
  • 30) 0.815 237 891 358 72 × 2 = 1 + 0.630 475 782 717 44;
  • 31) 0.630 475 782 717 44 × 2 = 1 + 0.260 951 565 434 88;
  • 32) 0.260 951 565 434 88 × 2 = 0 + 0.521 903 130 869 76;
  • 33) 0.521 903 130 869 76 × 2 = 1 + 0.043 806 261 739 52;
  • 34) 0.043 806 261 739 52 × 2 = 0 + 0.087 612 523 479 04;
  • 35) 0.087 612 523 479 04 × 2 = 0 + 0.175 225 046 958 08;
  • 36) 0.175 225 046 958 08 × 2 = 0 + 0.350 450 093 916 16;
  • 37) 0.350 450 093 916 16 × 2 = 0 + 0.700 900 187 832 32;
  • 38) 0.700 900 187 832 32 × 2 = 1 + 0.401 800 375 664 64;
  • 39) 0.401 800 375 664 64 × 2 = 0 + 0.803 600 751 329 28;
  • 40) 0.803 600 751 329 28 × 2 = 1 + 0.607 201 502 658 56;
  • 41) 0.607 201 502 658 56 × 2 = 1 + 0.214 403 005 317 12;
  • 42) 0.214 403 005 317 12 × 2 = 0 + 0.428 806 010 634 24;
  • 43) 0.428 806 010 634 24 × 2 = 0 + 0.857 612 021 268 48;
  • 44) 0.857 612 021 268 48 × 2 = 1 + 0.715 224 042 536 96;
  • 45) 0.715 224 042 536 96 × 2 = 1 + 0.430 448 085 073 92;
  • 46) 0.430 448 085 073 92 × 2 = 0 + 0.860 896 170 147 84;
  • 47) 0.860 896 170 147 84 × 2 = 1 + 0.721 792 340 295 68;
  • 48) 0.721 792 340 295 68 × 2 = 1 + 0.443 584 680 591 36;
  • 49) 0.443 584 680 591 36 × 2 = 0 + 0.887 169 361 182 72;
  • 50) 0.887 169 361 182 72 × 2 = 1 + 0.774 338 722 365 44;
  • 51) 0.774 338 722 365 44 × 2 = 1 + 0.548 677 444 730 88;
  • 52) 0.548 677 444 730 88 × 2 = 1 + 0.097 354 889 461 76;
  • 53) 0.097 354 889 461 76 × 2 = 0 + 0.194 709 778 923 52;
  • 54) 0.194 709 778 923 52 × 2 = 0 + 0.389 419 557 847 04;

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:


0.381 966 011 254 56(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0101 1001 1011 0111 00(2)

6. Positive number before normalization:

0.381 966 011 254 56(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0101 1001 1011 0111 00(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:


0.381 966 011 254 56(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0101 1001 1011 0111 00(2) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0101 1001 1011 0111 00(2) × 20 =

1.1000 0111 0010 0010 0001 1001 0001 1010 0001 0110 0110 1101 1100(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 1010 0001 0110 0110 1101 1100


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 1010 0001 0110 0110 1101 1100 =

1000 0111 0010 0010 0001 1001 0001 1010 0001 0110 0110 1101 1100


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 1010 0001 0110 0110 1101 1100


Decimal number -0.381 966 011 254 56 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1101 - 1000 0111 0010 0010 0001 1001 0001 1010 0001 0110 0110 1101 1100

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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