0.133 807 264 266 77 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.133 807 264 266 77(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
0.133 807 264 266 77(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: 0.133 807 264 266 77.

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) 0.133 807 264 266 77 × 2 = 0 + 0.267 614 528 533 54;
  • 2) 0.267 614 528 533 54 × 2 = 0 + 0.535 229 057 067 08;
  • 3) 0.535 229 057 067 08 × 2 = 1 + 0.070 458 114 134 16;
  • 4) 0.070 458 114 134 16 × 2 = 0 + 0.140 916 228 268 32;
  • 5) 0.140 916 228 268 32 × 2 = 0 + 0.281 832 456 536 64;
  • 6) 0.281 832 456 536 64 × 2 = 0 + 0.563 664 913 073 28;
  • 7) 0.563 664 913 073 28 × 2 = 1 + 0.127 329 826 146 56;
  • 8) 0.127 329 826 146 56 × 2 = 0 + 0.254 659 652 293 12;
  • 9) 0.254 659 652 293 12 × 2 = 0 + 0.509 319 304 586 24;
  • 10) 0.509 319 304 586 24 × 2 = 1 + 0.018 638 609 172 48;
  • 11) 0.018 638 609 172 48 × 2 = 0 + 0.037 277 218 344 96;
  • 12) 0.037 277 218 344 96 × 2 = 0 + 0.074 554 436 689 92;
  • 13) 0.074 554 436 689 92 × 2 = 0 + 0.149 108 873 379 84;
  • 14) 0.149 108 873 379 84 × 2 = 0 + 0.298 217 746 759 68;
  • 15) 0.298 217 746 759 68 × 2 = 0 + 0.596 435 493 519 36;
  • 16) 0.596 435 493 519 36 × 2 = 1 + 0.192 870 987 038 72;
  • 17) 0.192 870 987 038 72 × 2 = 0 + 0.385 741 974 077 44;
  • 18) 0.385 741 974 077 44 × 2 = 0 + 0.771 483 948 154 88;
  • 19) 0.771 483 948 154 88 × 2 = 1 + 0.542 967 896 309 76;
  • 20) 0.542 967 896 309 76 × 2 = 1 + 0.085 935 792 619 52;
  • 21) 0.085 935 792 619 52 × 2 = 0 + 0.171 871 585 239 04;
  • 22) 0.171 871 585 239 04 × 2 = 0 + 0.343 743 170 478 08;
  • 23) 0.343 743 170 478 08 × 2 = 0 + 0.687 486 340 956 16;
  • 24) 0.687 486 340 956 16 × 2 = 1 + 0.374 972 681 912 32;
  • 25) 0.374 972 681 912 32 × 2 = 0 + 0.749 945 363 824 64;
  • 26) 0.749 945 363 824 64 × 2 = 1 + 0.499 890 727 649 28;
  • 27) 0.499 890 727 649 28 × 2 = 0 + 0.999 781 455 298 56;
  • 28) 0.999 781 455 298 56 × 2 = 1 + 0.999 562 910 597 12;
  • 29) 0.999 562 910 597 12 × 2 = 1 + 0.999 125 821 194 24;
  • 30) 0.999 125 821 194 24 × 2 = 1 + 0.998 251 642 388 48;
  • 31) 0.998 251 642 388 48 × 2 = 1 + 0.996 503 284 776 96;
  • 32) 0.996 503 284 776 96 × 2 = 1 + 0.993 006 569 553 92;
  • 33) 0.993 006 569 553 92 × 2 = 1 + 0.986 013 139 107 84;
  • 34) 0.986 013 139 107 84 × 2 = 1 + 0.972 026 278 215 68;
  • 35) 0.972 026 278 215 68 × 2 = 1 + 0.944 052 556 431 36;
  • 36) 0.944 052 556 431 36 × 2 = 1 + 0.888 105 112 862 72;
  • 37) 0.888 105 112 862 72 × 2 = 1 + 0.776 210 225 725 44;
  • 38) 0.776 210 225 725 44 × 2 = 1 + 0.552 420 451 450 88;
  • 39) 0.552 420 451 450 88 × 2 = 1 + 0.104 840 902 901 76;
  • 40) 0.104 840 902 901 76 × 2 = 0 + 0.209 681 805 803 52;
  • 41) 0.209 681 805 803 52 × 2 = 0 + 0.419 363 611 607 04;
  • 42) 0.419 363 611 607 04 × 2 = 0 + 0.838 727 223 214 08;
  • 43) 0.838 727 223 214 08 × 2 = 1 + 0.677 454 446 428 16;
  • 44) 0.677 454 446 428 16 × 2 = 1 + 0.354 908 892 856 32;
  • 45) 0.354 908 892 856 32 × 2 = 0 + 0.709 817 785 712 64;
  • 46) 0.709 817 785 712 64 × 2 = 1 + 0.419 635 571 425 28;
  • 47) 0.419 635 571 425 28 × 2 = 0 + 0.839 271 142 850 56;
  • 48) 0.839 271 142 850 56 × 2 = 1 + 0.678 542 285 701 12;
  • 49) 0.678 542 285 701 12 × 2 = 1 + 0.357 084 571 402 24;
  • 50) 0.357 084 571 402 24 × 2 = 0 + 0.714 169 142 804 48;
  • 51) 0.714 169 142 804 48 × 2 = 1 + 0.428 338 285 608 96;
  • 52) 0.428 338 285 608 96 × 2 = 0 + 0.856 676 571 217 92;
  • 53) 0.856 676 571 217 92 × 2 = 1 + 0.713 353 142 435 84;
  • 54) 0.713 353 142 435 84 × 2 = 1 + 0.426 706 284 871 68;
  • 55) 0.426 706 284 871 68 × 2 = 0 + 0.853 412 569 743 36;

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:


0.133 807 264 266 77(10) =

0.0010 0010 0100 0001 0011 0001 0101 1111 1111 1110 0011 0101 1010 110(2)

5. Positive number before normalization:

0.133 807 264 266 77(10) =

0.0010 0010 0100 0001 0011 0001 0101 1111 1111 1110 0011 0101 1010 110(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:


0.133 807 264 266 77(10) =

0.0010 0010 0100 0001 0011 0001 0101 1111 1111 1110 0011 0101 1010 110(2) =

0.0010 0010 0100 0001 0011 0001 0101 1111 1111 1110 0011 0101 1010 110(2) × 20 =

1.0001 0010 0000 1001 1000 1010 1111 1111 1111 0001 1010 1101 0110(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 0001 1010 1101 0110


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 0001 1010 1101 0110 =

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


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 0001 1010 1101 0110


Decimal number 0.133 807 264 266 77 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 0001 1010 1101 0110

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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