0.133 807 264 267 57 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.133 807 264 267 57(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
0.133 807 264 267 57(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.133 807 264 267 57.

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) 0.133 807 264 267 57 × 2 = 0 + 0.267 614 528 535 14;
  • 2) 0.267 614 528 535 14 × 2 = 0 + 0.535 229 057 070 28;
  • 3) 0.535 229 057 070 28 × 2 = 1 + 0.070 458 114 140 56;
  • 4) 0.070 458 114 140 56 × 2 = 0 + 0.140 916 228 281 12;
  • 5) 0.140 916 228 281 12 × 2 = 0 + 0.281 832 456 562 24;
  • 6) 0.281 832 456 562 24 × 2 = 0 + 0.563 664 913 124 48;
  • 7) 0.563 664 913 124 48 × 2 = 1 + 0.127 329 826 248 96;
  • 8) 0.127 329 826 248 96 × 2 = 0 + 0.254 659 652 497 92;
  • 9) 0.254 659 652 497 92 × 2 = 0 + 0.509 319 304 995 84;
  • 10) 0.509 319 304 995 84 × 2 = 1 + 0.018 638 609 991 68;
  • 11) 0.018 638 609 991 68 × 2 = 0 + 0.037 277 219 983 36;
  • 12) 0.037 277 219 983 36 × 2 = 0 + 0.074 554 439 966 72;
  • 13) 0.074 554 439 966 72 × 2 = 0 + 0.149 108 879 933 44;
  • 14) 0.149 108 879 933 44 × 2 = 0 + 0.298 217 759 866 88;
  • 15) 0.298 217 759 866 88 × 2 = 0 + 0.596 435 519 733 76;
  • 16) 0.596 435 519 733 76 × 2 = 1 + 0.192 871 039 467 52;
  • 17) 0.192 871 039 467 52 × 2 = 0 + 0.385 742 078 935 04;
  • 18) 0.385 742 078 935 04 × 2 = 0 + 0.771 484 157 870 08;
  • 19) 0.771 484 157 870 08 × 2 = 1 + 0.542 968 315 740 16;
  • 20) 0.542 968 315 740 16 × 2 = 1 + 0.085 936 631 480 32;
  • 21) 0.085 936 631 480 32 × 2 = 0 + 0.171 873 262 960 64;
  • 22) 0.171 873 262 960 64 × 2 = 0 + 0.343 746 525 921 28;
  • 23) 0.343 746 525 921 28 × 2 = 0 + 0.687 493 051 842 56;
  • 24) 0.687 493 051 842 56 × 2 = 1 + 0.374 986 103 685 12;
  • 25) 0.374 986 103 685 12 × 2 = 0 + 0.749 972 207 370 24;
  • 26) 0.749 972 207 370 24 × 2 = 1 + 0.499 944 414 740 48;
  • 27) 0.499 944 414 740 48 × 2 = 0 + 0.999 888 829 480 96;
  • 28) 0.999 888 829 480 96 × 2 = 1 + 0.999 777 658 961 92;
  • 29) 0.999 777 658 961 92 × 2 = 1 + 0.999 555 317 923 84;
  • 30) 0.999 555 317 923 84 × 2 = 1 + 0.999 110 635 847 68;
  • 31) 0.999 110 635 847 68 × 2 = 1 + 0.998 221 271 695 36;
  • 32) 0.998 221 271 695 36 × 2 = 1 + 0.996 442 543 390 72;
  • 33) 0.996 442 543 390 72 × 2 = 1 + 0.992 885 086 781 44;
  • 34) 0.992 885 086 781 44 × 2 = 1 + 0.985 770 173 562 88;
  • 35) 0.985 770 173 562 88 × 2 = 1 + 0.971 540 347 125 76;
  • 36) 0.971 540 347 125 76 × 2 = 1 + 0.943 080 694 251 52;
  • 37) 0.943 080 694 251 52 × 2 = 1 + 0.886 161 388 503 04;
  • 38) 0.886 161 388 503 04 × 2 = 1 + 0.772 322 777 006 08;
  • 39) 0.772 322 777 006 08 × 2 = 1 + 0.544 645 554 012 16;
  • 40) 0.544 645 554 012 16 × 2 = 1 + 0.089 291 108 024 32;
  • 41) 0.089 291 108 024 32 × 2 = 0 + 0.178 582 216 048 64;
  • 42) 0.178 582 216 048 64 × 2 = 0 + 0.357 164 432 097 28;
  • 43) 0.357 164 432 097 28 × 2 = 0 + 0.714 328 864 194 56;
  • 44) 0.714 328 864 194 56 × 2 = 1 + 0.428 657 728 389 12;
  • 45) 0.428 657 728 389 12 × 2 = 0 + 0.857 315 456 778 24;
  • 46) 0.857 315 456 778 24 × 2 = 1 + 0.714 630 913 556 48;
  • 47) 0.714 630 913 556 48 × 2 = 1 + 0.429 261 827 112 96;
  • 48) 0.429 261 827 112 96 × 2 = 0 + 0.858 523 654 225 92;
  • 49) 0.858 523 654 225 92 × 2 = 1 + 0.717 047 308 451 84;
  • 50) 0.717 047 308 451 84 × 2 = 1 + 0.434 094 616 903 68;
  • 51) 0.434 094 616 903 68 × 2 = 0 + 0.868 189 233 807 36;
  • 52) 0.868 189 233 807 36 × 2 = 1 + 0.736 378 467 614 72;
  • 53) 0.736 378 467 614 72 × 2 = 1 + 0.472 756 935 229 44;
  • 54) 0.472 756 935 229 44 × 2 = 0 + 0.945 513 870 458 88;
  • 55) 0.945 513 870 458 88 × 2 = 1 + 0.891 027 740 917 76;

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).


4. 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:


0.133 807 264 267 57(10) =

0.0010 0010 0100 0001 0011 0001 0101 1111 1111 1111 0001 0110 1101 101(2)

5. Positive number before normalization:

0.133 807 264 267 57(10) =

0.0010 0010 0100 0001 0011 0001 0101 1111 1111 1111 0001 0110 1101 101(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.133 807 264 267 57(10) =

0.0010 0010 0100 0001 0011 0001 0101 1111 1111 1111 0001 0110 1101 101(2) =

0.0010 0010 0100 0001 0011 0001 0101 1111 1111 1111 0001 0110 1101 101(2) × 20 =

1.0001 0010 0000 1001 1000 1010 1111 1111 1111 1000 1011 0110 1101(2) × 2-3


7. 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 0 (a positive number)

Exponent (unadjusted): -3

Mantissa (not normalized):
1.0001 0010 0000 1001 1000 1010 1111 1111 1111 1000 1011 0110 1101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-3 + 2(11-1) - 1 =

(-3 + 1 023)(10) =

1 020(10)


9. 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;
  • 1 020 ÷ 2 = 510 + 0;
  • 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;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1020(10) =

011 1111 1100(2)


11. 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. 0001 0010 0000 1001 1000 1010 1111 1111 1111 1000 1011 0110 1101 =

0001 0010 0000 1001 1000 1010 1111 1111 1111 1000 1011 0110 1101


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 1100

Mantissa (52 bits) =
0001 0010 0000 1001 1000 1010 1111 1111 1111 1000 1011 0110 1101


Decimal number 0.133 807 264 267 57 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1100 - 0001 0010 0000 1001 1000 1010 1111 1111 1111 1000 1011 0110 1101

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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