0.133 807 264 267 09 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.133 807 264 267 09(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 09(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 09.

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 09 × 2 = 0 + 0.267 614 528 534 18;
  • 2) 0.267 614 528 534 18 × 2 = 0 + 0.535 229 057 068 36;
  • 3) 0.535 229 057 068 36 × 2 = 1 + 0.070 458 114 136 72;
  • 4) 0.070 458 114 136 72 × 2 = 0 + 0.140 916 228 273 44;
  • 5) 0.140 916 228 273 44 × 2 = 0 + 0.281 832 456 546 88;
  • 6) 0.281 832 456 546 88 × 2 = 0 + 0.563 664 913 093 76;
  • 7) 0.563 664 913 093 76 × 2 = 1 + 0.127 329 826 187 52;
  • 8) 0.127 329 826 187 52 × 2 = 0 + 0.254 659 652 375 04;
  • 9) 0.254 659 652 375 04 × 2 = 0 + 0.509 319 304 750 08;
  • 10) 0.509 319 304 750 08 × 2 = 1 + 0.018 638 609 500 16;
  • 11) 0.018 638 609 500 16 × 2 = 0 + 0.037 277 219 000 32;
  • 12) 0.037 277 219 000 32 × 2 = 0 + 0.074 554 438 000 64;
  • 13) 0.074 554 438 000 64 × 2 = 0 + 0.149 108 876 001 28;
  • 14) 0.149 108 876 001 28 × 2 = 0 + 0.298 217 752 002 56;
  • 15) 0.298 217 752 002 56 × 2 = 0 + 0.596 435 504 005 12;
  • 16) 0.596 435 504 005 12 × 2 = 1 + 0.192 871 008 010 24;
  • 17) 0.192 871 008 010 24 × 2 = 0 + 0.385 742 016 020 48;
  • 18) 0.385 742 016 020 48 × 2 = 0 + 0.771 484 032 040 96;
  • 19) 0.771 484 032 040 96 × 2 = 1 + 0.542 968 064 081 92;
  • 20) 0.542 968 064 081 92 × 2 = 1 + 0.085 936 128 163 84;
  • 21) 0.085 936 128 163 84 × 2 = 0 + 0.171 872 256 327 68;
  • 22) 0.171 872 256 327 68 × 2 = 0 + 0.343 744 512 655 36;
  • 23) 0.343 744 512 655 36 × 2 = 0 + 0.687 489 025 310 72;
  • 24) 0.687 489 025 310 72 × 2 = 1 + 0.374 978 050 621 44;
  • 25) 0.374 978 050 621 44 × 2 = 0 + 0.749 956 101 242 88;
  • 26) 0.749 956 101 242 88 × 2 = 1 + 0.499 912 202 485 76;
  • 27) 0.499 912 202 485 76 × 2 = 0 + 0.999 824 404 971 52;
  • 28) 0.999 824 404 971 52 × 2 = 1 + 0.999 648 809 943 04;
  • 29) 0.999 648 809 943 04 × 2 = 1 + 0.999 297 619 886 08;
  • 30) 0.999 297 619 886 08 × 2 = 1 + 0.998 595 239 772 16;
  • 31) 0.998 595 239 772 16 × 2 = 1 + 0.997 190 479 544 32;
  • 32) 0.997 190 479 544 32 × 2 = 1 + 0.994 380 959 088 64;
  • 33) 0.994 380 959 088 64 × 2 = 1 + 0.988 761 918 177 28;
  • 34) 0.988 761 918 177 28 × 2 = 1 + 0.977 523 836 354 56;
  • 35) 0.977 523 836 354 56 × 2 = 1 + 0.955 047 672 709 12;
  • 36) 0.955 047 672 709 12 × 2 = 1 + 0.910 095 345 418 24;
  • 37) 0.910 095 345 418 24 × 2 = 1 + 0.820 190 690 836 48;
  • 38) 0.820 190 690 836 48 × 2 = 1 + 0.640 381 381 672 96;
  • 39) 0.640 381 381 672 96 × 2 = 1 + 0.280 762 763 345 92;
  • 40) 0.280 762 763 345 92 × 2 = 0 + 0.561 525 526 691 84;
  • 41) 0.561 525 526 691 84 × 2 = 1 + 0.123 051 053 383 68;
  • 42) 0.123 051 053 383 68 × 2 = 0 + 0.246 102 106 767 36;
  • 43) 0.246 102 106 767 36 × 2 = 0 + 0.492 204 213 534 72;
  • 44) 0.492 204 213 534 72 × 2 = 0 + 0.984 408 427 069 44;
  • 45) 0.984 408 427 069 44 × 2 = 1 + 0.968 816 854 138 88;
  • 46) 0.968 816 854 138 88 × 2 = 1 + 0.937 633 708 277 76;
  • 47) 0.937 633 708 277 76 × 2 = 1 + 0.875 267 416 555 52;
  • 48) 0.875 267 416 555 52 × 2 = 1 + 0.750 534 833 111 04;
  • 49) 0.750 534 833 111 04 × 2 = 1 + 0.501 069 666 222 08;
  • 50) 0.501 069 666 222 08 × 2 = 1 + 0.002 139 332 444 16;
  • 51) 0.002 139 332 444 16 × 2 = 0 + 0.004 278 664 888 32;
  • 52) 0.004 278 664 888 32 × 2 = 0 + 0.008 557 329 776 64;
  • 53) 0.008 557 329 776 64 × 2 = 0 + 0.017 114 659 553 28;
  • 54) 0.017 114 659 553 28 × 2 = 0 + 0.034 229 319 106 56;
  • 55) 0.034 229 319 106 56 × 2 = 0 + 0.068 458 638 213 12;

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 09(10) =

0.0010 0010 0100 0001 0011 0001 0101 1111 1111 1110 1000 1111 1100 000(2)

5. Positive number before normalization:

0.133 807 264 267 09(10) =

0.0010 0010 0100 0001 0011 0001 0101 1111 1111 1110 1000 1111 1100 000(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 09(10) =

0.0010 0010 0100 0001 0011 0001 0101 1111 1111 1110 1000 1111 1100 000(2) =

0.0010 0010 0100 0001 0011 0001 0101 1111 1111 1110 1000 1111 1100 000(2) × 20 =

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


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 0100 0111 1110 0000 =

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


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 0100 0111 1110 0000


Decimal number 0.133 807 264 267 09 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 0100 0111 1110 0000

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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