0.133 807 264 266 29 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.133 807 264 266 29(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 29(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 29.

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 29 × 2 = 0 + 0.267 614 528 532 58;
  • 2) 0.267 614 528 532 58 × 2 = 0 + 0.535 229 057 065 16;
  • 3) 0.535 229 057 065 16 × 2 = 1 + 0.070 458 114 130 32;
  • 4) 0.070 458 114 130 32 × 2 = 0 + 0.140 916 228 260 64;
  • 5) 0.140 916 228 260 64 × 2 = 0 + 0.281 832 456 521 28;
  • 6) 0.281 832 456 521 28 × 2 = 0 + 0.563 664 913 042 56;
  • 7) 0.563 664 913 042 56 × 2 = 1 + 0.127 329 826 085 12;
  • 8) 0.127 329 826 085 12 × 2 = 0 + 0.254 659 652 170 24;
  • 9) 0.254 659 652 170 24 × 2 = 0 + 0.509 319 304 340 48;
  • 10) 0.509 319 304 340 48 × 2 = 1 + 0.018 638 608 680 96;
  • 11) 0.018 638 608 680 96 × 2 = 0 + 0.037 277 217 361 92;
  • 12) 0.037 277 217 361 92 × 2 = 0 + 0.074 554 434 723 84;
  • 13) 0.074 554 434 723 84 × 2 = 0 + 0.149 108 869 447 68;
  • 14) 0.149 108 869 447 68 × 2 = 0 + 0.298 217 738 895 36;
  • 15) 0.298 217 738 895 36 × 2 = 0 + 0.596 435 477 790 72;
  • 16) 0.596 435 477 790 72 × 2 = 1 + 0.192 870 955 581 44;
  • 17) 0.192 870 955 581 44 × 2 = 0 + 0.385 741 911 162 88;
  • 18) 0.385 741 911 162 88 × 2 = 0 + 0.771 483 822 325 76;
  • 19) 0.771 483 822 325 76 × 2 = 1 + 0.542 967 644 651 52;
  • 20) 0.542 967 644 651 52 × 2 = 1 + 0.085 935 289 303 04;
  • 21) 0.085 935 289 303 04 × 2 = 0 + 0.171 870 578 606 08;
  • 22) 0.171 870 578 606 08 × 2 = 0 + 0.343 741 157 212 16;
  • 23) 0.343 741 157 212 16 × 2 = 0 + 0.687 482 314 424 32;
  • 24) 0.687 482 314 424 32 × 2 = 1 + 0.374 964 628 848 64;
  • 25) 0.374 964 628 848 64 × 2 = 0 + 0.749 929 257 697 28;
  • 26) 0.749 929 257 697 28 × 2 = 1 + 0.499 858 515 394 56;
  • 27) 0.499 858 515 394 56 × 2 = 0 + 0.999 717 030 789 12;
  • 28) 0.999 717 030 789 12 × 2 = 1 + 0.999 434 061 578 24;
  • 29) 0.999 434 061 578 24 × 2 = 1 + 0.998 868 123 156 48;
  • 30) 0.998 868 123 156 48 × 2 = 1 + 0.997 736 246 312 96;
  • 31) 0.997 736 246 312 96 × 2 = 1 + 0.995 472 492 625 92;
  • 32) 0.995 472 492 625 92 × 2 = 1 + 0.990 944 985 251 84;
  • 33) 0.990 944 985 251 84 × 2 = 1 + 0.981 889 970 503 68;
  • 34) 0.981 889 970 503 68 × 2 = 1 + 0.963 779 941 007 36;
  • 35) 0.963 779 941 007 36 × 2 = 1 + 0.927 559 882 014 72;
  • 36) 0.927 559 882 014 72 × 2 = 1 + 0.855 119 764 029 44;
  • 37) 0.855 119 764 029 44 × 2 = 1 + 0.710 239 528 058 88;
  • 38) 0.710 239 528 058 88 × 2 = 1 + 0.420 479 056 117 76;
  • 39) 0.420 479 056 117 76 × 2 = 0 + 0.840 958 112 235 52;
  • 40) 0.840 958 112 235 52 × 2 = 1 + 0.681 916 224 471 04;
  • 41) 0.681 916 224 471 04 × 2 = 1 + 0.363 832 448 942 08;
  • 42) 0.363 832 448 942 08 × 2 = 0 + 0.727 664 897 884 16;
  • 43) 0.727 664 897 884 16 × 2 = 1 + 0.455 329 795 768 32;
  • 44) 0.455 329 795 768 32 × 2 = 0 + 0.910 659 591 536 64;
  • 45) 0.910 659 591 536 64 × 2 = 1 + 0.821 319 183 073 28;
  • 46) 0.821 319 183 073 28 × 2 = 1 + 0.642 638 366 146 56;
  • 47) 0.642 638 366 146 56 × 2 = 1 + 0.285 276 732 293 12;
  • 48) 0.285 276 732 293 12 × 2 = 0 + 0.570 553 464 586 24;
  • 49) 0.570 553 464 586 24 × 2 = 1 + 0.141 106 929 172 48;
  • 50) 0.141 106 929 172 48 × 2 = 0 + 0.282 213 858 344 96;
  • 51) 0.282 213 858 344 96 × 2 = 0 + 0.564 427 716 689 92;
  • 52) 0.564 427 716 689 92 × 2 = 1 + 0.128 855 433 379 84;
  • 53) 0.128 855 433 379 84 × 2 = 0 + 0.257 710 866 759 68;
  • 54) 0.257 710 866 759 68 × 2 = 0 + 0.515 421 733 519 36;
  • 55) 0.515 421 733 519 36 × 2 = 1 + 0.030 843 467 038 72;

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 29(10) =

0.0010 0010 0100 0001 0011 0001 0101 1111 1111 1101 1010 1110 1001 001(2)

5. Positive number before normalization:

0.133 807 264 266 29(10) =

0.0010 0010 0100 0001 0011 0001 0101 1111 1111 1101 1010 1110 1001 001(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 29(10) =

0.0010 0010 0100 0001 0011 0001 0101 1111 1111 1101 1010 1110 1001 001(2) =

0.0010 0010 0100 0001 0011 0001 0101 1111 1111 1101 1010 1110 1001 001(2) × 20 =

1.0001 0010 0000 1001 1000 1010 1111 1111 1110 1101 0111 0100 1001(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 1110 1101 0111 0100 1001


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 1110 1101 0111 0100 1001 =

0001 0010 0000 1001 1000 1010 1111 1111 1110 1101 0111 0100 1001


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 1110 1101 0111 0100 1001


Decimal number 0.133 807 264 266 29 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1100 - 0001 0010 0000 1001 1000 1010 1111 1111 1110 1101 0111 0100 1001

» Convert decimal 0.133 807 264 266 35 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