0.133 807 264 266 37 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.133 807 264 266 37(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 37(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 37.

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 37 × 2 = 0 + 0.267 614 528 532 74;
  • 2) 0.267 614 528 532 74 × 2 = 0 + 0.535 229 057 065 48;
  • 3) 0.535 229 057 065 48 × 2 = 1 + 0.070 458 114 130 96;
  • 4) 0.070 458 114 130 96 × 2 = 0 + 0.140 916 228 261 92;
  • 5) 0.140 916 228 261 92 × 2 = 0 + 0.281 832 456 523 84;
  • 6) 0.281 832 456 523 84 × 2 = 0 + 0.563 664 913 047 68;
  • 7) 0.563 664 913 047 68 × 2 = 1 + 0.127 329 826 095 36;
  • 8) 0.127 329 826 095 36 × 2 = 0 + 0.254 659 652 190 72;
  • 9) 0.254 659 652 190 72 × 2 = 0 + 0.509 319 304 381 44;
  • 10) 0.509 319 304 381 44 × 2 = 1 + 0.018 638 608 762 88;
  • 11) 0.018 638 608 762 88 × 2 = 0 + 0.037 277 217 525 76;
  • 12) 0.037 277 217 525 76 × 2 = 0 + 0.074 554 435 051 52;
  • 13) 0.074 554 435 051 52 × 2 = 0 + 0.149 108 870 103 04;
  • 14) 0.149 108 870 103 04 × 2 = 0 + 0.298 217 740 206 08;
  • 15) 0.298 217 740 206 08 × 2 = 0 + 0.596 435 480 412 16;
  • 16) 0.596 435 480 412 16 × 2 = 1 + 0.192 870 960 824 32;
  • 17) 0.192 870 960 824 32 × 2 = 0 + 0.385 741 921 648 64;
  • 18) 0.385 741 921 648 64 × 2 = 0 + 0.771 483 843 297 28;
  • 19) 0.771 483 843 297 28 × 2 = 1 + 0.542 967 686 594 56;
  • 20) 0.542 967 686 594 56 × 2 = 1 + 0.085 935 373 189 12;
  • 21) 0.085 935 373 189 12 × 2 = 0 + 0.171 870 746 378 24;
  • 22) 0.171 870 746 378 24 × 2 = 0 + 0.343 741 492 756 48;
  • 23) 0.343 741 492 756 48 × 2 = 0 + 0.687 482 985 512 96;
  • 24) 0.687 482 985 512 96 × 2 = 1 + 0.374 965 971 025 92;
  • 25) 0.374 965 971 025 92 × 2 = 0 + 0.749 931 942 051 84;
  • 26) 0.749 931 942 051 84 × 2 = 1 + 0.499 863 884 103 68;
  • 27) 0.499 863 884 103 68 × 2 = 0 + 0.999 727 768 207 36;
  • 28) 0.999 727 768 207 36 × 2 = 1 + 0.999 455 536 414 72;
  • 29) 0.999 455 536 414 72 × 2 = 1 + 0.998 911 072 829 44;
  • 30) 0.998 911 072 829 44 × 2 = 1 + 0.997 822 145 658 88;
  • 31) 0.997 822 145 658 88 × 2 = 1 + 0.995 644 291 317 76;
  • 32) 0.995 644 291 317 76 × 2 = 1 + 0.991 288 582 635 52;
  • 33) 0.991 288 582 635 52 × 2 = 1 + 0.982 577 165 271 04;
  • 34) 0.982 577 165 271 04 × 2 = 1 + 0.965 154 330 542 08;
  • 35) 0.965 154 330 542 08 × 2 = 1 + 0.930 308 661 084 16;
  • 36) 0.930 308 661 084 16 × 2 = 1 + 0.860 617 322 168 32;
  • 37) 0.860 617 322 168 32 × 2 = 1 + 0.721 234 644 336 64;
  • 38) 0.721 234 644 336 64 × 2 = 1 + 0.442 469 288 673 28;
  • 39) 0.442 469 288 673 28 × 2 = 0 + 0.884 938 577 346 56;
  • 40) 0.884 938 577 346 56 × 2 = 1 + 0.769 877 154 693 12;
  • 41) 0.769 877 154 693 12 × 2 = 1 + 0.539 754 309 386 24;
  • 42) 0.539 754 309 386 24 × 2 = 1 + 0.079 508 618 772 48;
  • 43) 0.079 508 618 772 48 × 2 = 0 + 0.159 017 237 544 96;
  • 44) 0.159 017 237 544 96 × 2 = 0 + 0.318 034 475 089 92;
  • 45) 0.318 034 475 089 92 × 2 = 0 + 0.636 068 950 179 84;
  • 46) 0.636 068 950 179 84 × 2 = 1 + 0.272 137 900 359 68;
  • 47) 0.272 137 900 359 68 × 2 = 0 + 0.544 275 800 719 36;
  • 48) 0.544 275 800 719 36 × 2 = 1 + 0.088 551 601 438 72;
  • 49) 0.088 551 601 438 72 × 2 = 0 + 0.177 103 202 877 44;
  • 50) 0.177 103 202 877 44 × 2 = 0 + 0.354 206 405 754 88;
  • 51) 0.354 206 405 754 88 × 2 = 0 + 0.708 412 811 509 76;
  • 52) 0.708 412 811 509 76 × 2 = 1 + 0.416 825 623 019 52;
  • 53) 0.416 825 623 019 52 × 2 = 0 + 0.833 651 246 039 04;
  • 54) 0.833 651 246 039 04 × 2 = 1 + 0.667 302 492 078 08;
  • 55) 0.667 302 492 078 08 × 2 = 1 + 0.334 604 984 156 16;

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 37(10) =

0.0010 0010 0100 0001 0011 0001 0101 1111 1111 1101 1100 0101 0001 011(2)

5. Positive number before normalization:

0.133 807 264 266 37(10) =

0.0010 0010 0100 0001 0011 0001 0101 1111 1111 1101 1100 0101 0001 011(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 37(10) =

0.0010 0010 0100 0001 0011 0001 0101 1111 1111 1101 1100 0101 0001 011(2) =

0.0010 0010 0100 0001 0011 0001 0101 1111 1111 1101 1100 0101 0001 011(2) × 20 =

1.0001 0010 0000 1001 1000 1010 1111 1111 1110 1110 0010 1000 1011(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 1110 0010 1000 1011


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 1110 0010 1000 1011 =

0001 0010 0000 1001 1000 1010 1111 1111 1110 1110 0010 1000 1011


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 1110 0010 1000 1011


Decimal number 0.133 807 264 266 37 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 1110 0010 1000 1011

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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